Prove that the function f(x)=5 x-3 is continuous at x=0, at x=-3 and at x=5.

Definition: f is continuous at x=c if f(c) is defined, lim x→ cf(x) exists, and lim x→ cf(x)=f(c). Here f(x)=5x-3 is defined for every real x. At x=0: f(0)=5(0)-3=-3, and by the algebra of limits lim x→ 0(5x-3)=5(0)-3=-3. Since lim x→ 0f(x)=f(0), f is continuous at x=0. At x=-3: f(-3)=5(-3)-3=-18, and lim x→ -3(5x-3)=5(-3)-3=-18. The two agree, so f is continuous at x=-3. At x=5: f(5)=5(5)-3=22, and lim x→ 5(5x-3)=5(5)-3=22. The two agree, so f is continuous at x=5. Hence f(x)=5x-3 is continuous at x=0, at x=-3 and at x=5.

Examine the continuity of the function f(x)=2 x^2-1 at x=3.

f is continuous at x=3. Test the definition of continuity at x=3: compare lim x→ 3f(x) with f(3). The function f(x)=2x^2-1 is defined for all real x, so f(3)=2(3)^2-1=18-1=17. By the algebra of limits (limit of a product and of a sum), lim x→ 3(2x^2-1)=2( lim x→ 3x)^2-1=2(3)^2-1=17. Since lim x→ 3f(x)=17=f(3), the function f is continuous at x=3.

Examine the following functions for continuity. (a) f(x)=x-5 (b) f(x)=1/x-5, x ≠ 5 (c) f(x)=(x^2-25)/(x+5), x ≠-5 (d) f(x)= x-5

(a), (b), (c) and (d) are all continuous functions. In each part, take an arbitrary point c of the domain and compare lim x→ cf(x) with f(c). A point outside the domain is not a point of discontinuity, since continuity is only discussed at points where the function is defined. (a) f(x)=x-5, domain R. For any real c, lim x→ c(x-5)=c-5=f(c). Hence f is continuous at every real point; there is no point of discontinuity. (b) f(x)=1/x-5, x≠ 5; domain R - 5. Let c≠ 5. Then c-5≠ 0, so by the quotient rule for limits lim x→ c1/x-5=1/c-5=f(c). Hence f is continuous at every point of its domain. (At x=5 the function is not defined, so it is not a point of discontinuity.) (c) f(x)=(x^2-25)/(x+5), x≠ -5; domain R - -5. Let c≠ -5; then c+5≠ 0 and lim x→ c(x^2-25)/(x+5)=(c^2-25)/(c+5)=f(c). Hence f is continuous at every point of its domain. (Cancelling gives f(x)=x-5 for x≠-5, which is the same conclusion.) (d) f(x)= x-5, domain R. Write f(x)=5-x for x<5 and f(x)=x-5 for x≥ 5. For c<5 and for c 5, f agrees near c with a polynomial, so it is continuous there. At c=5: LHL = lim x→ 5^- (5-x)=0, RHL = lim x→ 5^+ (x-5)=0, and f(5)=0. All three are equal, so f is continuous at x=5. Thus x-5 is continuous at every real point.

Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.

Here f(x)=x^n with n a positive integer, defined for all real x. Value at the point: f(n)=n^n. Limit at the point: x^n is the product of n copies of x, so by the product rule for limits applied n-1 times, lim x→ nx^n=( lim x→ nx)^n=n^n. Since lim x→ nf(x)=n^n=f(n), the function f(x)=x^n is continuous at x=n. (The same argument at any real c shows x^n, and hence every polynomial, is continuous on R.)

Is the function f defined by f(x)= cases x, & if x ≤ 1 5, & if x 1 cases continuous at x=0? At x=1? At x=2? Find all points of discontinuity of f, where f is defined by

f is continuous at x=0 and x=2; Not continuous at x=1. The rule changes only at x=1: f(x)=x for x≤ 1 and f(x)=5 for x 1. At x=0: all points near 0 satisfy x≤ 1, so f(x)=x there. Hence lim x→ 0f(x)=0 and f(0)=0. They are equal, so f is continuous at x=0. At x=1: the two one-sided limits must be computed from different rules. lim x→ 1^- f(x)= lim x→ 1^- x=1, lim x→ 1^+ f(x)= lim x→ 1^+ 5=5. Since LHL ≠ RHL, lim x→ 1f(x) does not exist, so f is discontinuous at x=1. At x=2: all points near 2 satisfy x 1, so f(x)=5 there. Hence lim x→ 2f(x)=5=f(2), and f is continuous at x=2. Answer: f is continuous at x=0 and at x=2, and discontinuous at x=1.

Discontinuous at x=2. f(x)=2x+3 for x≤ 2 and f(x)=2x-3 for x 2. On each of (-∞,2) and (2,∞) the function is a polynomial, hence continuous; only the joining point x=2 needs testing. At x=2: lim x→ 2^- f(x)= lim x→ 2^- (2x+3)=7, lim x→ 2^+ f(x)= lim x→ 2^+ (2x-3)=1, and f(2)=2(2)+3=7. Since LHL =7≠ 1= RHL, the limit at x=2 does not exist. Answer: x=2 is the only point of discontinuity of f.

Discontinuous at x=3. For x≤ -3, x =-x, so f(x)=-x+3 there. Thus f(x)= cases -x+3,&x≤ -3 -2x,&-3<x<3 6x+2,&x≥ 3. cases Each piece is a polynomial, so f is continuous at every point of (-∞,-3), (-3,3) and (3,∞). Only x=-3 and x=3 need testing. At x=-3: lim x→ -3^- (-x+3)=3+3=6, lim x→ -3^+ (-2x)=6, f(-3)= -3 +3=6. All three agree, so f is continuous at x=-3. At x=3: lim x→ 3^- (-2x)=-6, lim x→ 3^+ (6x+2)=20, f(3)=6(3)+2=20. Since LHL =-6≠ 20= RHL, the limit does not exist at x=3. Answer: x=3 is the only point of discontinuity.

Discontinuous at x=0. Remove the modulus: for x 0, x =x so ( x )/x=1; for x<0, x =-x so ( x )/x=-1. Hence f(x)= cases -1,&x<0 0,&x=0 1,&x 0. cases On (-∞,0) and on (0,∞), f is constant, hence continuous. Only x=0 needs testing. At x=0: lim x→ 0^- f(x)=-1, lim x→ 0^+ f(x)=1, f(0)=0. Since LHL ≠ RHL, lim x→ 0f(x) does not exist. Answer: x=0 is the only point of discontinuity of f.

f(x)= cases x/( x ), & if x<0 -1, & if x ≥ 0 cases

No point of discontinuity. Simplify the first branch. For x<0 we have x =-x, so x/( x )=x/-x=-1. Therefore f(x)=-1 for x<0 and f(x)=-1 for x≥ 0; that is, f(x)=-1 for every real x. A constant function is continuous: for any real c, lim x→ cf(x)=-1=f(c). In particular at the joining point x=0, LHL = RHL =f(0)=-1. Answer: f is continuous on R; it has no point of discontinuity.

f(x)= cases x+1, & if x ≥ 1 x^2+1, if x<1 cases

No point of discontinuity. f(x)=x^2+1 for x<1 and f(x)=x+1 for x≥ 1. Each piece is a polynomial, so f is continuous on (-∞,1) and on (1,∞); only x=1 needs testing. At x=1: lim x→ 1^- f(x)= lim x→ 1^- (x^2+1)=1+1=2, lim x→ 1^+ f(x)= lim x→ 1^+ (x+1)=2, and f(1)=1+1=2. Since LHL = RHL =f(1)=2, f is continuous at x=1 as well. Answer: f is continuous on R; there is no point of discontinuity.

f(x)= cases x^3-3, & if x ≤ 2 x^2+1, & if x 2 cases

No point of discontinuity. f(x)=x^3-3 for x≤ 2 and f(x)=x^2+1 for x 2. Both pieces are polynomials, hence continuous on (-∞,2) and (2,∞) respectively; test the joining point x=2. At x=2: lim x→ 2^- (x^3-3)=8-3=5, lim x→ 2^+ (x^2+1)=4+1=5, f(2)=2^3-3=5. All three are equal, so f is continuous at x=2. Answer: f is continuous on R; it has no point of discontinuity.

f(x)= cases x^10-1, & if x ≤ 1 x^2, & if x 1 cases

f is discontinuous at x=1. f(x)=x^10-1 for x≤ 1 and f(x)=x^2 for x 1. Both pieces are polynomials, hence continuous away from the joining point; test x=1. At x=1: lim x→ 1^- (x^10-1)=1-1=0, lim x→ 1^+ x^2=1, f(1)=1^10-1=0. Since LHL =0≠ 1= RHL, lim x→ 1f(x) does not exist. Answer: x=1 is the only point of discontinuity of f.

Is the function defined by f(x)= cases x+5, & if x ≤ 1 x-5, & if x 1 cases a continuous function? Discuss the continuity of the function f, where f is defined by

f is not continuous at x=1. f(x)=x+5 for x≤ 1 and f(x)=x-5 for x 1. Each piece is a polynomial, so f is continuous at every point of (-∞,1) and of (1,∞). The only doubtful point is x=1. At x=1: lim x→ 1^- (x+5)=6, lim x→ 1^+ (x-5)=-4, f(1)=1+5=6. Since LHL =6≠ -4= RHL, the limit at x=1 does not exist, so f is discontinuous at x=1. Answer: f is not a continuous function; it is continuous at every real point except x=1.

f is not continuous at x=1 and x=3. The function is defined on the domain [0,10], and is constant on each of [0,1], (1,3) and [3,10]. A constant function is continuous, so f is continuous at every point of (0,1), (1,3) and (3,10), and one-sidedly continuous at the end points x=0 and x=10. Only x=1 and x=3 need testing. At x=1: lim x→ 1^- f(x)=3, lim x→ 1^+ f(x)=4, f(1)=3. LHL ≠ RHL, so f is discontinuous at x=1. At x=3: lim x→ 3^- f(x)=4, lim x→ 3^+ f(x)=5, f(3)=5. LHL ≠ RHL, so f is discontinuous at x=3. Answer: f is continuous on [0,10] except at x=1 and x=3, which are its only points of discontinuity.

f(x)= cases 2 x, & if x<0 0, & if 0 ≤ x ≤ 1 4 x, & if x 1 cases

x=1 is the only point of discontinuity. f(x)=2x for x<0, f(x)=0 for 0≤ x≤ 1, and f(x)=4x for x 1. Each piece is a polynomial, so continuity can fail only at the joining points x=0 and x=1. At x=0: lim x→ 0^- 2x=0, lim x→ 0^+ 0=0, f(0)=0. All three agree, so f is continuous at x=0. At x=1: lim x→ 1^- f(x)=0, lim x→ 1^+ 4x=4, f(1)=0. Since LHL =0≠ 4= RHL, f is discontinuous at x=1. Answer: f is continuous everywhere except at x=1, which is its only point of discontinuity.

f(x)= cases -2, & if x ≤-1 2 x, & if -1<x ≤ 1 2, & if x 1 cases

Continuous. f(x)=-2 for x≤ -1, f(x)=2x for -1<x≤ 1, and f(x)=2 for x 1. Each piece is a constant or a polynomial, so only the joining points x=-1 and x=1 need testing. At x=-1: lim x→ -1^- f(x)=-2, lim x→ -1^+ 2x=2(-1)=-2, f(-1)=-2. All three agree, so f is continuous at x=-1. At x=1: lim x→ 1^- 2x=2, lim x→ 1^+ f(x)=2, f(1)=2(1)=2. All three agree, so f is continuous at x=1. Answer: f is continuous on R; it has no point of discontinuity.

Find the relationship between a and b so that the function f defined by f(x)= cases a x+1, & if x ≤ 3 b x+3, & if x 3 cases is continuous at x=3.

Continuity at x=3 requires LHL = RHL =f(3). Value: f(3)=a(3)+1=3a+1 (the point x=3 belongs to the first branch). Left-hand limit: lim x→ 3^- (ax+1)=3a+1. Right-hand limit: lim x→ 3^+ (bx+3)=3b+3. The left-hand limit already equals f(3), so the single condition is 3a+1=3b+3 ⇒ 3a-3b=2 ⇒ a-b=2/3. Answer: f is continuous at x=3 precisely when a=b+2/3, i.e. a-b=2/3.

For what value of λ is the function defined by f(x)= cases λ(x^2-2 x), & if x ≤ 0 4 x+1, & if x 0 cases continuous at x=0? What about continuity at x=1?

For no value of λ, f is continuous at x=0 but f is continuous at x=1 for any value of λ.. At x=0, compare the two one-sided limits with f(0). Value: f(0)=λ(0^2-2· 0)=0. Left-hand limit: lim x→ 0^- λ(x^2-2x)=λ(0-0)=0. Right-hand limit: lim x→ 0^+ (4x+1)=1. The right-hand limit is 1 and does not involve λ at all, while the left-hand limit is 0 for every λ. Since 0≠ 1, the limit at x=0 never exists. Answer: there is no value of λ for which f is continuous at x=0. At x=1: every point near 1 satisfies x 0, so f(x)=4x+1 throughout a neighbourhood of 1. Hence lim x→ 1f(x)=5=f(1), and f is continuous at x=1 for every value of λ. (Likewise f is continuous at every point other than x=0.)

Show that the function defined by g(x)=x-[x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Let c be any integer. Recall the definition of the greatest integer function: [x] is the greatest integer ≤ x. Hence c-1≤ x<c ⇒ [x]=c-1, c≤ x<c+1 ⇒ [x]=c. Left-hand limit at x=c: for x just below c we have [x]=c-1, so lim x→ c^- g(x)= lim x→ c^- (x-(c-1))=c-c+1=1. Right-hand limit at x=c: for x just above c we have [x]=c, so lim x→ c^+ g(x)= lim x→ c^+ (x-c)=0. Value: g(c)=c-[c]=c-c=0. Since LHL =1≠ 0= RHL, lim x→ cg(x) does not exist, so g is discontinuous at x=c. As c was an arbitrary integer, g(x)=x-[x] is discontinuous at all integral points.

Is the function defined by f(x)=x^2- sin x+5 continuous at x=π?

f is continuous at x=π. Use the algebra of continuous functions: if g and h are continuous at c, then g± h is continuous at c. Here f(x)=x^2- sin x+5, where x x^2 is a polynomial (continuous on R ), x sin x is continuous on R, and x 5 is constant. Hence f is continuous at every real point, in particular at x=π. Checking directly at x=π, put x=π+h so that h→ 0: lim x→ (x)= lim h→ 0[(π+h)^2- sin (π+h)+5]= lim h→ 0[(π+h)^2+ sin h+5]=π^2+0+5. And f(π)=π^2- sin π+5=π^2+5. Since lim x→ (x)=π^2+5=f(π), the answer is yes: f is continuous at x=π.

Discuss the continuity of the following functions: (a) f(x)= sin x+ cos x (b) f(x)= sin x- cos x (c) f(x)= sin x · cos x

(a), (b) and (c) are all continuous. Standard facts used: sin x and cos x are continuous at every real point, and by the algebra of continuous functions the sum, difference and product of two functions continuous at c are continuous at c. That sin is continuous: at any c, putting x=c+h, lim h→ 0 sin (c+h)= lim h→ 0( sin c cos h+ cos c sin h)= sin c· 1+ cos c· 0= sin c, and similarly lim h→ 0 cos (c+h)= cos c cos h- sin c sin h→ cos c. (a) f(x)= sin x+ cos x is the sum of two functions continuous on R, hence continuous at every real point. (b) f(x)= sin x- cos x is their difference, hence continuous at every real point. (c) f(x)= sin x· cos x is their product, hence continuous at every real point. Answer: all three functions are continuous on R; none has a point of discontinuity.

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Cosine function is continuous for all x ∈ R; cosecant is continuous except for x=n π, n ∈ Z; secant is continuous except for x=(2 n+1) (π)/2, n ∈ Z and cotangent function is continuous except for x=n π, n ∈ Z. Take as known that sin x and cos x are continuous on R, and use the quotient rule: if g,h are continuous at c and h(c)≠ 0, then g/h is continuous at c. Cosine. At any real c, with x=c+h, lim h→ 0 cos (c+h)= lim h→ 0( cos c cos h- sin c sin h)= cos c. So cos is continuous on R (domain R, no point of discontinuity). Cosecant. cosec x=1/( sin x), defined where sin x≠ 0, i.e. on R - nπ: n∈ Z. At any c in this domain, sin c≠ 0, so the quotient rule gives continuity. Hence cosec is continuous on its whole domain R - nπ. Secant. sec x=1/( cos x), defined where cos x≠ 0, i.e. on R - (2n+1)(π)/2: n∈ Z. By the quotient rule it is continuous at every point of that domain. Cotangent. cot x=( cos x)/( sin x), defined on R - nπ: n∈ Z, and continuous there by the quotient rule. Answer: each of cos, cosec, sec and cot is continuous at every point of its own domain; the excluded points nπ (for cosec, cot ) and (2n+1)π/2 (for sec ) are not points of discontinuity, because the functions are not defined there.

Find all points of discontinuity of f, where f(x)= cases ( sin x)/x, & if x<0 x+1, & if x ≥ 0 cases

There is no point of discontinuity.. For x<0: f(x)=( sin x)/x is a quotient of two functions continuous on R with denominator x≠ 0, so f is continuous at every c<0. For x 0: f(x)=x+1 is a polynomial, so f is continuous at every c 0. Only x=0 remains. Using the standard limit lim x→ 0( sin x)/x=1, lim x→ 0^- f(x)= lim x→ 0^- ( sin x)/x=1, lim x→ 0^+ f(x)= lim x→ 0^+ (x+1)=1, and f(0)=0+1=1. Since LHL = RHL =f(0)=1, f is continuous at x=0 too. Answer: f has no point of discontinuity; it is continuous on R.

Determine if f defined by f(x)= cases x^2 sin 1/x, & if x ≠ 0 0, & if x=0 cases is a continuous function?

Yes, f is continuous for all x ∈ R. For x≠ 0: x x^2 is continuous, x 1/x is continuous where x≠ 0, and sin is continuous on R; so sin 1/x is continuous at every c≠ 0 (composition of continuous functions), and the product x^2 sin 1/x is continuous at every c≠ 0. At x=0 the factor sin 1x has no limit, so use the sandwich (squeeze) theorem instead. For all x≠ 0, -1≤ sin 1/x≤ 1 ⇒ -x^2≤ x^2 sin 1/x≤ x^2. Since lim x→ 0(-x^2)=0= lim x→ 0x^2, the sandwich theorem gives lim x→ 0x^2 sin 1/x=0. And f(0)=0, so lim x→ 0f(x)=f(0). Answer: yes, f is continuous at every real point, i.e. f is a continuous function.

Examine the continuity of f, where f is defined by f(x)= cases sin x- cos x, & if x ≠ 0 -1, & if x=0 cases

f is continuous for all x ∈ R. For x≠ 0: f(x)= sin x- cos x is the difference of two functions continuous on R, hence continuous at every c≠ 0. At x=0: using continuity of sin and cos, lim x→ 0f(x)= lim x→ 0( sin x- cos x)= sin 0- cos 0=0-1=-1, while the definition gives f(0)=-1. Since lim x→ 0f(x)=-1=f(0), f is continuous at x=0 as well. (The second branch simply supplies the value the first branch was heading towards, so the definition creates no break.) Answer: f is continuous at every point of R.

k=6. Continuity at x=(π)/2 requires lim x→ π/2f(x)=f ((π)/2)=3. The expression (k cos x)/(π-2x) is of the form 0/0 at x=π/2, so substitute x=(π)/2+h, where h→ 0 as x→ π/2. Then cos x= cos ((π)/2+h)=- sin h, π-2x=π-π-2h=-2h. Hence lim x→ π/2(k cos x)/(π-2x)= lim h→ 0(k(- sin h))/-2h=k/2 lim h→ 0( sin h)/h=k/2· 1=k/2. Setting this equal to the value 3: k/2=3 ⇒ k=6. Answer: k=6.

k=3/4. Continuity at x=2 requires LHL = RHL =f(2). Value: f(2)=k(2)^2=4k (the point x=2 lies in the branch x≤ 2 ). Left-hand limit: lim x→ 2^- kx^2=4k. Right-hand limit: lim x→ 2^+ 3=3. Since LHL already equals f(2), the condition reduces to 4k=3 ⇒ k=3/4. Answer: k=3/4.

k=-2/(π). Continuity at x=π requires LHL = RHL =f(π). Value: f(π)=kπ+1 (the point x=π lies in the branch x≤π ). Left-hand limit: lim x→ π^- (kx+1)=kπ+1. Right-hand limit: lim x→ π^+ cos x= cos π=-1 (using continuity of cos ). Equating, kπ+1=-1 ⇒ kπ=-2 ⇒ k=-2/(π). Answer: k=-2/(π).

k=9/5. Continuity at x=5 requires LHL = RHL =f(5). Value: f(5)=5k+1 (the point x=5 lies in the branch x≤ 5 ). Left-hand limit: lim x→ 5^- (kx+1)=5k+1. Right-hand limit: lim x→ 5^+ (3x-5)=15-5=10. Equating, 5k+1=10 ⇒ 5k=9 ⇒ k=9/5. Answer: k=9/5.

Find the values of a and b such that the function defined by f(x)= cases 5, & if x ≤ 2 a x+b, & if 2<x<10 21, & if x ≥ 10 cases is a continuous function.

a=2, b=1. Each branch (a constant, a linear function, a constant) is continuous on the open interval where it applies, so f can fail to be continuous only at the joining points x=2 and x=10. Impose continuity at both. At x=2: f(2)=5, LHL = lim x→ 2^- 5=5, RHL = lim x→ 2^+ (ax+b)=2a+b. Continuity gives 2a+b=5. (1) At x=10: f(10)=21, LHL = lim x→ 10^- (ax+b)=10a+b, RHL = lim x→ 10^+ 21=21. Continuity gives 10a+b=21. (2) Subtract (1) from (2): 8a=16 ⇒ a=2, and then from (1), b=5-2a=5-4=1. Answer: a=2 and b=1 (so the middle branch is 2x+1 ), and with these values f is continuous on R.

Show that the function defined by f(x)= cos (x^2) is a continuous function.

Use the theorem on composition: if g is continuous at c and h is continuous at g(c), then h° g is continuous at c. Write f=h° g with g(x)=x^2, h(y)= cos y, (h° g)(x)= cos (x^2)=f(x). g(x)=x^2 is a polynomial, hence continuous at every real c. h(y)= cos y is continuous at every real y: with y=d+t, lim t→ 0 cos (d+t)= lim t→ 0( cos d cos t- sin d sin t)= cos d. In particular h is continuous at g(c)=c^2. Therefore h° g is continuous at every real c; that is, f(x)= cos (x^2) is a continuous function on R.

Show that the function defined by f(x)= cos x is a continuous function.

Use the composition theorem again. Write f=h° g with g(x)= cos x, h(y)= y, (h° g)(x)= cos x =f(x). g(x)= cos x is continuous at every real c. h(y)= y is continuous at every real y: for d 0, h(y)=y near d; for d<0, h(y)=-y near d; and at d=0, lim y→ 0^- (-y)=0= lim y→ 0^+ y=h(0). So h is continuous everywhere, in particular at y=g(c)= cos c. Since g is continuous at c and h is continuous at g(c), the composite h° g is continuous at c; c being arbitrary, f(x)= cos x is a continuous function on R. (Note that the corners of the graph, at x=(2n+1)π/2 where cos x=0, are points where f is continuous but not differentiable.)

Examine that sin x is a continuous function.

Take f=h° g with g(x)= x, h(y)= sin y, (h° g)(x)= sin x. g(x)= x is continuous at every real c: for c 0 it agrees with x near c and for c<0 with -x near c; at c=0, lim x→ 0^- (-x)=0, lim x→ 0^+ x=0, g(0)=0, so all three agree. h(y)= sin y is continuous at every real y, since lim t→ 0 sin (d+t)= lim t→ 0( sin d cos t+ cos d sin t)= sin d. By the theorem on the composition of continuous functions, h° g is continuous at every real c. Hence sin x is a continuous function on R.

Find all the points of discontinuity of f defined by f(x)= x - x+1.

There is no point of discontinuity.. Remove both moduli by splitting at the points where the expressions inside change sign, namely x=-1 and x=0. For x<-1: x =-x and x+1 =-(x+1), so f(x)=-x+(x+1)=1. For -1≤ x<0: x =-x and x+1 =x+1, so f(x)=-x-(x+1)=-2x-1. For x≥ 0: x =x and x+1 =x+1, so f(x)=x-(x+1)=-1. Thus f(x)= cases 1,&x<-1 -2x-1,&-1≤ x<0 -1,&x≥ 0. cases Each piece is a polynomial, so only x=-1 and x=0 need testing. At x=-1: lim x→ -1^- f(x)=1, lim x→ -1^+ (-2x-1)=2-1=1, and f(-1)=-2(-1)-1=1. All equal, so f is continuous at x=-1. At x=0: lim x→ 0^- (-2x-1)=-1, lim x→ 0^+ f(x)=-1, and f(0)=-1. All equal, so f is continuous at x=0. (The same conclusion follows at once from the algebra of continuous functions: x and x+1 are continuous on R, so their difference is too.) Answer: f has no point of discontinuity; it is continuous on the whole of R.

2 x cos (x^2+5). Differentiate with respect to x by the chain rule: if y=f(u) and u=g(x) are both differentiable, then dy/dx=dy/du /dx. Put u=x^2+5, so that y= sin u. Then dy/du= cos u, du/dx=2x. The step most often dropped is the second factor du/dx; cos (x^2+5) alone is not the derivative. Multiplying, dy/dx= cos (x^2+5)· 2x. Answer: dy/dx=2x cos (x^2+5), valid for all x∈ R.

- cos x sin ( sin x). Differentiate with respect to x by the chain rule, dy/dx=dy/du /dx. Let u= sin x, so y= cos u. Then dy/du=- sin u=- sin ( sin x), du/dx= cos x. Note the two sines play different roles: the outer one comes from differentiating cos, and its argument stays sin x untouched. Hence dy/dx=- sin ( sin x)· cos x. Answer: dy/dx=- cos x sin ( sin x), valid for all x∈ R.

a cos (a x+b). Differentiate with respect to x by the chain rule, treating a and b as constants. Let u=ax+b, so y= sin u. Then dy/du= cos u= cos (ax+b), du/dx=a. Therefore dy/dx= cos (ax+b)· a. Answer: dy/dx=a cos (ax+b), valid for all x∈ R.

( sec ( tan √ x ) · tan ( tan √ x ) · sec^2 √ x )/(2 √ x ). This is a chain of three functions, so the chain rule is applied twice: with y= sec v, v= tan u, u=√ x, dy/dx=dy/dv /du /dx. The three factors are dy/dv= sec v tan v= sec ( tan √ x ) tan ( tan √ x ), dv/du= sec^2u= sec^2√ x, du/dx=1/(2√ x ). The factor most often forgotten is the innermost 1/(2√ x ), obtained from d/dxx^1/2= 12 x^-1/2. Multiplying the three, dy/dx=( sec ( tan √ x ) tan ( tan √ x ) sec^2 (√ x ))/(2√ x ). Answer: dy/dx=( sec ( tan √ x ) tan ( tan √ x ) sec^2(√ x ))/(2√ x ), for x 0 (with √ x and tan √ x avoiding the odd multiples of (π)/2 ); at x=0 the expression 1/(2√ x ) is undefined.

( sin (a x+b))/( cos (c x+d))

a cos (a x+b) sec (c x+d)+c sin (a x+b) tan (c x+d) sec (c x+d). This is a quotient, so use the quotient rule d/dx(u/v)=(vdu/dx-udv/dx)/(v^2), with each of u and v differentiated by the chain rule. Take u= sin (ax+b) and v= cos (cx+d). Then du/dx=a cos (ax+b), dv/dx=-c sin (cx+d). The sign is the trap here: dv/dx is already negative, and the quotient rule subtracts it, so the second term comes out positive. Substituting, dy/dx=( cos (cx+d)· a cos (ax+b)- sin (ax+b)· (-c sin (cx+d) ))/( cos^2(cx+d)), dy/dx=(a cos (ax+b) cos (cx+d)+c sin (ax+b) sin (cx+d))/( cos^2(cx+d)). Answer: dy/dx=(a cos (ax+b) cos (cx+d)+c sin (ax+b) sin (cx+d))/( cos^2(cx+d)), valid wherever cos (cx+d)≠ 0.

cos x^3 · sin^2(x^5)

10 x^4 sin x^5 cos x^5 cos x^3-3 x^2 sin x^3 sin^2 x^5. Read the expression as y= cos (x^3)· ( sin (x^5) )^2: it is a product, so use the product rule d/dx(uv)=u'v+uv', with each factor differentiated by the chain rule. Let u= cos (x^3) and v= sin^2(x^5). For u: du/dx=- sin (x^3)· 3x^2=-3x^2 sin (x^3). For v, the chain rule is needed twice — once for the square and once for x^5: dv/dx=2 sin (x^5)· cos (x^5)· 5x^4=10x^4 sin (x^5) cos (x^5). Hence dy/dx=-3x^2 sin (x^3) sin^2(x^5)+10x^4 cos (x^3) sin (x^5) cos (x^5). Using 2 sin θ cos θ= sin 2θ with θ=x^5, the second term is 5x^4 cos (x^3) sin (2x^5). Answer: dy/dx=10x^4 cos (x^3) sin (x^5) cos (x^5)-3x^2 sin (x^3) sin^2(x^5), equivalently 5x^4 cos (x^3) sin (2x^5)-3x^2 sin (x^3) sin^2(x^5).

-2 √ 2 x sin x^2 √ sin 2 x^2. Write y=2√ cot (x^2) =2 ( cot (x^2) )^1/2 and apply the chain rule twice, using d/dt√ t =1/(2√ t ) and d/du cot u=- cosec^2u. With t= cot u and u=x^2, dy/dx=2· 1 2√ cot (x^2) ·(- cosec^2(x^2))· 2x. The minus sign comes from cot, and the last factor 2x from the inner x^2; dropping either is the usual error. Simplifying the numerical factors, dy/dx= -2x cosec^2(x^2) √ cot (x^2). This may be recast in sines and cosines: since cosec^2(x^2)=1/( sin^2x^2) and √ cot x^2 = √ cos x^2 √ sin x^2, dy/dx= -2x sin^3/2(x^2)√ cos (x^2) = -2√ 2 x sin (x^2)√ sin (2x^2). Answer: dy/dx= -2x cosec^2(x^2) √ cot (x^2), valid where cot (x^2) 0.

-( sin √ x )/(2 √ x ). Differentiate with respect to x by the chain rule, with u=√ x and y= cos u. dy/du=- sin u=- sin √ x, du/dx=d/dxx^1/2=1/(2√ x ). Multiplying, dy/dx=- sin (√ x )·1/(2√ x ). Answer: dy/dx=-( sin √ x )/(2√ x ), for x 0; the function is defined at x=0 but this derivative formula is not, since 1/(2√ x )→∞ as x→ 0^+.

Prove that the function f given by f(x)= x-1, x ∈ R is not differentiable at x=1.

By definition, f is differentiable at x=c only if the limit lim h→ 0(f(c+h)-f(c))/h exists, i.e. only if the left-hand derivative and the right-hand derivative at c both exist and are equal. So compute the two one-sided limits at c=1. Here f(x)= x-1, so f(1)=0 and f(1+h)= 1+h-1 = h. The whole proof turns on unpacking h by sign, which is the case split students skip. Left-hand derivative ( h<0, so h =-h ): Lf'(1)= lim h→ 0^- (f(1+h)-f(1))/h= lim h→ 0^- ( h -0)/h= lim h→ 0^- -h/h=-1. Right-hand derivative ( h 0, so h =h ): Rf'(1)= lim h→ 0^+ ( h -0)/h= lim h→ 0^+ h/h=1. Since Lf'(1)=-1≠ 1=Rf'(1), the two one-sided limits differ, so lim h→ 0(f(1+h)-f(1))/h does not exist. Hence f(x)= x-1 is not differentiable at x=1. (It is continuous there — continuity does not imply differentiability; the graph has a corner at (1,0).)

Prove that the greatest integer function defined by f(x)=[x], 0<x<3 is not differentiable at x=1 and x=2.

Let f(x)=[x] on 0<x<3, where [x] is the greatest integer not exceeding x. Differentiability at c requires that lim h→ 0(f(c+h)-f(c))/h exist, so test the one-sided derivatives. The key step is evaluating [c+h] separately for h<0 and h 0, because the bracket function jumps exactly at integers. At x=1. Here f(1)=[1]=1. For small h 0 (say 0<h<1 ), 1<1+h<2, so [1+h]=1; for small h<0 (say -1<h<0 ), 0<1+h<1, so [1+h]=0. Left-hand derivative: Lf'(1)= lim h→ 0^- ([1+h]-[1])/h= lim h→ 0^- 0-1/h= lim h→ 0^- -1/h, and as h→ 0^- the quotient -1/h→+∞; the limit does not exist (it is not a finite number). Right-hand derivative: Rf'(1)= lim h→ 0^+ 1-1/h=0. Since Lf'(1) does not exist, f is not differentiable at x=1. At x=2. Here f(2)=[2]=2. For 0<h<1, [2+h]=2; for -1<h<0, 1<2+h<2 so [2+h]=1. Lf'(2)= lim h→ 0^- 1-2/h= lim h→ 0^- -1/h→+∞ ( does not exist ), Rf'(2)= lim h→ 0^+ 2-2/h=0. So f is not differentiable at x=2 either. Alternative one-line argument: every function differentiable at a point is continuous there. Since lim x→ 1^- [x]=0≠ 1=f(1) and lim x→ 2^- [x]=1≠ 2=f(2), f is discontinuous at x=1 and x=2, and therefore cannot be differentiable at those points. Hence the greatest integer function f(x)=[x], 0<x<3, is not differentiable at x=1 and x=2.

( cos x-2)/3. Implicit differentiation: differentiate every term with respect to x, remembering that y is a function of x, so d/dx(3y)=3dy/dx by the chain rule. d/dx(2x)+d/dx(3y)=d/dx( sin x) 2+3dy/dx= cos x Solve for the derivative: 3dy/dx= cos x-2 dy/dx=( cos x-2)/3

2/( cos y-3). Here the right-hand side is sin y, a function of y which is itself a function of x, so it must be differentiated by the chain rule as cos y /dx - this is the step most often dropped. d/dx(2x)+d/dx(3y)=d/dx( sin y) 2+3dy/dx= cos y dy/dx Collect the dy/dx terms on one side: 2=( cos y-3)dy/dx Since cos y ≤ 1 for every y, the factor cos y-3 is never zero (it lies in [-4,-2] ), so division is always legitimate. dy/dx=2/( cos y-3)

-a/(2 b y+ sin y). a and b are constants. Differentiate implicitly with respect to x: the term by^2 gives 2bydy/dx and cos y gives - sin ydy/dx, both by the chain rule. a+2bydy/dx=- sin y dy/dx Bring the derivative terms together: 2bydy/dx+ sin y dy/dx=-a (2by+ sin y)dy/dx=-a dy/dx=-a/(2by+ sin y), 2by+ sin y 0.

( sec^2 x-y)/(x+2 y-1). The term xy is a product of two functions of x, so it needs the product rule d/dx(uv)=u'v+uv'; everything containing y also needs the chain rule. d/dx(xy)+d/dx(y^2)=d/dx( tan x)+d/dx(y) (y· 1+xdy/dx)+2ydy/dx= sec^2x+dy/dx Collect all dy/dx terms on the left and the rest on the right: xdy/dx+2ydy/dx-dy/dx= sec^2x-y (x+2y-1)dy/dx= sec^2x-y dy/dx=( sec^2x-y)/(x+2y-1), x+2y 1.

-((2 x+y))/((x+2 y)). Differentiate implicitly with respect to x; the middle term xy requires the product rule and the constant 100 differentiates to 0. d/dx(x^2)+d/dx(xy)+d/dx(y^2)=d/dx(100) 2x+(y+xdy/dx)+2ydy/dx=0 Group the derivative terms: (x+2y)dy/dx=-(2x+y) dy/dx=-(2x+y)/(x+2y), x+2y 0.

x^3+x^2 y+x y^2+y^3=81

-((3 x^2+2 x y+y^2))/((x^2+2 x y+3 y^2)). Differentiate term by term with respect to x. Both x^2y and xy^2 are products, so each needs the product rule together with the chain rule on the y -factor. d/dx(x^2y)=2xy+x^2dy/dx, d/dx(xy^2)=y^2+2xydy/dx So the whole equation gives 3x^2+(2xy+x^2dy/dx)+(y^2+2xydy/dx)+3y^2dy/dx=0 Collect the coefficient of dy/dx: (x^2+2xy+3y^2)dy/dx=-(3x^2+2xy+y^2) dy/dx=-(3x^2+2xy+y^2)/(x^2+2xy+3y^2), x^2+2xy+3y^2 0.

(y sin x y)/( sin 2 y-x sin x y). Here the right-hand side is a constant, so it differentiates to 0. Two chain rules are needed - one for sin^2y=( sin y)^2 and one for cos (xy) - and inside the second, d/dx(xy) needs the product rule. d/dx( sin^2y)=2 sin y cos y dy/dx= sin 2y dy/dx d/dx( cos xy)=- sin (xy) /dx(xy)=- sin (xy)(y+xdy/dx) Adding and equating to 0: sin 2y dy/dx-y sin (xy)-x sin (xy)dy/dx=0 Collect dy/dx: ( sin 2y-x sin (xy))dy/dx=y sin (xy) dy/dx=(y sin (xy))/( sin 2y-x sin (xy)), sin 2y x sin (xy).

SolveItNCERT · CBSE Boards

NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

137 exercises · 137 still being checked

EXERCISE 5.1 1–10 (part 1 of 15)

Exercise 1
Prove that the function $\displaystyle f(x)=5 x-3$ is continuous at $\displaystyle x=0$, at $\displaystyle x=-3$ and at $\displaystyle x=5$.
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Definition: $\displaystyle f$ is continuous at $\displaystyle x=c$ if $\displaystyle f(c)$ is defined, $\displaystyle \lim_{x\to c}f(x)$ exists, and $\displaystyle \lim_{x\to c}f(x)=f(c)$.Here $\displaystyle f(x)=5x-3$ is defined for every real $\displaystyle x$.At $\displaystyle x=0$: $\displaystyle f(0)=5(0)-3=-3$, and by the algebra of limits $\displaystyle \lim_{x\to 0}(5x-3)=5(0)-3=-3$. Since $\displaystyle \lim_{x\to 0}f(x)=f(0)$, $\displaystyle f$ is continuous at $\displaystyle x=0$.At $\displaystyle x=-3$: $\displaystyle f(-3)=5(-3)-3=-18$, and $\displaystyle \lim_{x\to -3}(5x-3)=5(-3)-3=-18$. The two agree, so $\displaystyle f$ is continuous at $\displaystyle x=-3$.At $\displaystyle x=5$: $\displaystyle f(5)=5(5)-3=22$, and $\displaystyle \lim_{x\to 5}(5x-3)=5(5)-3=22$. The two agree, so $\displaystyle f$ is continuous at $\displaystyle x=5$.Hence $\displaystyle f(x)=5x-3$ is continuous at $\displaystyle x=0$, at $\displaystyle x=-3$ and at $\displaystyle x=5$.
Exercise 2
Examine the continuity of the function $\displaystyle f(x)=2 x^{2}-1$ at $\displaystyle x=3$.
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NCERT’s answer
$\displaystyle f$ is continuous at $\displaystyle x=3$
Test the definition of continuity at $\displaystyle x=3$: compare $\displaystyle \lim_{x\to 3}f(x)$ with $\displaystyle f(3)$.The function $\displaystyle f(x)=2x^{2}-1$ is defined for all real $\displaystyle x$, so $\displaystyle f(3)=2(3)^{2}-1=18-1=17$.By the algebra of limits (limit of a product and of a sum), \[\lim_{x\to 3}\left(2x^{2}-1\right)=2\left(\lim_{x\to 3}x\right)^{2}-1=2(3)^{2}-1=17.\]Since $\displaystyle \lim_{x\to 3}f(x)=17=f(3)$, the function $\displaystyle f$ is continuous at $\displaystyle x=3$.
Exercise 3
Examine the following functions for continuity.
(a)
$\displaystyle f(x)=x-5$
(b)
$\displaystyle f(x)=\frac{1}{x-5}, x \neq 5$
(c)
$\displaystyle f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$
(d)
$\displaystyle f(x)=|x-5|$
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NCERT’s answer
(a)
, (b), (c) and (d) are all continuous functions
In each part, take an arbitrary point $\displaystyle c$ of the domain and compare $\displaystyle \lim_{x\to c}f(x)$ with $\displaystyle f(c)$. A point outside the domain is not a point of discontinuity, since continuity is only discussed at points where the function is defined.
(a)
$\displaystyle f(x)=x-5$, domain $\displaystyle \mathbf{R}$. For any real $\displaystyle c$, $\displaystyle \lim_{x\to c}(x-5)=c-5=f(c)$. Hence $\displaystyle f$ is continuous at every real point; there is no point of discontinuity.
(b)
$\displaystyle f(x)=\dfrac{1}{x-5}$, $\displaystyle x\neq 5$; domain $\displaystyle \mathbf{R}-\{5\}$. Let $\displaystyle c\neq 5$. Then $\displaystyle c-5\neq 0$, so by the quotient rule for limits
\[\lim_{x\to c}\frac{1}{x-5}=\frac{1}{c-5}=f(c).\]
Hence $\displaystyle f$ is continuous at every point of its domain. (At $\displaystyle x=5$ the function is not defined, so it is not a point of discontinuity.)
(c)
$\displaystyle f(x)=\dfrac{x^{2}-25}{x+5}$, $\displaystyle x\neq -5$; domain $\displaystyle \mathbf{R}-\{-5\}$. Let $\displaystyle c\neq -5$; then $\displaystyle c+5\neq 0$ and
\[\lim_{x\to c}\frac{x^{2}-25}{x+5}=\frac{c^{2}-25}{c+5}=f(c).\]
Hence $\displaystyle f$ is continuous at every point of its domain. (Cancelling gives $\displaystyle f(x)=x-5$ for $\displaystyle x\neq-5$, which is the same conclusion.)
(d)
$\displaystyle f(x)=|x-5|$, domain $\displaystyle \mathbf{R}$. Write $\displaystyle f(x)=5-x$ for $\displaystyle x<5$ and $\displaystyle f(x)=x-5$ for $\displaystyle x\ge 5$.
For $\displaystyle c<5$ and for $\displaystyle c>5$, $\displaystyle f$ agrees near $\displaystyle c$ with a polynomial, so it is continuous there.
At $\displaystyle c=5$: LHL $\displaystyle =\lim_{x\to 5^{-}}(5-x)=0$, RHL $\displaystyle =\lim_{x\to 5^{+}}(x-5)=0$, and $\displaystyle f(5)=0$. All three are equal, so $\displaystyle f$ is continuous at $\displaystyle x=5$.
Thus $\displaystyle |x-5|$ is continuous at every real point.
Exercise 4
Prove that the function $\displaystyle f(x)=x^{n}$ is continuous at $\displaystyle x=n$, where $\displaystyle n$ is a positive integer.
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Here $\displaystyle f(x)=x^{n}$ with $\displaystyle n$ a positive integer, defined for all real $\displaystyle x$.Value at the point: $\displaystyle f(n)=n^{n}$.Limit at the point: $\displaystyle x^{n}$ is the product of $\displaystyle n$ copies of $\displaystyle x$, so by the product rule for limits applied $\displaystyle n-1$ times, \[\lim_{x\to n}x^{n}=\left(\lim_{x\to n}x\right)^{n}=n^{n}.\]Since $\displaystyle \lim_{x\to n}f(x)=n^{n}=f(n)$, the function $\displaystyle f(x)=x^{n}$ is continuous at $\displaystyle x=n$.(The same argument at any real $\displaystyle c$ shows $\displaystyle x^{n}$, and hence every polynomial, is continuous on $\displaystyle \mathbf{R}$.)
Exercise 5
Is the function $\displaystyle f$ defined by \[f(x)= \begin{cases}x, & \text { if } x \leq 1 \\ 5, & \text { if } x>1\end{cases} \] continuous at $\displaystyle x=0$ ? At $\displaystyle x=1$ ? At $\displaystyle x=2$ ? Find all points of discontinuity of $\displaystyle f$, where $\displaystyle f$ is defined by
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NCERT’s answer
$\displaystyle f$ is continuous at $\displaystyle x=0$ and $\displaystyle x=2$; Not continuous at $\displaystyle x=1$
The rule changes only at $\displaystyle x=1$: $\displaystyle f(x)=x$ for $\displaystyle x\le 1$ and $\displaystyle f(x)=5$ for $\displaystyle x>1$.At $\displaystyle x=0$: all points near $\displaystyle 0$ satisfy $\displaystyle x\le 1$, so $\displaystyle f(x)=x$ there. Hence $\displaystyle \lim_{x\to 0}f(x)=0$ and $\displaystyle f(0)=0$. They are equal, so $\displaystyle f$ is continuous at $\displaystyle x=0$.At $\displaystyle x=1$: the two one-sided limits must be computed from different rules. \[\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}x=1,\qquad \lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}5=5.\] Since LHL $\displaystyle \neq$ RHL, $\displaystyle \lim_{x\to 1}f(x)$ does not exist, so $\displaystyle f$ is discontinuous at $\displaystyle x=1$.At $\displaystyle x=2$: all points near $\displaystyle 2$ satisfy $\displaystyle x>1$, so $\displaystyle f(x)=5$ there. Hence $\displaystyle \lim_{x\to 2}f(x)=5=f(2)$, and $\displaystyle f$ is continuous at $\displaystyle x=2$.Answer: $\displaystyle f$ is continuous at $\displaystyle x=0$ and at $\displaystyle x=2$, and discontinuous at $\displaystyle x=1$.
Exercise 6
$\displaystyle f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.$
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NCERT’s answer
Discontinuous at $\displaystyle x=2$
$\displaystyle f(x)=2x+3$ for $\displaystyle x\le 2$ and $\displaystyle f(x)=2x-3$ for $\displaystyle x>2$. On each of $\displaystyle (-\infty,2)$ and $\displaystyle (2,\infty)$ the function is a polynomial, hence continuous; only the joining point $\displaystyle x=2$ needs testing.At $\displaystyle x=2$: \[\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{-}}(2x+3)=7,\qquad \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}(2x-3)=1,\] and $\displaystyle f(2)=2(2)+3=7$.Since LHL $\displaystyle =7\neq 1=$ RHL, the limit at $\displaystyle x=2$ does not exist.Answer: $\displaystyle x=2$ is the only point of discontinuity of $\displaystyle f$.
Exercise 7
$\displaystyle f(x)=\left\{\begin{array}{cl}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 \\ 6 x+2, & \text { if } x \geq 3\end{array}\right.$
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NCERT’s answer
Discontinuous at $\displaystyle x=3$
For $\displaystyle x\le -3$, $\displaystyle |x|=-x$, so $\displaystyle f(x)=-x+3$ there. Thus \[f(x)=\begin{cases}-x+3,&x\le -3\\ -2x,&-3<x<3\\ 6x+2,&x\ge 3.\end{cases}\] Each piece is a polynomial, so $\displaystyle f$ is continuous at every point of $\displaystyle (-\infty,-3)$, $\displaystyle (-3,3)$ and $\displaystyle (3,\infty)$. Only $\displaystyle x=-3$ and $\displaystyle x=3$ need testing.At $\displaystyle x=-3$: \[\lim_{x\to -3^{-}}(-x+3)=3+3=6,\qquad \lim_{x\to -3^{+}}(-2x)=6,\qquad f(-3)=|-3|+3=6.\] All three agree, so $\displaystyle f$ is continuous at $\displaystyle x=-3$.At $\displaystyle x=3$: \[\lim_{x\to 3^{-}}(-2x)=-6,\qquad \lim_{x\to 3^{+}}(6x+2)=20,\qquad f(3)=6(3)+2=20.\] Since LHL $\displaystyle =-6\neq 20=$ RHL, the limit does not exist at $\displaystyle x=3$.Answer: $\displaystyle x=3$ is the only point of discontinuity.
Exercise 8
$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$
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NCERT’s answer
Discontinuous at $\displaystyle x=0$
Remove the modulus: for $\displaystyle x>0$, $\displaystyle |x|=x$ so $\displaystyle \dfrac{|x|}{x}=1$; for $\displaystyle x<0$, $\displaystyle |x|=-x$ so $\displaystyle \dfrac{|x|}{x}=-1$. Hence \[f(x)=\begin{cases}-1,&x<0\\ 0,&x=0\\ 1,&x>0.\end{cases}\] On $\displaystyle (-\infty,0)$ and on $\displaystyle (0,\infty)$, $\displaystyle f$ is constant, hence continuous. Only $\displaystyle x=0$ needs testing.At $\displaystyle x=0$: \[\lim_{x\to 0^{-}}f(x)=-1,\qquad \lim_{x\to 0^{+}}f(x)=1,\qquad f(0)=0.\] Since LHL $\displaystyle \neq$ RHL, $\displaystyle \lim_{x\to 0}f(x)$ does not exist.Answer: $\displaystyle x=0$ is the only point of discontinuity of $\displaystyle f$.
Exercise 9
$\displaystyle f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0\end{cases}$
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NCERT’s answer
No point of discontinuity
Simplify the first branch. For $\displaystyle x<0$ we have $\displaystyle |x|=-x$, so \[\frac{x}{|x|}=\frac{x}{-x}=-1.\] Therefore $\displaystyle f(x)=-1$ for $\displaystyle x<0$ and $\displaystyle f(x)=-1$ for $\displaystyle x\ge 0$; that is, $\displaystyle f(x)=-1$ for every real $\displaystyle x$.A constant function is continuous: for any real $\displaystyle c$, $\displaystyle \lim_{x\to c}f(x)=-1=f(c)$. In particular at the joining point $\displaystyle x=0$, LHL $\displaystyle =$ RHL $\displaystyle =f(0)=-1$.Answer: $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$; it has no point of discontinuity.
Exercise 10
$\displaystyle f(x)= \begin{cases}x+1, & \text { if } x \geq 1 \\ x^{2}+1, \text { if } x<1\end{cases}$
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NCERT’s answer
No point of discontinuity
$\displaystyle f(x)=x^{2}+1$ for $\displaystyle x<1$ and $\displaystyle f(x)=x+1$ for $\displaystyle x\ge 1$. Each piece is a polynomial, so $\displaystyle f$ is continuous on $\displaystyle (-\infty,1)$ and on $\displaystyle (1,\infty)$; only $\displaystyle x=1$ needs testing.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}\left(x^{2}+1\right)=1+1=2,\qquad \lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}(x+1)=2,\] and $\displaystyle f(1)=1+1=2$.Since LHL $\displaystyle =$ RHL $\displaystyle =f(1)=2$, $\displaystyle f$ is continuous at $\displaystyle x=1$ as well.Answer: $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$; there is no point of discontinuity.

EXERCISE 5.1 11–20 (part 2 of 15)

Exercise 11
$\displaystyle f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \\ x^{2}+1, & \text { if } x>2\end{cases}$
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NCERT’s answer
No point of discontinuity
$\displaystyle f(x)=x^{3}-3$ for $\displaystyle x\le 2$ and $\displaystyle f(x)=x^{2}+1$ for $\displaystyle x>2$. Both pieces are polynomials, hence continuous on $\displaystyle (-\infty,2)$ and $\displaystyle (2,\infty)$ respectively; test the joining point $\displaystyle x=2$.At $\displaystyle x=2$: \[\lim_{x\to 2^{-}}\left(x^{3}-3\right)=8-3=5,\qquad \lim_{x\to 2^{+}}\left(x^{2}+1\right)=4+1=5,\qquad f(2)=2^{3}-3=5.\]All three are equal, so $\displaystyle f$ is continuous at $\displaystyle x=2$.Answer: $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$; it has no point of discontinuity.
Exercise 12
$\displaystyle f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^{2}, & \text { if } x>1\end{cases}$
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NCERT’s answer
$\displaystyle f$ is discontinuous at $\displaystyle x=1$
$\displaystyle f(x)=x^{10}-1$ for $\displaystyle x\le 1$ and $\displaystyle f(x)=x^{2}$ for $\displaystyle x>1$. Both pieces are polynomials, hence continuous away from the joining point; test $\displaystyle x=1$.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}\left(x^{10}-1\right)=1-1=0,\qquad \lim_{x\to 1^{+}}x^{2}=1,\qquad f(1)=1^{10}-1=0.\]Since LHL $\displaystyle =0\neq 1=$ RHL, $\displaystyle \lim_{x\to 1}f(x)$ does not exist.Answer: $\displaystyle x=1$ is the only point of discontinuity of $\displaystyle f$.
Exercise 13
Is the function defined by \[f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1\end{cases} \] a continuous function? Discuss the continuity of the function $\displaystyle f$, where $\displaystyle f$ is defined by
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NCERT’s answer
$\displaystyle f$ is not continuous at $\displaystyle x=1$
$\displaystyle f(x)=x+5$ for $\displaystyle x\le 1$ and $\displaystyle f(x)=x-5$ for $\displaystyle x>1$. Each piece is a polynomial, so $\displaystyle f$ is continuous at every point of $\displaystyle (-\infty,1)$ and of $\displaystyle (1,\infty)$. The only doubtful point is $\displaystyle x=1$.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}(x+5)=6,\qquad \lim_{x\to 1^{+}}(x-5)=-4,\qquad f(1)=1+5=6.\]Since LHL $\displaystyle =6\neq -4=$ RHL, the limit at $\displaystyle x=1$ does not exist, so $\displaystyle f$ is discontinuous at $\displaystyle x=1$.Answer: $\displaystyle f$ is not a continuous function; it is continuous at every real point except $\displaystyle x=1$.
Exercise 14
$\displaystyle f(x)=\left\{\begin{array}{l}3, \text { if } 0 \leq x \leq 1 \\ 4, \text { if } 1<x<3 \\ 5, \text { if } 3 \leq x \leq 10\end{array}\right.$
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NCERT’s answer
$\displaystyle f$ is not continuous at $\displaystyle x=1$ and $\displaystyle x=3$
The function is defined on the domain $\displaystyle [0,10]$, and is constant on each of $\displaystyle [0,1]$, $\displaystyle (1,3)$ and $\displaystyle [3,10]$. A constant function is continuous, so $\displaystyle f$ is continuous at every point of $\displaystyle (0,1)$, $\displaystyle (1,3)$ and $\displaystyle (3,10)$, and one-sidedly continuous at the end points $\displaystyle x=0$ and $\displaystyle x=10$. Only $\displaystyle x=1$ and $\displaystyle x=3$ need testing.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}f(x)=3,\qquad \lim_{x\to 1^{+}}f(x)=4,\qquad f(1)=3.\] LHL $\displaystyle \neq$ RHL, so $\displaystyle f$ is discontinuous at $\displaystyle x=1$.At $\displaystyle x=3$: \[\lim_{x\to 3^{-}}f(x)=4,\qquad \lim_{x\to 3^{+}}f(x)=5,\qquad f(3)=5.\] LHL $\displaystyle \neq$ RHL, so $\displaystyle f$ is discontinuous at $\displaystyle x=3$.Answer: $\displaystyle f$ is continuous on $\displaystyle [0,10]$ except at $\displaystyle x=1$ and $\displaystyle x=3$, which are its only points of discontinuity.
Exercise 15
$\displaystyle f(x)= \begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}$
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NCERT’s answer
$\displaystyle x=1$ is the only point of discontinuity
$\displaystyle f(x)=2x$ for $\displaystyle x<0$, $\displaystyle f(x)=0$ for $\displaystyle 0\le x\le 1$, and $\displaystyle f(x)=4x$ for $\displaystyle x>1$. Each piece is a polynomial, so continuity can fail only at the joining points $\displaystyle x=0$ and $\displaystyle x=1$.At $\displaystyle x=0$: \[\lim_{x\to 0^{-}}2x=0,\qquad \lim_{x\to 0^{+}}0=0,\qquad f(0)=0.\] All three agree, so $\displaystyle f$ is continuous at $\displaystyle x=0$.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}f(x)=0,\qquad \lim_{x\to 1^{+}}4x=4,\qquad f(1)=0.\] Since LHL $\displaystyle =0\neq 4=$ RHL, $\displaystyle f$ is discontinuous at $\displaystyle x=1$.Answer: $\displaystyle f$ is continuous everywhere except at $\displaystyle x=1$, which is its only point of discontinuity.
Exercise 16
$\displaystyle f(x)= \begin{cases}-2, & \text { if } x \leq-1 \\ 2 x, & \text { if }-1<x \leq 1 \\ 2, & \text { if } x>1\end{cases}$
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NCERT’s answer
Continuous
$\displaystyle f(x)=-2$ for $\displaystyle x\le -1$, $\displaystyle f(x)=2x$ for $\displaystyle -1<x\le 1$, and $\displaystyle f(x)=2$ for $\displaystyle x>1$. Each piece is a constant or a polynomial, so only the joining points $\displaystyle x=-1$ and $\displaystyle x=1$ need testing.At $\displaystyle x=-1$: \[\lim_{x\to -1^{-}}f(x)=-2,\qquad \lim_{x\to -1^{+}}2x=2(-1)=-2,\qquad f(-1)=-2.\] All three agree, so $\displaystyle f$ is continuous at $\displaystyle x=-1$.At $\displaystyle x=1$: \[\lim_{x\to 1^{-}}2x=2,\qquad \lim_{x\to 1^{+}}f(x)=2,\qquad f(1)=2(1)=2.\] All three agree, so $\displaystyle f$ is continuous at $\displaystyle x=1$.Answer: $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$; it has no point of discontinuity.
Exercise 17
Find the relationship between $\displaystyle a$ and $\displaystyle b$ so that the function $\displaystyle f$ defined by \[f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \\ b x+3, & \text { if } x>3\end{cases} \] is continuous at $\displaystyle x=3$.
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Continuity at $\displaystyle x=3$ requires LHL $\displaystyle =$ RHL $\displaystyle =f(3)$.Value: $\displaystyle f(3)=a(3)+1=3a+1$ (the point $\displaystyle x=3$ belongs to the first branch).Left-hand limit: $\displaystyle \displaystyle\lim_{x\to 3^{-}}(ax+1)=3a+1$.Right-hand limit: $\displaystyle \displaystyle\lim_{x\to 3^{+}}(bx+3)=3b+3$.The left-hand limit already equals $\displaystyle f(3)$, so the single condition is \[3a+1=3b+3\ \Longrightarrow\ 3a-3b=2\ \Longrightarrow\ a-b=\frac{2}{3}.\]Answer: $\displaystyle f$ is continuous at $\displaystyle x=3$ precisely when $\displaystyle a=b+\dfrac{2}{3}$, i.e. $\displaystyle a-b=\dfrac{2}{3}$.
Exercise 18
For what value of $\displaystyle \lambda$ is the function defined by \[f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases} \] continuous at $\displaystyle x=0$ ? What about continuity at $\displaystyle x=1$ ?
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NCERT’s answer
For no value of $\displaystyle \lambda$, $\displaystyle f$ is continuous at $\displaystyle x=0$ but $\displaystyle f$ is continuous at $\displaystyle x=1$ for any \end{itemize} value of $\displaystyle \lambda$.
At $\displaystyle x=0$, compare the two one-sided limits with $\displaystyle f(0)$.Value: $\displaystyle f(0)=\lambda\left(0^{2}-2\cdot 0\right)=0$.Left-hand limit: $\displaystyle \displaystyle\lim_{x\to 0^{-}}\lambda\left(x^{2}-2x\right)=\lambda(0-0)=0$.Right-hand limit: $\displaystyle \displaystyle\lim_{x\to 0^{+}}(4x+1)=1$.The right-hand limit is $\displaystyle 1$ and does not involve $\displaystyle \lambda$ at all, while the left-hand limit is $\displaystyle 0$ for every $\displaystyle \lambda$. Since $\displaystyle 0\neq 1$, the limit at $\displaystyle x=0$ never exists.Answer: there is no value of $\displaystyle \lambda$ for which $\displaystyle f$ is continuous at $\displaystyle x=0$.At $\displaystyle x=1$: every point near $\displaystyle 1$ satisfies $\displaystyle x>0$, so $\displaystyle f(x)=4x+1$ throughout a neighbourhood of $\displaystyle 1$. Hence $\displaystyle \lim_{x\to 1}f(x)=5=f(1)$, and $\displaystyle f$ is continuous at $\displaystyle x=1$ for every value of $\displaystyle \lambda$. (Likewise $\displaystyle f$ is continuous at every point other than $\displaystyle x=0$.)
Exercise 19
Show that the function defined by $\displaystyle g(x)=x-[x]$ is discontinuous at all integral points. Here $\displaystyle [x]$ denotes the greatest integer less than or equal to $\displaystyle x$.
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Let $\displaystyle c$ be any integer. Recall the definition of the greatest integer function: $\displaystyle [x]$ is the greatest integer $\displaystyle \le x$. Hence \[c-1\le x<c\ \Rightarrow\ [x]=c-1,\qquad c\le x<c+1\ \Rightarrow\ [x]=c.\]Left-hand limit at $\displaystyle x=c$: for $\displaystyle x$ just below $\displaystyle c$ we have $\displaystyle [x]=c-1$, so \[\lim_{x\to c^{-}}g(x)=\lim_{x\to c^{-}}\left(x-(c-1)\right)=c-c+1=1.\]Right-hand limit at $\displaystyle x=c$: for $\displaystyle x$ just above $\displaystyle c$ we have $\displaystyle [x]=c$, so \[\lim_{x\to c^{+}}g(x)=\lim_{x\to c^{+}}(x-c)=0.\]Value: $\displaystyle g(c)=c-[c]=c-c=0$.Since LHL $\displaystyle =1\neq 0=$ RHL, $\displaystyle \lim_{x\to c}g(x)$ does not exist, so $\displaystyle g$ is discontinuous at $\displaystyle x=c$.As $\displaystyle c$ was an arbitrary integer, $\displaystyle g(x)=x-[x]$ is discontinuous at all integral points.
Exercise 20
Is the function defined by $\displaystyle f(x)=x^{2}-\sin x+5$ continuous at $\displaystyle x=\pi$ ?
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NCERT’s answer
$\displaystyle f$ is continuous at $\displaystyle x=\pi$
Use the algebra of continuous functions: if $\displaystyle g$ and $\displaystyle h$ are continuous at $\displaystyle c$, then $\displaystyle g\pm h$ is continuous at $\displaystyle c$.Here $\displaystyle f(x)=x^{2}-\sin x+5$, where $\displaystyle x\mapsto x^{2}$ is a polynomial (continuous on $\displaystyle \mathbf{R}$), $\displaystyle x\mapsto \sin x$ is continuous on $\displaystyle \mathbf{R}$, and $\displaystyle x\mapsto 5$ is constant. Hence $\displaystyle f$ is continuous at every real point, in particular at $\displaystyle x=\pi$.Checking directly at $\displaystyle x=\pi$, put $\displaystyle x=\pi+h$ so that $\displaystyle h\to 0$: \[\lim_{x\to\pi}f(x)=\lim_{h\to 0}\left[(\pi+h)^{2}-\sin(\pi+h)+5\right]=\lim_{h\to 0}\left[(\pi+h)^{2}+\sin h+5\right]=\pi^{2}+0+5.\] And $\displaystyle f(\pi)=\pi^{2}-\sin\pi+5=\pi^{2}+5$.Since $\displaystyle \lim_{x\to\pi}f(x)=\pi^{2}+5=f(\pi)$, the answer is yes: $\displaystyle f$ is continuous at $\displaystyle x=\pi$.

EXERCISE 5.1 21–30 (part 3 of 15)

Exercise 21
Discuss the continuity of the following functions:
(a)
$\displaystyle f(x)=\sin x+\cos x$
(b)
$\displaystyle f(x)=\sin x-\cos x$
(c)
$\displaystyle f(x)=\sin x \cdot \cos x$
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NCERT’s answer
(a)
, (b) and (c) are all continuous
Standard facts used: $\displaystyle \sin x$ and $\displaystyle \cos x$ are continuous at every real point, and by the algebra of continuous functions the sum, difference and product of two functions continuous at $\displaystyle c$ are continuous at $\displaystyle c$.
That $\displaystyle \sin$ is continuous: at any $\displaystyle c$, putting $\displaystyle x=c+h$,
\[\lim_{h\to 0}\sin(c+h)=\lim_{h\to 0}\left(\sin c\cos h+\cos c\sin h\right)=\sin c\cdot 1+\cos c\cdot 0=\sin c,\]
and similarly $\displaystyle \lim_{h\to 0}\cos(c+h)=\cos c\cos h-\sin c\sin h\to\cos c$.
(a)
$\displaystyle f(x)=\sin x+\cos x$ is the sum of two functions continuous on $\displaystyle \mathbf{R}$, hence continuous at every real point.
(b)
$\displaystyle f(x)=\sin x-\cos x$ is their difference, hence continuous at every real point.
(c)
$\displaystyle f(x)=\sin x\cdot\cos x$ is their product, hence continuous at every real point.
Answer: all three functions are continuous on $\displaystyle \mathbf{R}$; none has a point of discontinuity.
Exercise 22
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
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NCERT’s answer
Cosine function is continuous for all $\displaystyle x \in \mathbf{R}$; cosecant is continuous except for $\displaystyle x=n \pi, n \in \mathbf{Z}$; secant is continuous except for $\displaystyle x=(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}$ and cotangent function is continuous except for $\displaystyle x=n \pi, n \in \mathbf{Z}$
Take as known that $\displaystyle \sin x$ and $\displaystyle \cos x$ are continuous on $\displaystyle \mathbf{R}$, and use the quotient rule: if $\displaystyle g,h$ are continuous at $\displaystyle c$ and $\displaystyle h(c)\neq 0$, then $\displaystyle g/h$ is continuous at $\displaystyle c$.Cosine. At any real $\displaystyle c$, with $\displaystyle x=c+h$, \[\lim_{h\to 0}\cos(c+h)=\lim_{h\to 0}\left(\cos c\cos h-\sin c\sin h\right)=\cos c.\] So $\displaystyle \cos$ is continuous on $\displaystyle \mathbf{R}$ (domain $\displaystyle \mathbf{R}$, no point of discontinuity).Cosecant. $\displaystyle \mathrm{cosec}\,x=\dfrac{1}{\sin x}$, defined where $\displaystyle \sin x\neq 0$, i.e. on $\displaystyle \mathbf{R}-\{n\pi: n\in\mathbf{Z}\}$. At any $\displaystyle c$ in this domain, $\displaystyle \sin c\neq 0$, so the quotient rule gives continuity. Hence $\displaystyle \mathrm{cosec}$ is continuous on its whole domain $\displaystyle \mathbf{R}-\{n\pi\}$.Secant. $\displaystyle \sec x=\dfrac{1}{\cos x}$, defined where $\displaystyle \cos x\neq 0$, i.e. on $\displaystyle \mathbf{R}-\left\{(2n+1)\dfrac{\pi}{2}: n\in\mathbf{Z}\right\}$. By the quotient rule it is continuous at every point of that domain.Cotangent. $\displaystyle \cot x=\dfrac{\cos x}{\sin x}$, defined on $\displaystyle \mathbf{R}-\{n\pi: n\in\mathbf{Z}\}$, and continuous there by the quotient rule.Answer: each of $\displaystyle \cos$, $\displaystyle \mathrm{cosec}$, $\displaystyle \sec$ and $\displaystyle \cot$ is continuous at every point of its own domain; the excluded points $\displaystyle n\pi$ (for $\displaystyle \mathrm{cosec}$, $\displaystyle \cot$) and $\displaystyle (2n+1)\pi/2$ (for $\displaystyle \sec$) are not points of discontinuity, because the functions are not defined there.
Exercise 23
Find all points of discontinuity of $\displaystyle f$, where \[f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, & \text { if } x \geq 0\end{cases} \]
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NCERT’s answer
There is no point of discontinuity.
For $\displaystyle x<0$: $\displaystyle f(x)=\dfrac{\sin x}{x}$ is a quotient of two functions continuous on $\displaystyle \mathbf{R}$ with denominator $\displaystyle x\neq 0$, so $\displaystyle f$ is continuous at every $\displaystyle c<0$.For $\displaystyle x>0$: $\displaystyle f(x)=x+1$ is a polynomial, so $\displaystyle f$ is continuous at every $\displaystyle c>0$.Only $\displaystyle x=0$ remains. Using the standard limit $\displaystyle \displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$, \[\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(x+1)=1,\] and $\displaystyle f(0)=0+1=1$.Since LHL $\displaystyle =$ RHL $\displaystyle =f(0)=1$, $\displaystyle f$ is continuous at $\displaystyle x=0$ too.Answer: $\displaystyle f$ has no point of discontinuity; it is continuous on $\displaystyle \mathbf{R}$.
Exercise 24
Determine if $\displaystyle f$ defined by \[f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases} \] is a continuous function?
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NCERT’s answer
Yes, $\displaystyle f$ is continuous for all $\displaystyle x \in \mathbf{R}$
For $\displaystyle x\neq 0$: $\displaystyle x\mapsto x^{2}$ is continuous, $\displaystyle x\mapsto \dfrac{1}{x}$ is continuous where $\displaystyle x\neq 0$, and $\displaystyle \sin$ is continuous on $\displaystyle \mathbf{R}$; so $\displaystyle \sin\dfrac{1}{x}$ is continuous at every $\displaystyle c\neq 0$ (composition of continuous functions), and the product $\displaystyle x^{2}\sin\dfrac{1}{x}$ is continuous at every $\displaystyle c\neq 0$.At $\displaystyle x=0$ the factor $\displaystyle \sin\dfrac1x$ has no limit, so use the sandwich (squeeze) theorem instead. For all $\displaystyle x\neq 0$, \[-1\le \sin\frac{1}{x}\le 1\ \Longrightarrow\ -x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}.\] Since $\displaystyle \lim_{x\to 0}\left(-x^{2}\right)=0=\lim_{x\to 0}x^{2}$, the sandwich theorem gives \[\lim_{x\to 0}x^{2}\sin\frac{1}{x}=0.\] And $\displaystyle f(0)=0$, so $\displaystyle \lim_{x\to 0}f(x)=f(0)$.Answer: yes, $\displaystyle f$ is continuous at every real point, i.e. $\displaystyle f$ is a continuous function.
Exercise 25
Examine the continuity of $\displaystyle f$, where $\displaystyle f$ is defined by \[f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1, & \text { if } x=0\end{cases} \]
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NCERT’s answer
$\displaystyle f$ is continuous for all $\displaystyle x \in \mathbf{R}$
For $\displaystyle x\neq 0$: $\displaystyle f(x)=\sin x-\cos x$ is the difference of two functions continuous on $\displaystyle \mathbf{R}$, hence continuous at every $\displaystyle c\neq 0$.At $\displaystyle x=0$: using continuity of $\displaystyle \sin$ and $\displaystyle \cos$, \[\lim_{x\to 0}f(x)=\lim_{x\to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1,\] while the definition gives $\displaystyle f(0)=-1$.Since $\displaystyle \lim_{x\to 0}f(x)=-1=f(0)$, $\displaystyle f$ is continuous at $\displaystyle x=0$ as well. (The second branch simply supplies the value the first branch was heading towards, so the definition creates no break.)Answer: $\displaystyle f$ is continuous at every point of $\displaystyle \mathbf{R}$.
Find the values of $\displaystyle k$ so that the function $\displaystyle f$ is continuous at the indicated point in Exercises $\displaystyle 26$ to 29.
Exercise 26
$\displaystyle f(x)=\left\{\begin{array}{ll}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{array} \quad\right.$ at $\displaystyle x=\frac{\pi}{2}$
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NCERT’s answer
$\displaystyle k=6$
Continuity at $\displaystyle x=\dfrac{\pi}{2}$ requires $\displaystyle \displaystyle\lim_{x\to \pi/2}f(x)=f\!\left(\frac{\pi}{2}\right)=3$.The expression $\displaystyle \dfrac{k\cos x}{\pi-2x}$ is of the form $\displaystyle \dfrac{0}{0}$ at $\displaystyle x=\pi/2$, so substitute $\displaystyle x=\dfrac{\pi}{2}+h$, where $\displaystyle h\to 0$ as $\displaystyle x\to \pi/2$. Then \[\cos x=\cos\!\left(\frac{\pi}{2}+h\right)=-\sin h,\qquad \pi-2x=\pi-\pi-2h=-2h.\] Hence \[\lim_{x\to \pi/2}\frac{k\cos x}{\pi-2x}=\lim_{h\to 0}\frac{k(-\sin h)}{-2h}=\frac{k}{2}\lim_{h\to 0}\frac{\sin h}{h}=\frac{k}{2}\cdot 1=\frac{k}{2}.\]Setting this equal to the value $\displaystyle 3$: \[\frac{k}{2}=3\ \Longrightarrow\ k=6.\]Answer: $\displaystyle k=6$.
Exercise 27
$\displaystyle f(x)=\left\{\begin{array}{ll}k x^{2}, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{array} \quad\right.$ at $\displaystyle x=2$
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NCERT’s answer
$\displaystyle k=\frac{3}{4}$
Continuity at $\displaystyle x=2$ requires LHL $\displaystyle =$ RHL $\displaystyle =f(2)$.Value: $\displaystyle f(2)=k(2)^{2}=4k$ (the point $\displaystyle x=2$ lies in the branch $\displaystyle x\le 2$).Left-hand limit: $\displaystyle \displaystyle\lim_{x\to 2^{-}}kx^{2}=4k$.Right-hand limit: $\displaystyle \displaystyle\lim_{x\to 2^{+}}3=3$.Since LHL already equals $\displaystyle f(2)$, the condition reduces to \[4k=3\ \Longrightarrow\ k=\frac{3}{4}.\]Answer: $\displaystyle k=\dfrac{3}{4}$.
Exercise 28
$\displaystyle f(x)=\left\{\begin{array}{ll}k x+1, & \text { if } x \leq \pi \\ \cos x, & \text { if } x>\pi\end{array} \quad\right.$ at $\displaystyle x=\pi$
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NCERT’s answer
$\displaystyle k=\frac{-2}{\pi}$
Continuity at $\displaystyle x=\pi$ requires LHL $\displaystyle =$ RHL $\displaystyle =f(\pi)$.Value: $\displaystyle f(\pi)=k\pi+1$ (the point $\displaystyle x=\pi$ lies in the branch $\displaystyle x\le\pi$).Left-hand limit: $\displaystyle \displaystyle\lim_{x\to \pi^{-}}(kx+1)=k\pi+1$.Right-hand limit: $\displaystyle \displaystyle\lim_{x\to \pi^{+}}\cos x=\cos\pi=-1$ (using continuity of $\displaystyle \cos$).Equating, \[k\pi+1=-1\ \Longrightarrow\ k\pi=-2\ \Longrightarrow\ k=-\frac{2}{\pi}.\]Answer: $\displaystyle k=-\dfrac{2}{\pi}$.
Exercise 29
$\displaystyle f(x)=\left\{\begin{array}{ll}k x+1, & \text { if } x \leq 5 \\ 3 x-5, & \text { if } x>5\end{array} \quad\right.$ at $\displaystyle x=5$
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NCERT’s answer
$\displaystyle k=\frac{9}{5}$
Continuity at $\displaystyle x=5$ requires LHL $\displaystyle =$ RHL $\displaystyle =f(5)$.Value: $\displaystyle f(5)=5k+1$ (the point $\displaystyle x=5$ lies in the branch $\displaystyle x\le 5$).Left-hand limit: $\displaystyle \displaystyle\lim_{x\to 5^{-}}(kx+1)=5k+1$.Right-hand limit: $\displaystyle \displaystyle\lim_{x\to 5^{+}}(3x-5)=15-5=10$.Equating, \[5k+1=10\ \Longrightarrow\ 5k=9\ \Longrightarrow\ k=\frac{9}{5}.\]Answer: $\displaystyle k=\dfrac{9}{5}$.
Exercise 30
Find the values of $\displaystyle a$ and $\displaystyle b$ such that the function defined by \[f(x)= \begin{cases}5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2<x<10 \\ 21, & \text { if } x \geq 10\end{cases} \] is a continuous function.
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NCERT’s answer
$\displaystyle a=2, b=1$
Each branch (a constant, a linear function, a constant) is continuous on the open interval where it applies, so $\displaystyle f$ can fail to be continuous only at the joining points $\displaystyle x=2$ and $\displaystyle x=10$. Impose continuity at both.At $\displaystyle x=2$: $\displaystyle f(2)=5$, LHL $\displaystyle =\displaystyle\lim_{x\to 2^{-}}5=5$, RHL $\displaystyle =\displaystyle\lim_{x\to 2^{+}}(ax+b)=2a+b$. Continuity gives \[2a+b=5.\qquad (1)\]At $\displaystyle x=10$: $\displaystyle f(10)=21$, LHL $\displaystyle =\displaystyle\lim_{x\to 10^{-}}(ax+b)=10a+b$, RHL $\displaystyle =\displaystyle\lim_{x\to 10^{+}}21=21$. Continuity gives \[10a+b=21.\qquad (2)\]Subtract ($\displaystyle 1$) from ($\displaystyle 2$): \[8a=16\ \Longrightarrow\ a=2,\] and then from ($\displaystyle 1$), $\displaystyle b=5-2a=5-4=1$.Answer: $\displaystyle a=2$ and $\displaystyle b=1$ (so the middle branch is $\displaystyle 2x+1$), and with these values $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$.

EXERCISE 5.1 31–34 (part 4 of 15)

Exercise 31
Show that the function defined by $\displaystyle f(x)=\cos \left(x^{2}\right)$ is a continuous function.
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Use the theorem on composition: if $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle h$ is continuous at $\displaystyle g(c)$, then $\displaystyle h\circ g$ is continuous at $\displaystyle c$.Write $\displaystyle f=h\circ g$ with \[g(x)=x^{2},\qquad h(y)=\cos y,\qquad (h\circ g)(x)=\cos\left(x^{2}\right)=f(x).\]$\displaystyle g(x)=x^{2}$ is a polynomial, hence continuous at every real $\displaystyle c$.$\displaystyle h(y)=\cos y$ is continuous at every real $\displaystyle y$: with $\displaystyle y=d+t$, \[\lim_{t\to 0}\cos(d+t)=\lim_{t\to 0}\left(\cos d\cos t-\sin d\sin t\right)=\cos d.\] In particular $\displaystyle h$ is continuous at $\displaystyle g(c)=c^{2}$.Therefore $\displaystyle h\circ g$ is continuous at every real $\displaystyle c$; that is, $\displaystyle f(x)=\cos\left(x^{2}\right)$ is a continuous function on $\displaystyle \mathbf{R}$.
Exercise 32
Show that the function defined by $\displaystyle f(x)=|\cos x|$ is a continuous function.
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Use the composition theorem again. Write $\displaystyle f=h\circ g$ with \[g(x)=\cos x,\qquad h(y)=|y|,\qquad (h\circ g)(x)=|\cos x|=f(x).\]$\displaystyle g(x)=\cos x$ is continuous at every real $\displaystyle c$.$\displaystyle h(y)=|y|$ is continuous at every real $\displaystyle y$: for $\displaystyle d>0$, $\displaystyle h(y)=y$ near $\displaystyle d$; for $\displaystyle d<0$, $\displaystyle h(y)=-y$ near $\displaystyle d$; and at $\displaystyle d=0$, \[\lim_{y\to 0^{-}}(-y)=0=\lim_{y\to 0^{+}}y=h(0).\] So $\displaystyle h$ is continuous everywhere, in particular at $\displaystyle y=g(c)=\cos c$.Since $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle h$ is continuous at $\displaystyle g(c)$, the composite $\displaystyle h\circ g$ is continuous at $\displaystyle c$; $\displaystyle c$ being arbitrary, $\displaystyle f(x)=|\cos x|$ is a continuous function on $\displaystyle \mathbf{R}$.(Note that the corners of the graph, at $\displaystyle x=(2n+1)\pi/2$ where $\displaystyle \cos x=0$, are points where $\displaystyle f$ is continuous but not differentiable.)
Exercise 33
Examine that $\displaystyle \sin |x|$ is a continuous function.
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Take $\displaystyle f=h\circ g$ with \[g(x)=|x|,\qquad h(y)=\sin y,\qquad (h\circ g)(x)=\sin|x|.\]$\displaystyle g(x)=|x|$ is continuous at every real $\displaystyle c$: for $\displaystyle c>0$ it agrees with $\displaystyle x$ near $\displaystyle c$ and for $\displaystyle c<0$ with $\displaystyle -x$ near $\displaystyle c$; at $\displaystyle c=0$, \[\lim_{x\to 0^{-}}(-x)=0,\qquad \lim_{x\to 0^{+}}x=0,\qquad g(0)=0,\] so all three agree.$\displaystyle h(y)=\sin y$ is continuous at every real $\displaystyle y$, since $\displaystyle \lim_{t\to 0}\sin(d+t)=\lim_{t\to 0}\left(\sin d\cos t+\cos d\sin t\right)=\sin d$.By the theorem on the composition of continuous functions, $\displaystyle h\circ g$ is continuous at every real $\displaystyle c$.Hence $\displaystyle \sin|x|$ is a continuous function on $\displaystyle \mathbf{R}$.
Exercise 34
Find all the points of discontinuity of $\displaystyle f$ defined by $\displaystyle f(x)=|x|-|x+1|$.
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NCERT’s answer
There is no point of discontinuity.
Remove both moduli by splitting at the points where the expressions inside change sign, namely $\displaystyle x=-1$ and $\displaystyle x=0$.For $\displaystyle x<-1$: $\displaystyle |x|=-x$ and $\displaystyle |x+1|=-(x+1)$, so $\displaystyle f(x)=-x+(x+1)=1$.For $\displaystyle -1\le x<0$: $\displaystyle |x|=-x$ and $\displaystyle |x+1|=x+1$, so $\displaystyle f(x)=-x-(x+1)=-2x-1$.For $\displaystyle x\ge 0$: $\displaystyle |x|=x$ and $\displaystyle |x+1|=x+1$, so $\displaystyle f(x)=x-(x+1)=-1$.Thus \[f(x)=\begin{cases}1,&x<-1\\ -2x-1,&-1\le x<0\\ -1,&x\ge 0.\end{cases}\] Each piece is a polynomial, so only $\displaystyle x=-1$ and $\displaystyle x=0$ need testing.At $\displaystyle x=-1$: $\displaystyle \displaystyle\lim_{x\to -1^{-}}f(x)=1$, $\displaystyle \displaystyle\lim_{x\to -1^{+}}(-2x-1)=2-1=1$, and $\displaystyle f(-1)=-2(-1)-1=1$. All equal, so $\displaystyle f$ is continuous at $\displaystyle x=-1$.At $\displaystyle x=0$: $\displaystyle \displaystyle\lim_{x\to 0^{-}}(-2x-1)=-1$, $\displaystyle \displaystyle\lim_{x\to 0^{+}}f(x)=-1$, and $\displaystyle f(0)=-1$. All equal, so $\displaystyle f$ is continuous at $\displaystyle x=0$.(The same conclusion follows at once from the algebra of continuous functions: $\displaystyle |x|$ and $\displaystyle |x+1|$ are continuous on $\displaystyle \mathbf{R}$, so their difference is too.)Answer: $\displaystyle f$ has no point of discontinuity; it is continuous on the whole of $\displaystyle \mathbf{R}$.

EXERCISE 5.2 1–10 (part 5 of 15)

Differentiate the functions with respect to $\displaystyle x$ in Exercises $\displaystyle 1$ to 8.
Exercise 1
$\displaystyle \sin \left(x^{2}+5\right)$
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NCERT’s answer
$\displaystyle 2 x \cos \left(x^{2}+5\right)$
Differentiate with respect to $\displaystyle x$ by the chain rule: if $\displaystyle y=f(u)$ and $\displaystyle u=g(x)$ are both differentiable, then $\displaystyle \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$.Put $\displaystyle u=x^{2}+5$, so that $\displaystyle y=\sin u$. Then \[\frac{dy}{du}=\cos u,\qquad \frac{du}{dx}=2x.\] The step most often dropped is the second factor $\displaystyle \dfrac{du}{dx}$; $\displaystyle \cos\left(x^{2}+5\right)$ alone is not the derivative. Multiplying, \[\frac{dy}{dx}=\cos\left(x^{2}+5\right)\cdot 2x.\]Answer: $\displaystyle \dfrac{dy}{dx}=2x\cos\left(x^{2}+5\right)$, valid for all $\displaystyle x\in\mathbf{R}$.
Exercise 2
$\displaystyle \cos (\sin x)$
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NCERT’s answer
$\displaystyle -\cos x \sin (\sin x)$
Differentiate with respect to $\displaystyle x$ by the chain rule, $\displaystyle \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$.Let $\displaystyle u=\sin x$, so $\displaystyle y=\cos u$. Then \[\frac{dy}{du}=-\sin u=-\sin(\sin x),\qquad \frac{du}{dx}=\cos x.\] Note the two sines play different roles: the outer one comes from differentiating $\displaystyle \cos$, and its argument stays $\displaystyle \sin x$ untouched. Hence \[\frac{dy}{dx}=-\sin(\sin x)\cdot\cos x.\]Answer: $\displaystyle \dfrac{dy}{dx}=-\cos x\,\sin(\sin x)$, valid for all $\displaystyle x\in\mathbf{R}$.
Exercise 3
$\displaystyle \sin (a x+b)$
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NCERT’s answer
$\displaystyle a \cos (a x+b)$
Differentiate with respect to $\displaystyle x$ by the chain rule, treating $\displaystyle a$ and $\displaystyle b$ as constants.Let $\displaystyle u=ax+b$, so $\displaystyle y=\sin u$. Then \[\frac{dy}{du}=\cos u=\cos(ax+b),\qquad \frac{du}{dx}=a.\] Therefore \[\frac{dy}{dx}=\cos(ax+b)\cdot a.\]Answer: $\displaystyle \dfrac{dy}{dx}=a\cos(ax+b)$, valid for all $\displaystyle x\in\mathbf{R}$.
Exercise 4
$\displaystyle \sec (\tan (\sqrt{x}))$
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NCERT’s answer
$\displaystyle \frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2} \sqrt{x}}{2 \sqrt{x}}$
This is a chain of three functions, so the chain rule is applied twice: with $\displaystyle y=\sec v$, $\displaystyle v=\tan u$, $\displaystyle u=\sqrt{x}$, \[\frac{dy}{dx}=\frac{dy}{dv}\cdot\frac{dv}{du}\cdot\frac{du}{dx}.\]The three factors are \[\frac{dy}{dv}=\sec v\tan v=\sec\big(\tan\sqrt{x}\big)\tan\big(\tan\sqrt{x}\big),\qquad \frac{dv}{du}=\sec^{2}u=\sec^{2}\sqrt{x},\qquad \frac{du}{dx}=\frac{1}{2\sqrt{x}}.\] The factor most often forgotten is the innermost $\displaystyle \dfrac{1}{2\sqrt{x}}$, obtained from $\displaystyle \dfrac{d}{dx}x^{1/2}=\tfrac12 x^{-1/2}$. Multiplying the three, \[\frac{dy}{dx}=\frac{\sec\big(\tan\sqrt{x}\big)\,\tan\big(\tan\sqrt{x}\big)\,\sec^{2}\big(\sqrt{x}\big)}{2\sqrt{x}}.\]Answer: $\displaystyle \dfrac{dy}{dx}=\dfrac{\sec\left(\tan\sqrt{x}\right)\tan\left(\tan\sqrt{x}\right)\sec^{2}\left(\sqrt{x}\right)}{2\sqrt{x}}$, for $\displaystyle x>0$ (with $\displaystyle \sqrt{x}$ and $\displaystyle \tan\sqrt{x}$ avoiding the odd multiples of $\displaystyle \tfrac{\pi}{2}$); at $\displaystyle x=0$ the expression $\displaystyle \dfrac{1}{2\sqrt{x}}$ is undefined.
Exercise 5
$\displaystyle \frac{\sin (a x+b)}{\cos (c x+d)}$
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NCERT’s answer
$\displaystyle a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) \tan (c x+d) \sec (c x+d)$
This is a quotient, so use the quotient rule \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}},\] with each of $\displaystyle u$ and $\displaystyle v$ differentiated by the chain rule.Take $\displaystyle u=\sin(ax+b)$ and $\displaystyle v=\cos(cx+d)$. Then \[\frac{du}{dx}=a\cos(ax+b),\qquad \frac{dv}{dx}=-c\sin(cx+d).\] The sign is the trap here: $\displaystyle \dfrac{dv}{dx}$ is already negative, and the quotient rule subtracts it, so the second term comes out positive. Substituting, \[\frac{dy}{dx}=\frac{\cos(cx+d)\cdot a\cos(ax+b)-\sin(ax+b)\cdot\big(-c\sin(cx+d)\big)}{\cos^{2}(cx+d)},\] \[\frac{dy}{dx}=\frac{a\cos(ax+b)\cos(cx+d)+c\sin(ax+b)\sin(cx+d)}{\cos^{2}(cx+d)}.\]Answer: $\displaystyle \dfrac{dy}{dx}=\dfrac{a\cos(ax+b)\cos(cx+d)+c\sin(ax+b)\sin(cx+d)}{\cos^{2}(cx+d)}$, valid wherever $\displaystyle \cos(cx+d)\neq 0$.
Exercise 6
$\displaystyle \cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$
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NCERT’s answer
$\displaystyle 10 x^{4} \sin x^{5} \cos x^{5} \cos x^{3}-3 x^{2} \sin x^{3} \sin ^{2} x^{5}$
Read the expression as $\displaystyle y=\cos\left(x^{3}\right)\cdot\big(\sin\left(x^{5}\right)\big)^{2}$: it is a product, so use the product rule $\displaystyle \dfrac{d}{dx}(uv)=u'v+uv'$, with each factor differentiated by the chain rule.Let $\displaystyle u=\cos\left(x^{3}\right)$ and $\displaystyle v=\sin^{2}\left(x^{5}\right)$.For $\displaystyle u$: $\displaystyle \dfrac{du}{dx}=-\sin\left(x^{3}\right)\cdot 3x^{2}=-3x^{2}\sin\left(x^{3}\right)$.For $\displaystyle v$, the chain rule is needed twice — once for the square and once for $\displaystyle x^{5}$: \[\frac{dv}{dx}=2\sin\left(x^{5}\right)\cdot\cos\left(x^{5}\right)\cdot 5x^{4}=10x^{4}\sin\left(x^{5}\right)\cos\left(x^{5}\right).\] Hence \[\frac{dy}{dx}=-3x^{2}\sin\left(x^{3}\right)\sin^{2}\left(x^{5}\right)+10x^{4}\cos\left(x^{3}\right)\sin\left(x^{5}\right)\cos\left(x^{5}\right).\] Using $\displaystyle 2\sin\theta\cos\theta=\sin 2\theta$ with $\displaystyle \theta=x^{5}$, the second term is $\displaystyle 5x^{4}\cos\left(x^{3}\right)\sin\left(2x^{5}\right)$.Answer: $\displaystyle \dfrac{dy}{dx}=10x^{4}\cos\left(x^{3}\right)\sin\left(x^{5}\right)\cos\left(x^{5}\right)-3x^{2}\sin\left(x^{3}\right)\sin^{2}\left(x^{5}\right)$, equivalently $\displaystyle 5x^{4}\cos\left(x^{3}\right)\sin\left(2x^{5}\right)-3x^{2}\sin\left(x^{3}\right)\sin^{2}\left(x^{5}\right)$.
Exercise 7
$\displaystyle 2 \sqrt{\cot \left(x^{2}\right)}$
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NCERT’s answer
$\displaystyle \frac{-2 \sqrt{2} x}{\sin x^{2} \sqrt{\sin 2 x^{2}}}$
Write $\displaystyle y=2\sqrt{\cot\left(x^{2}\right)}=2\big(\cot\left(x^{2}\right)\big)^{1/2}$ and apply the chain rule twice, using $\displaystyle \dfrac{d}{dt}\sqrt{t}=\dfrac{1}{2\sqrt{t}}$ and $\displaystyle \dfrac{d}{du}\cot u=-\mathrm{cosec}^{2}u$.With $\displaystyle t=\cot u$ and $\displaystyle u=x^{2}$, \[\frac{dy}{dx}=2\cdot\frac{1}{2\sqrt{\cot\left(x^{2}\right)}}\cdot\left(-\mathrm{cosec}^{2}\left(x^{2}\right)\right)\cdot 2x.\] The minus sign comes from $\displaystyle \cot$, and the last factor $\displaystyle 2x$ from the inner $\displaystyle x^{2}$; dropping either is the usual error. Simplifying the numerical factors, \[\frac{dy}{dx}=\frac{-2x\,\mathrm{cosec}^{2}\left(x^{2}\right)}{\sqrt{\cot\left(x^{2}\right)}}.\] This may be recast in sines and cosines: since $\displaystyle \mathrm{cosec}^{2}\left(x^{2}\right)=\dfrac{1}{\sin^{2}x^{2}}$ and $\displaystyle \sqrt{\cot x^{2}}=\dfrac{\sqrt{\cos x^{2}}}{\sqrt{\sin x^{2}}}$, \[\frac{dy}{dx}=\frac{-2x}{\sin^{3/2}\left(x^{2}\right)\sqrt{\cos\left(x^{2}\right)}}=\frac{-2\sqrt{2}\,x}{\sin\left(x^{2}\right)\sqrt{\sin\left(2x^{2}\right)}}.\]Answer: $\displaystyle \dfrac{dy}{dx}=\dfrac{-2x\,\mathrm{cosec}^{2}\left(x^{2}\right)}{\sqrt{\cot\left(x^{2}\right)}}$, valid where $\displaystyle \cot\left(x^{2}\right)>0$.
Exercise 8
$\displaystyle \cos (\sqrt{x})$
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NCERT’s answer
$\displaystyle -\frac{\sin \sqrt{x}}{2 \sqrt{x}}$
Differentiate with respect to $\displaystyle x$ by the chain rule, with $\displaystyle u=\sqrt{x}$ and $\displaystyle y=\cos u$.\[\frac{dy}{du}=-\sin u=-\sin\sqrt{x},\qquad \frac{du}{dx}=\frac{d}{dx}x^{1/2}=\frac{1}{2\sqrt{x}}.\] Multiplying, \[\frac{dy}{dx}=-\sin\left(\sqrt{x}\right)\cdot\frac{1}{2\sqrt{x}}.\]Answer: $\displaystyle \dfrac{dy}{dx}=-\dfrac{\sin\sqrt{x}}{2\sqrt{x}}$, for $\displaystyle x>0$; the function is defined at $\displaystyle x=0$ but this derivative formula is not, since $\displaystyle \dfrac{1}{2\sqrt{x}}\to\infty$ as $\displaystyle x\to 0^{+}$.
Exercise 9
Prove that the function $\displaystyle f$ given by \[f(x)=|x-1|, x \in \mathbf{R} \] is not differentiable at $\displaystyle x=1$.
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By definition, $\displaystyle f$ is differentiable at $\displaystyle x=c$ only if the limit \[\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}\] exists, i.e. only if the left-hand derivative and the right-hand derivative at $\displaystyle c$ both exist and are equal. So compute the two one-sided limits at $\displaystyle c=1$.Here $\displaystyle f(x)=|x-1|$, so $\displaystyle f(1)=0$ and $\displaystyle f(1+h)=|1+h-1|=|h|$. The whole proof turns on unpacking $\displaystyle |h|$ by sign, which is the case split students skip.Left-hand derivative ($\displaystyle h<0$, so $\displaystyle |h|=-h$): \[Lf'(1)=\lim_{h\to 0^{-}}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^{-}}\frac{|h|-0}{h}=\lim_{h\to 0^{-}}\frac{-h}{h}=-1.\]Right-hand derivative ($\displaystyle h>0$, so $\displaystyle |h|=h$): \[Rf'(1)=\lim_{h\to 0^{+}}\frac{|h|-0}{h}=\lim_{h\to 0^{+}}\frac{h}{h}=1.\]Since $\displaystyle Lf'(1)=-1\neq 1=Rf'(1)$, the two one-sided limits differ, so $\displaystyle \lim_{h\to 0}\dfrac{f(1+h)-f(1)}{h}$ does not exist.Hence $\displaystyle f(x)=|x-1|$ is not differentiable at $\displaystyle x=1$. (It is continuous there — continuity does not imply differentiability; the graph has a corner at $\displaystyle (1,0)$.)
Exercise 10
Prove that the greatest integer function defined by \[f(x)=[x], 0<x<3 \] is not differentiable at $\displaystyle x=1$ and $\displaystyle x=2$.
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Let $\displaystyle f(x)=[x]$ on $\displaystyle 0<x<3$, where $\displaystyle [x]$ is the greatest integer not exceeding $\displaystyle x$. Differentiability at $\displaystyle c$ requires that $\displaystyle \lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ exist, so test the one-sided derivatives. The key step is evaluating $\displaystyle [c+h]$ separately for $\displaystyle h<0$ and $\displaystyle h>0$, because the bracket function jumps exactly at integers.At $\displaystyle x=1$. Here $\displaystyle f(1)=[1]=1$. For small $\displaystyle h>0$ (say $\displaystyle 0<h<1$), $\displaystyle 1<1+h<2$, so $\displaystyle [1+h]=1$; for small $\displaystyle h<0$ (say $\displaystyle -1<h<0$), $\displaystyle 0<1+h<1$, so $\displaystyle [1+h]=0$.Left-hand derivative: \[Lf'(1)=\lim_{h\to 0^{-}}\frac{[1+h]-[1]}{h}=\lim_{h\to 0^{-}}\frac{0-1}{h}=\lim_{h\to 0^{-}}\frac{-1}{h},\] and as $\displaystyle h\to 0^{-}$ the quotient $\displaystyle \dfrac{-1}{h}\to+\infty$; the limit does not exist (it is not a finite number).Right-hand derivative: \[Rf'(1)=\lim_{h\to 0^{+}}\frac{1-1}{h}=0.\]Since $\displaystyle Lf'(1)$ does not exist, $\displaystyle f$ is not differentiable at $\displaystyle x=1$.At $\displaystyle x=2$. Here $\displaystyle f(2)=[2]=2$. For $\displaystyle 0<h<1$, $\displaystyle [2+h]=2$; for $\displaystyle -1<h<0$, $\displaystyle 1<2+h<2$ so $\displaystyle [2+h]=1$. \[Lf'(2)=\lim_{h\to 0^{-}}\frac{1-2}{h}=\lim_{h\to 0^{-}}\frac{-1}{h}\to+\infty\ \ (\text{does not exist}),\qquad Rf'(2)=\lim_{h\to 0^{+}}\frac{2-2}{h}=0.\] So $\displaystyle f$ is not differentiable at $\displaystyle x=2$ either.Alternative one-line argument: every function differentiable at a point is continuous there. Since $\displaystyle \lim_{x\to 1^{-}}[x]=0\neq 1=f(1)$ and $\displaystyle \lim_{x\to 2^{-}}[x]=1\neq 2=f(2)$, $\displaystyle f$ is discontinuous at $\displaystyle x=1$ and $\displaystyle x=2$, and therefore cannot be differentiable at those points.Hence the greatest integer function $\displaystyle f(x)=[x],\ 0<x<3$, is not differentiable at $\displaystyle x=1$ and $\displaystyle x=2$.

EXERCISE 5.3 1–10 (part 6 of 15)

Find $\displaystyle \frac{d y}{d x}$ in the following:
Exercise 1
$\displaystyle 2 x+3 y=\sin x$
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NCERT’s answer
$\displaystyle \frac{\cos x-2}{3}$
Implicit differentiation: differentiate every term with respect to $\displaystyle x$, remembering that $\displaystyle y$ is a function of $\displaystyle x$, so $\displaystyle \frac{d}{dx}(3y)=3\frac{dy}{dx}$ by the chain rule.\[\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin x) \] \[2+3\frac{dy}{dx}=\cos x \]Solve for the derivative: \[3\frac{dy}{dx}=\cos x-2 \]\[\boxed{\ \frac{dy}{dx}=\frac{\cos x-2}{3}\ } \]
Exercise 2
$\displaystyle 2 x+3 y=\sin y$
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NCERT’s answer
$\displaystyle \frac{2}{\cos y-3}$
Here the right-hand side is $\displaystyle \sin y$, a function of $\displaystyle y$ which is itself a function of $\displaystyle x$, so it must be differentiated by the chain rule as $\displaystyle \cos y\cdot\frac{dy}{dx}$ $\displaystyle -$ this is the step most often dropped.\[\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin y) \] \[2+3\frac{dy}{dx}=\cos y\,\frac{dy}{dx} \]Collect the $\displaystyle \frac{dy}{dx}$ terms on one side: \[2=(\cos y-3)\frac{dy}{dx} \]Since $\displaystyle |\cos y|\le 1$ for every $\displaystyle y$, the factor $\displaystyle \cos y-3$ is never zero (it lies in $\displaystyle [-4,-2]$), so division is always legitimate.\[\boxed{\ \frac{dy}{dx}=\frac{2}{\cos y-3}\ } \]
Exercise 3
$\displaystyle a x+b y^{2}=\cos y$
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NCERT’s answer
$\displaystyle -\frac{a}{2 b y+\sin y}$
$\displaystyle a$ and $\displaystyle b$ are constants. Differentiate implicitly with respect to $\displaystyle x$: the term $\displaystyle by^{2}$ gives $\displaystyle 2by\frac{dy}{dx}$ and $\displaystyle \cos y$ gives $\displaystyle -\sin y\frac{dy}{dx}$, both by the chain rule.\[a+2by\frac{dy}{dx}=-\sin y\,\frac{dy}{dx} \]Bring the derivative terms together: \[2by\frac{dy}{dx}+\sin y\,\frac{dy}{dx}=-a \] \[(2by+\sin y)\frac{dy}{dx}=-a \]\[\boxed{\ \frac{dy}{dx}=\frac{-a}{2by+\sin y}\ },\qquad 2by+\sin y\ne 0. \]
Exercise 4
$\displaystyle x y+y^{2}=\tan x+y$
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NCERT’s answer
$\displaystyle \frac{\sec ^{2} x-y}{x+2 y-1}$
The term $\displaystyle xy$ is a product of two functions of $\displaystyle x$, so it needs the product rule $\displaystyle \frac{d}{dx}(uv)=u'v+uv'$; everything containing $\displaystyle y$ also needs the chain rule.\[\frac{d}{dx}(xy)+\frac{d}{dx}(y^{2})=\frac{d}{dx}(\tan x)+\frac{d}{dx}(y) \] \[\left(y\cdot 1+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=\sec^{2}x+\frac{dy}{dx} \]Collect all $\displaystyle \frac{dy}{dx}$ terms on the left and the rest on the right: \[x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=\sec^{2}x-y \] \[(x+2y-1)\frac{dy}{dx}=\sec^{2}x-y \]\[\boxed{\ \frac{dy}{dx}=\frac{\sec^{2}x-y}{x+2y-1}\ },\qquad x+2y\ne 1. \]
Exercise 5
$\displaystyle x^{2}+x y+y^{2}=100$
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NCERT’s answer
$\displaystyle -\frac{(2 x+y)}{(x+2 y)}$
Differentiate implicitly with respect to $\displaystyle x$; the middle term $\displaystyle xy$ requires the product rule and the constant $\displaystyle 100$ differentiates to $\displaystyle 0$.\[\frac{d}{dx}(x^{2})+\frac{d}{dx}(xy)+\frac{d}{dx}(y^{2})=\frac{d}{dx}(100) \] \[2x+\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0 \]Group the derivative terms: \[(x+2y)\frac{dy}{dx}=-(2x+y) \]\[\boxed{\ \frac{dy}{dx}=-\frac{2x+y}{x+2y}\ },\qquad x+2y\ne 0. \]
Exercise 6
$\displaystyle x^{3}+x^{2} y+x y^{2}+y^{3}=81$
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NCERT’s answer
$\displaystyle -\frac{\left(3 x^{2}+2 x y+y^{2}\right)}{\left(x^{2}+2 x y+3 y^{2}\right)}$
Differentiate term by term with respect to $\displaystyle x$. Both $\displaystyle x^{2}y$ and $\displaystyle xy^{2}$ are products, so each needs the product rule together with the chain rule on the $\displaystyle y$-factor.\[\frac{d}{dx}(x^{2}y)=2xy+x^{2}\frac{dy}{dx},\qquad \frac{d}{dx}(xy^{2})=y^{2}+2xy\frac{dy}{dx} \]So the whole equation gives \[3x^{2}+\left(2xy+x^{2}\frac{dy}{dx}\right)+\left(y^{2}+2xy\frac{dy}{dx}\right)+3y^{2}\frac{dy}{dx}=0 \]Collect the coefficient of $\displaystyle \frac{dy}{dx}$: \[\left(x^{2}+2xy+3y^{2}\right)\frac{dy}{dx}=-\left(3x^{2}+2xy+y^{2}\right) \]\[\boxed{\ \frac{dy}{dx}=-\frac{3x^{2}+2xy+y^{2}}{x^{2}+2xy+3y^{2}}\ },\qquad x^{2}+2xy+3y^{2}\ne 0. \]
Exercise 7
$\displaystyle \sin ^{2} y+\cos x y=\kappa$
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NCERT’s answer
$\displaystyle \frac{y \sin x y}{\sin 2 y-x \sin x y}$
Here the right-hand side is a constant, so it differentiates to $\displaystyle 0$. Two chain rules are needed $\displaystyle -$ one for $\displaystyle \sin^{2}y=(\sin y)^{2}$ and one for $\displaystyle \cos(xy)$ $\displaystyle -$ and inside the second, $\displaystyle \frac{d}{dx}(xy)$ needs the product rule.\[\frac{d}{dx}\left(\sin^{2}y\right)=2\sin y\cos y\,\frac{dy}{dx}=\sin 2y\,\frac{dy}{dx} \] \[\frac{d}{dx}\left(\cos xy\right)=-\sin(xy)\cdot\frac{d}{dx}(xy)=-\sin(xy)\left(y+x\frac{dy}{dx}\right) \]Adding and equating to $\displaystyle 0$: \[\sin 2y\,\frac{dy}{dx}-y\sin(xy)-x\sin(xy)\frac{dy}{dx}=0 \]Collect $\displaystyle \frac{dy}{dx}$: \[\left(\sin 2y-x\sin (xy)\right)\frac{dy}{dx}=y\sin (xy) \]\[\boxed{\ \frac{dy}{dx}=\frac{y\sin (xy)}{\sin 2y-x\sin (xy)}\ },\qquad \sin 2y\ne x\sin (xy). \]
Exercise 8
$\displaystyle \sin ^{2} x+\cos ^{2} y=1$
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NCERT’s answer
$\displaystyle \frac{\sin 2 x}{\sin 2 y}$
Differentiate both sides with respect to $\displaystyle x$, using the chain rule on each squared trigonometric term; the right side is the constant $\displaystyle 1$.\[2\sin x\cos x-2\cos y\sin y\,\frac{dy}{dx}=0 \]Using $\displaystyle 2\sin\theta\cos\theta=\sin 2\theta$, \[\sin 2x-\sin 2y\,\frac{dy}{dx}=0 \]\[\boxed{\ \frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}\ },\qquad \sin 2y\ne 0. \]A check worth doing: the given relation says $\displaystyle \cos^{2}y=1-\sin^{2}x=\cos^{2}x$, so $\displaystyle y=\pm x+n\pi$ and hence $\displaystyle \frac{dy}{dx}=\pm 1$. This agrees with the answer above, because $\displaystyle \sin 2y=\sin(\pm 2x+2n\pi)=\pm\sin 2x$.
Exercise 9
$\displaystyle y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
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NCERT’s answer
$\displaystyle \frac{2}{1+x^{2}}$
Substitution. Put $\displaystyle x=\tan\theta$, so $\displaystyle \theta=\tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Then \[\frac{2x}{1+x^{2}}=\frac{2\tan\theta}{1+\tan^{2}\theta}=\sin 2\theta,\qquad\text{so}\qquad y=\sin^{-1}(\sin 2\theta). \]The step that must not be skipped: $\displaystyle \sin^{-1}(\sin\alpha)=\alpha$ only when $\displaystyle \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Here $\displaystyle \alpha=2\theta$, and $\displaystyle 2\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ exactly when $\displaystyle \theta\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$, i.e. when $\displaystyle |x|\le 1$. No restriction on $\displaystyle x$ is given, so the cases must be separated.Case $\displaystyle 1$: $\displaystyle -1<x<1$. Then $\displaystyle 2\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so $\displaystyle y=2\theta=2\tan^{-1}x$ and \[\frac{dy}{dx}=\frac{2}{1+x^{2}}. \]Case $\displaystyle 2$: $\displaystyle x>1$. Then $\displaystyle \theta\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$, so $\displaystyle 2\theta\in\left(\frac{\pi}{2},\pi\right)$ and $\displaystyle \sin^{-1}(\sin 2\theta)=\pi-2\theta$. Thus $\displaystyle y=\pi-2\tan^{-1}x$ and \[\frac{dy}{dx}=-\frac{2}{1+x^{2}}. \]Case $\displaystyle 3$: $\displaystyle x<-1$. Then $\displaystyle 2\theta\in(-\pi,-\frac{\pi}{2})$, so $\displaystyle \sin^{-1}(\sin 2\theta)=-\pi-2\theta$, giving $\displaystyle y=-\pi-2\tan^{-1}x$ and \[\frac{dy}{dx}=-\frac{2}{1+x^{2}}. \]At $\displaystyle x=\pm 1$ the two one-sided derivatives are $\displaystyle +1$ and $\displaystyle -1$, so $\displaystyle y$ is not differentiable there.\[\boxed{\ \frac{dy}{dx}=\frac{2}{1+x^{2}}\ \text{for }|x|<1,\qquad \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ \text{for }|x|>1;\ \text{not differentiable at }x=\pm1.\ } \]
Exercise 10
$\displaystyle y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$
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NCERT’s answer
$\displaystyle \frac{3}{1+x^{2}}$
Substitution. Put $\displaystyle x=\tan\theta$, i.e. $\displaystyle \theta=\tan^{-1}x$. The given restriction $\displaystyle -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$ means \[-\frac{\pi}{6}<\theta<\frac{\pi}{6}. \]By the triple-angle identity $\displaystyle \tan 3\theta=\dfrac{3\tan\theta-\tan^{3}\theta}{1-3\tan^{2}\theta}$, the argument becomes \[\frac{3x-x^{3}}{1-3x^{2}}=\tan 3\theta,\qquad\text{so}\qquad y=\tan^{-1}(\tan 3\theta). \]Check the principal branch before cancelling: $\displaystyle \tan^{-1}(\tan\alpha)=\alpha$ requires $\displaystyle \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. From $\displaystyle -\frac{\pi}{6}<\theta<\frac{\pi}{6}$ we get $\displaystyle -\frac{\pi}{2}<3\theta<\frac{\pi}{2}$, which is exactly what the restriction on $\displaystyle x$ was there to guarantee. Hence \[y=3\theta=3\tan^{-1}x. \]Differentiate, using $\displaystyle \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}$: \[\boxed{\ \frac{dy}{dx}=\frac{3}{1+x^{2}}\ } \]

EXERCISE 5.3 11–15 (part 7 of 15)

Find $\displaystyle \frac{d y}{d x}$ in the following:
Exercise 11
$\displaystyle y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$
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NCERT’s answer
$\displaystyle \frac{2}{1+x^{2}}$
Substitution. Put $\displaystyle x=\tan\theta$, so $\displaystyle \theta=\tan^{-1}x$. With $\displaystyle 0<x<1$, \[0<\theta<\frac{\pi}{4}. \]Using $\displaystyle \dfrac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta$, \[y=\cos^{-1}(\cos 2\theta). \]Branch check: $\displaystyle \cos^{-1}(\cos\alpha)=\alpha$ only for $\displaystyle \alpha\in[0,\pi]$. Here $\displaystyle 2\theta\in\left(0,\frac{\pi}{2}\right)\subset[0,\pi]$, so the cancellation is valid and \[y=2\theta=2\tan^{-1}x. \]Differentiating, \[\boxed{\ \frac{dy}{dx}=\frac{2}{1+x^{2}}\ } \]
Exercise 12
$\displaystyle y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$
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NCERT’s answer
$\displaystyle \frac{-2}{1+x^{2}}$
Substitution. Put $\displaystyle x=\tan\theta$, $\displaystyle \theta=\tan^{-1}x$; with $\displaystyle 0<x<1$ we get $\displaystyle 0<\theta<\frac{\pi}{4}$, so $\displaystyle 0<2\theta<\frac{\pi}{2}$.As in the previous question the argument is $\displaystyle \cos 2\theta$, but here it sits inside $\displaystyle \sin^{-1}$, so convert the cosine to a sine using $\displaystyle \cos 2\theta=\sin\left(\frac{\pi}{2}-2\theta\right)$: \[y=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right). \]Branch check: $\displaystyle \sin^{-1}(\sin\alpha)=\alpha$ requires $\displaystyle \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since $\displaystyle 0<2\theta<\frac{\pi}{2}$, we have $\displaystyle \frac{\pi}{2}-2\theta\in\left(0,\frac{\pi}{2}\right)$, so the cancellation is valid: \[y=\frac{\pi}{2}-2\theta=\frac{\pi}{2}-2\tan^{-1}x. \]Differentiating (the constant $\displaystyle \frac{\pi}{2}$ contributes nothing, and the minus sign survives): \[\boxed{\ \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ } \]
Exercise 13
$\displaystyle y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$
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NCERT’s answer
$\displaystyle \frac{-2}{1+x^{2}}$
Substitution. Put $\displaystyle x=\tan\theta$, $\displaystyle \theta=\tan^{-1}x$; with $\displaystyle -1<x<1$, \[-\frac{\pi}{4}<\theta<\frac{\pi}{4},\qquad\text{so}\qquad -\frac{\pi}{2}<2\theta<\frac{\pi}{2}. \]Then $\displaystyle \dfrac{2x}{1+x^{2}}=\sin 2\theta$, which must be turned into a cosine because the outer function is $\displaystyle \cos^{-1}$: \[\sin 2\theta=\cos\left(\frac{\pi}{2}-2\theta\right),\qquad y=\cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right). \]Branch check: $\displaystyle \cos^{-1}(\cos\alpha)=\alpha$ needs $\displaystyle \alpha\in[0,\pi]$. From $\displaystyle -\frac{\pi}{2}<2\theta<\frac{\pi}{2}$ we get $\displaystyle \frac{\pi}{2}-2\theta\in(0,\pi)$, so the cancellation holds and \[y=\frac{\pi}{2}-2\tan^{-1}x. \]Differentiating, \[\boxed{\ \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ } \]
Exercise 14
$\displaystyle y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
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NCERT’s answer
$\displaystyle \frac{2}{\sqrt{1-x^{2}}}$
Substitution. The factor $\displaystyle \sqrt{1-x^{2}}$ suggests a sine substitution: put $\displaystyle x=\sin\theta$, so $\displaystyle \theta=\sin^{-1}x$. The restriction $\displaystyle -\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$ gives \[-\frac{\pi}{4}<\theta<\frac{\pi}{4}. \]On this interval $\displaystyle \cos\theta>0$, so $\displaystyle \sqrt{1-x^{2}}=\sqrt{1-\sin^{2}\theta}=|\cos\theta|=\cos\theta$ $\displaystyle -$ the positive root is the one to take, and this is where the restriction earns its keep. Hence \[2x\sqrt{1-x^{2}}=2\sin\theta\cos\theta=\sin 2\theta,\qquad y=\sin^{-1}(\sin 2\theta). \]Branch check: $\displaystyle 2\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, which lies in the principal range of $\displaystyle \sin^{-1}$, so \[y=2\theta=2\sin^{-1}x. \]Differentiating with $\displaystyle \frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}$: \[\boxed{\ \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}\ } \]
Exercise 15
$\displaystyle y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}$
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NCERT’s answer
$\displaystyle -\frac{2}{\sqrt{1-x^{2}}}$
Substitution. Since $\displaystyle \mathrm{sec}^{-1}$ undoes a secant, first make the argument a secant. Put $\displaystyle x=\cos\theta$, so $\displaystyle \theta=\cos^{-1}x$. The restriction $\displaystyle 0<x<\frac{1}{\sqrt{2}}$ gives \[\frac{\pi}{4}<\theta<\frac{\pi}{2}. \]Using $\displaystyle 2\cos^{2}\theta-1=\cos 2\theta$, \[\frac{1}{2x^{2}-1}=\frac{1}{\cos 2\theta}=\sec 2\theta,\qquad y=\mathrm{sec}^{-1}(\sec 2\theta). \]Branch check: $\displaystyle \mathrm{sec}^{-1}(\sec\alpha)=\alpha$ requires $\displaystyle \alpha\in[0,\pi]$ with $\displaystyle \alpha\ne\frac{\pi}{2}$. From $\displaystyle \frac{\pi}{4}<\theta<\frac{\pi}{2}$ we get $\displaystyle 2\theta\in\left(\frac{\pi}{2},\pi\right)$, which is inside the principal range (and note $\displaystyle \cos 2\theta<0$ there, consistent with $\displaystyle 2x^{2}-1<0$ for $\displaystyle x<\frac{1}{\sqrt{2}}$). Hence \[y=2\theta=2\cos^{-1}x. \]Differentiating with $\displaystyle \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^{2}}}$: \[\boxed{\ \frac{dy}{dx}=-\frac{2}{\sqrt{1-x^{2}}}\ } \]

EXERCISE 5.4 1–10 (part 8 of 15)

Differentiate the following w.r.t. $\displaystyle x$ :
Exercise 1
$\displaystyle \frac{e^{x}}{\sin x}$
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NCERT’s answer
$\displaystyle \frac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}, x \neq n \pi, n \in \mathbf{Z}$
Use the quotient rule: if $\displaystyle y=\dfrac{u}{v}$, then $\displaystyle \dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}$, valid wherever $\displaystyle v\neq 0$.Here $\displaystyle u=e^{x}$ and $\displaystyle v=\sin x$, so $\displaystyle u'=e^{x}$ and $\displaystyle v'=\cos x$. Hence \[\frac{d}{dx}\left(\frac{e^{x}}{\sin x}\right)=\frac{e^{x}\sin x-e^{x}\cos x}{\sin^{2}x}=\frac{e^{x}\left(\sin x-\cos x\right)}{\sin^{2}x}.\]The function itself is undefined where $\displaystyle \sin x=0$, so the derivative is \[\frac{e^{x}(\sin x-\cos x)}{\sin^{2}x},\qquad x\neq n\pi,\ n\in\mathbf{Z}.\]
Exercise 2
$\displaystyle e^{\sin ^{-1} x}$
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NCERT’s answer
$\displaystyle \frac{e^{\sin -1} x}{\sqrt{1-x^{2}}}, x \in(-1,1)$
Chain rule with the outer function $\displaystyle e^{t}$ and inner $\displaystyle t=\sin^{-1}x$: \[\frac{d}{dx}\left(e^{t}\right)=e^{t}\cdot\frac{dt}{dx}.\]The standard derivative of the inverse sine is $\displaystyle \dfrac{d}{dx}\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^{2}}}$, which itself holds only for $\displaystyle -1<x<1$ (at $\displaystyle x=\pm 1$ the tangent is vertical). Therefore \[\frac{d}{dx}\left(e^{\sin^{-1}x}\right)=e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^{2}}}.\]So the derivative is $\displaystyle \dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}$, for $\displaystyle x\in(-1,1)$.
Exercise 3
$\displaystyle e^{x^{3}}$
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NCERT’s answer
$\displaystyle 3 x^{2} e^{x^{3}}$
Chain rule, outer function $\displaystyle e^{t}$ with $\displaystyle t=x^{3}$: \[\frac{d}{dx}\left(e^{x^{3}}\right)=e^{x^{3}}\cdot\frac{d}{dx}\left(x^{3}\right)=e^{x^{3}}\cdot 3x^{2}.\]Note the exponent is $\displaystyle x^{3}$, not $\displaystyle 3x$; the factor brought down is $\displaystyle 3x^{2}$, not $\displaystyle 3$.Hence $\displaystyle \dfrac{d}{dx}\left(e^{x^{3}}\right)=3x^{2}e^{x^{3}}$, for all $\displaystyle x\in\mathbf{R}$.
Exercise 4
$\displaystyle \sin \left(\tan ^{-1} e^{-x}\right)$
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NCERT’s answer
$\displaystyle -\frac{e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}}$
This is a chain of three functions: $\displaystyle \sin$ of $\displaystyle \tan^{-1}$ of $\displaystyle e^{-x}$. Differentiate from the outside in, using $\displaystyle \dfrac{d}{dx}\left(\tan^{-1}t\right)=\dfrac{1}{1+t^{2}}$.With $\displaystyle y=\sin\left(\tan^{-1}e^{-x}\right)$, \[\frac{dy}{dx}=\cos\left(\tan^{-1}e^{-x}\right)\cdot\frac{1}{1+\left(e^{-x}\right)^{2}}\cdot\frac{d}{dx}\left(e^{-x}\right).\]The innermost derivative carries the minus sign: $\displaystyle \dfrac{d}{dx}\left(e^{-x}\right)=-e^{-x}$. So \[\frac{dy}{dx}=\frac{-e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}.\]One may simplify the cosine: if $\displaystyle \theta=\tan^{-1}e^{-x}$ then $\displaystyle \tan\theta=e^{-x}$ with $\displaystyle \theta\in\left(0,\tfrac{\pi}{2}\right)$, so $\displaystyle \cos\theta=\dfrac{1}{\sqrt{1+e^{-2x}}}$, giving the equivalent form \[\frac{dy}{dx}=\frac{-e^{-x}}{\left(1+e^{-2x}\right)^{3/2}},\qquad x\in\mathbf{R}.\]
Exercise 5
$\displaystyle \log \left(\cos e^{x}\right)$
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NCERT’s answer
$\displaystyle -e^{x} \tan e^{x}, e^{x} \neq(2 n+1) \frac{\pi}{2}, n \in \mathbf{N}$
Chain rule twice, with $\displaystyle \dfrac{d}{dx}\left(\log t\right)=\dfrac{1}{t}$ (natural logarithm).Let $\displaystyle y=\log\left(\cos e^{x}\right)$. Then \[\frac{dy}{dx}=\frac{1}{\cos e^{x}}\cdot\frac{d}{dx}\left(\cos e^{x}\right)=\frac{1}{\cos e^{x}}\cdot\left(-\sin e^{x}\right)\cdot\frac{d}{dx}\left(e^{x}\right).\]The last factor is $\displaystyle e^{x}$ itself, so \[\frac{dy}{dx}=-\frac{e^{x}\sin e^{x}}{\cos e^{x}}=-e^{x}\tan\left(e^{x}\right).\]Thus the derivative is $\displaystyle -e^{x}\tan\left(e^{x}\right)$, valid on the set where the function is defined, i.e. where $\displaystyle \cos e^{x}>0$.
Exercise 6
$\displaystyle e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}$
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NCERT’s answer
$\displaystyle e^{x}+2 x^{e^{x^{2}}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}$
Differentiate term by term (the derivative of a finite sum is the sum of the derivatives), each term by the chain rule $\displaystyle \dfrac{d}{dx}\left(e^{x^{n}}\right)=e^{x^{n}}\cdot nx^{n-1}$.Let $\displaystyle y=e^{x}+e^{x^{2}}+e^{x^{3}}+e^{x^{4}}+e^{x^{5}}$. Then \[\frac{dy}{dx}=e^{x}\cdot 1+e^{x^{2}}\cdot 2x+e^{x^{3}}\cdot 3x^{2}+e^{x^{4}}\cdot 4x^{3}+e^{x^{5}}\cdot 5x^{4}.\]So \[\frac{dy}{dx}=e^{x}+2x\,e^{x^{2}}+3x^{2}e^{x^{3}}+4x^{3}e^{x^{4}}+5x^{4}e^{x^{5}},\qquad x\in\mathbf{R}.\]
Exercise 7
$\displaystyle \sqrt{e^{\sqrt{x}}}, x>0$
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NCERT’s answer
$\displaystyle \frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0$
First rewrite the surd as a single exponential, using $\displaystyle \sqrt{a}=a^{1/2}$ and $\displaystyle \left(e^{t}\right)^{1/2}=e^{t/2}$: \[y=\sqrt{e^{\sqrt{x}}}=e^{\frac{\sqrt{x}}{2}}=e^{\frac{1}{2}x^{1/2}}.\]Now the chain rule with $\displaystyle t=\tfrac{1}{2}x^{1/2}$, for which $\displaystyle \dfrac{dt}{dx}=\tfrac{1}{2}\cdot\tfrac{1}{2}x^{-1/2}=\dfrac{1}{4\sqrt{x}}$ (this is the step most often dropped — the $\displaystyle \tfrac12$ from the outer square root and the $\displaystyle \tfrac12$ from $\displaystyle \sqrt{x}$ both appear): \[\frac{dy}{dx}=e^{\frac{\sqrt{x}}{2}}\cdot\frac{1}{4\sqrt{x}}.\]Hence \[\frac{dy}{dx}=\frac{e^{\sqrt{x}/2}}{4\sqrt{x}}=\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}},\qquad x>0.\]
Exercise 8
$\displaystyle \log (\log x), x>1$
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NCERT’s answer
$\displaystyle \frac{1}{x \log x}, x>1$
Chain rule with the outer function $\displaystyle \log t$, $\displaystyle t=\log x$, using $\displaystyle \dfrac{d}{dx}\left(\log x\right)=\dfrac{1}{x}$.Let $\displaystyle y=\log\left(\log x\right)$. Then \[\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{d}{dx}\left(\log x\right)=\frac{1}{\log x}\cdot\frac{1}{x}.\]The restriction $\displaystyle x>1$ is what makes $\displaystyle \log x>0$, so that the outer logarithm is defined and $\displaystyle \log x\neq 0$ in the denominator.Therefore \[\frac{dy}{dx}=\frac{1}{x\log x},\qquad x>1.\]
Exercise 9
$\displaystyle \frac{\cos x}{\log x}, x>0$
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NCERT’s answer
$\displaystyle -\frac{(x \sin x \cdot \log x+\cos x)}{x(\log x)^{2}}, x>0$
Quotient rule with $\displaystyle u=\cos x$, $\displaystyle v=\log x$, so $\displaystyle u'=-\sin x$ and $\displaystyle v'=\dfrac{1}{x}$: \[\frac{d}{dx}\left(\frac{\cos x}{\log x}\right)=\frac{u'v-uv'}{v^{2}}=\frac{\left(-\sin x\right)\log x-\cos x\cdot\dfrac{1}{x}}{\left(\log x\right)^{2}}.\]Multiply numerator and denominator by $\displaystyle x$ to clear the internal fraction: \[=\frac{-x\sin x\,\log x-\cos x}{x\left(\log x\right)^{2}}=-\frac{x\log x\,\sin x+\cos x}{x\left(\log x\right)^{2}}.\]So the derivative is \[-\frac{x\log x\,\sin x+\cos x}{x\left(\log x\right)^{2}},\qquad x>0,\ x\neq 1\ \text{(since }\log 1=0\text{)}.\]
Exercise 10
$\displaystyle \cos \left(\log x+e^{x}\right), x>0$
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NCERT’s answer
$\displaystyle -\frac{1}{x}+e^{x} \sin \left(\log x+e^{x}\right), x>0$
Chain rule with the outer function $\displaystyle \cos t$, $\displaystyle t=\log x+e^{x}$, so that $\displaystyle \dfrac{dy}{dx}=-\sin t\cdot\dfrac{dt}{dx}$.Differentiate the inner sum term by term: \[\frac{dt}{dx}=\frac{d}{dx}\left(\log x\right)+\frac{d}{dx}\left(e^{x}\right)=\frac{1}{x}+e^{x}.\]Hence \[\frac{d}{dx}\Big(\cos\left(\log x+e^{x}\right)\Big)=-\sin\left(\log x+e^{x}\right)\left(\frac{1}{x}+e^{x}\right),\qquad x>0,\] the restriction $\displaystyle x>0$ being needed for $\displaystyle \log x$ to exist.

EXERCISE 5.5 1–10 (part 9 of 15)

Differentiate the functions given in Exercises $\displaystyle 1$ to $\displaystyle 11$ w.r.t. $\displaystyle x$.
Exercise 1
$\displaystyle \cos x \cdot \cos 2 x \cdot \cos 3 x$
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NCERT’s answer
$\displaystyle -\cos x \cos 2 x \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x]$
A product of three factors: logarithmic differentiation turns it into a sum.Let $\displaystyle y=\cos x\cdot\cos 2x\cdot\cos 3x$. Taking logarithms of the modulus of both sides (valid wherever no factor vanishes), \[\log|y|=\log|\cos x|+\log|\cos 2x|+\log|\cos 3x|.\]Differentiate with respect to $\displaystyle x$, using $\displaystyle \dfrac{d}{dx}\log|f(x)|=\dfrac{f'(x)}{f(x)}$ and the chain rule on the inner functions $\displaystyle 2x$ and $\displaystyle 3x$: \[\frac{1}{y}\frac{dy}{dx}=\frac{-\sin x}{\cos x}+\frac{-2\sin 2x}{\cos 2x}+\frac{-3\sin 3x}{\cos 3x}.\]The inner-derivative factors $\displaystyle 2$ and $\displaystyle 3$ are the step most often dropped. Multiplying through by $\displaystyle y$, \[\frac{dy}{dx}=-\cos x\cdot\cos 2x\cdot\cos 3x\,\bigl(\tan x+2\tan 2x+3\tan 3x\bigr),\] for every $\displaystyle x$ with $\displaystyle \cos x\cos 2x\cos 3x\neq 0$.
Exercise 2
$\displaystyle \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
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A quotient inside a square root: take logarithms so the root becomes a factor $\displaystyle \tfrac12$ and the quotient becomes a difference.Let \[y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}},\] working on a set where the radicand is positive, so $\displaystyle y>0$. Then \[\log y=\frac{1}{2}\Bigl[\log|x-1|+\log|x-2|-\log|x-3|-\log|x-4|-\log|x-5|\Bigr].\]The three denominator factors carry minus signs; that sign change is the point of the step. Differentiating, \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right].\]Hence \[\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\;\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right],\] valid at every interior point of the domain (in particular $\displaystyle x\neq 1,2,3,4,5$).
Exercise 3
$\displaystyle (\log x)^{\cos x}$
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NCERT’s answer
$\displaystyle (\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]$
The base $\displaystyle \log x$ and the exponent $\displaystyle \cos x$ both vary, so neither the power rule nor the rule for $\displaystyle a^{x}$ applies. Take logarithms.Let $\displaystyle y=(\log x)^{\cos x}$. For the base to be positive we need $\displaystyle \log x>0$, i.e. $\displaystyle x>1$. Then \[\log y=\cos x\cdot\log(\log x).\]Differentiate both sides: product rule on the right, and the chain rule on $\displaystyle \log(\log x)$, whose derivative is $\displaystyle \dfrac{1}{\log x}\cdot\dfrac{1}{x}$: \[\frac{1}{y}\frac{dy}{dx}=-\sin x\cdot\log(\log x)+\cos x\cdot\frac{1}{x\log x}.\]Therefore \[\frac{dy}{dx}=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\,\log(\log x)\right],\qquad x>1.\]
Exercise 4
$\displaystyle x^{x}-2^{\sin x}$
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NCERT’s answer
$\displaystyle x^{x}(1+\log x)-2^{\sin x} \cos x \log 2$
Differentiate the two terms separately; each has a variable in the exponent, but for different reasons.Let $\displaystyle y=u-v$ with $\displaystyle u=x^{x}$ and $\displaystyle v=2^{\sin x}$, and take $\displaystyle x>0$ so that $\displaystyle x^{x}$ is defined.First term (variable base and variable exponent). $\displaystyle \log u=x\log x$, so by the product rule \[\frac{1}{u}\frac{du}{dx}=\log x+x\cdot\frac{1}{x}=1+\log x,\qquad \frac{du}{dx}=x^{x}(1+\log x).\]Second term (constant base). $\displaystyle \log v=\sin x\cdot\log 2$, so \[\frac{1}{v}\frac{dv}{dx}=\cos x\,\log 2,\qquad \frac{dv}{dx}=2^{\sin x}\cos x\,\log 2.\] Here $\displaystyle \log 2$ is a constant and must not be differentiated away; dropping it is the usual error.Hence \[\frac{dy}{dx}=x^{x}(1+\log x)-2^{\sin x}\cos x\,\log 2,\qquad x>0.\]
Exercise 5
$\displaystyle (x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$
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NCERT’s answer
$\displaystyle (x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133\right)$
Logarithmic differentiation converts the triple product into a sum of three simple terms.Let $\displaystyle y=(x+3)^{2}(x+4)^{3}(x+5)^{4}$. Then \[\log|y|=2\log|x+3|+3\log|x+4|+4\log|x+5|,\] the exponents coming down as multipliers. Differentiating, \[\frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5},\] so \[\frac{dy}{dx}=(x+3)^{2}(x+4)^{3}(x+5)^{4}\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right].\]Cancelling one factor against each denominator gives a polynomial form, which also shows the answer is valid at $\displaystyle x=-3,-4,-5$ where the logarithms were not: \[\frac{dy}{dx}=(x+3)(x+4)^{2}(x+5)^{3}\Bigl[2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)\Bigr]\] \[=(x+3)(x+4)^{2}(x+5)^{3}\,\bigl(9x^{2}+70x+133\bigr).\]
Exercise 6
$\displaystyle \left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
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NCERT’s answer
$\displaystyle \left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{1+\frac{1}{x}}\left(\frac{x+1-\log x}{x^{2}}\right)$
Two terms, each with a variable exponent; treat them separately. Let $\displaystyle y=u+v$, where \[u=\left(x+\frac{1}{x}\right)^{x},\qquad v=x^{\left(1+\frac{1}{x}\right)},\qquad x>0.\]First term. $\displaystyle \log u=x\log\!\left(x+\dfrac{1}{x}\right)$, so by the product rule and chain rule \[\frac{1}{u}\frac{du}{dx}=\log\!\left(x+\frac{1}{x}\right)+x\cdot\frac{1-\dfrac{1}{x^{2}}}{\,x+\dfrac{1}{x}\,}.\] Simplify that fraction before going further — this is where the algebra usually goes wrong: \[\frac{1-\dfrac{1}{x^{2}}}{x+\dfrac{1}{x}}=\frac{(x^{2}-1)/x^{2}}{(x^{2}+1)/x}=\frac{x^{2}-1}{x(x^{2}+1)},\] so the second term is $\displaystyle x\cdot\dfrac{x^{2}-1}{x(x^{2}+1)}=\dfrac{x^{2}-1}{x^{2}+1}$. Hence \[\frac{du}{dx}=\left(x+\frac{1}{x}\right)^{x}\left[\log\!\left(x+\frac{1}{x}\right)+\frac{x^{2}-1}{x^{2}+1}\right].\]Second term. $\displaystyle \log v=\left(1+\dfrac{1}{x}\right)\log x$, so \[\frac{1}{v}\frac{dv}{dx}=-\frac{1}{x^{2}}\log x+\left(1+\frac{1}{x}\right)\frac{1}{x}=\frac{x+1-\log x}{x^{2}},\] and therefore $\displaystyle \dfrac{dv}{dx}=x^{\left(1+\frac{1}{x}\right)}\cdot\dfrac{x+1-\log x}{x^{2}}$.Adding, \[\frac{dy}{dx}=\left(x+\frac{1}{x}\right)^{x}\left[\log\!\left(x+\frac{1}{x}\right)+\frac{x^{2}-1}{x^{2}+1}\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right],\qquad x>0.\]
Exercise 7
$\displaystyle (\log x)^{x}+x^{\log x}$
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NCERT’s answer
$\displaystyle (\log x)^{x-1}[1+\log x . \log (\log x)]+2 x^{\log x-1} . \log x$
Let $\displaystyle y=u+v$ with $\displaystyle u=(\log x)^{x}$ and $\displaystyle v=x^{\log x}$. The first needs $\displaystyle \log x>0$, so take $\displaystyle x>1$.First term. $\displaystyle \log u=x\log(\log x)$. Differentiating by the product rule, with $\displaystyle \dfrac{d}{dx}\log(\log x)=\dfrac{1}{x\log x}$, \[\frac{1}{u}\frac{du}{dx}=\log(\log x)+x\cdot\frac{1}{x\log x}=\log(\log x)+\frac{1}{\log x},\] so \[\frac{du}{dx}=(\log x)^{x}\left[\log(\log x)+\frac{1}{\log x}\right].\]Second term. $\displaystyle \log v=\log x\cdot\log x=(\log x)^{2}$. Note the exponent is itself $\displaystyle \log x$, so the logarithm of $\displaystyle v$ is a square, not a product of two different things: \[\frac{1}{v}\frac{dv}{dx}=2\log x\cdot\frac{1}{x},\qquad \frac{dv}{dx}=x^{\log x}\cdot\frac{2\log x}{x}=2x^{\log x-1}\log x.\]Hence \[\frac{dy}{dx}=(\log x)^{x}\left[\log(\log x)+\frac{1}{\log x}\right]+2x^{\log x-1}\log x,\qquad x>1.\]
Exercise 8
$\displaystyle (\sin x)^{x}+\sin ^{-1} \sqrt{x}$
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NCERT’s answer
$\displaystyle (\sin x)^{x}(x \cot x+\log \sin x)+\frac{1}{2} \frac{1}{\sqrt{x-x^{2}}}$
Let $\displaystyle y=u+v$ with $\displaystyle u=(\sin x)^{x}$ and $\displaystyle v=\sin^{-1}\sqrt{x}$. The base $\displaystyle \sin x$ must be positive and $\displaystyle \sqrt{x}$ must lie in $\displaystyle [-1,1]$, so work on $\displaystyle 0<x<1$ (where indeed $\displaystyle \sin x>0$).First term, by logarithmic differentiation. $\displaystyle \log u=x\log\sin x$, so \[\frac{1}{u}\frac{du}{dx}=\log\sin x+x\cdot\frac{\cos x}{\sin x}=\log\sin x+x\cot x,\] \[\frac{du}{dx}=(\sin x)^{x}\bigl(x\cot x+\log\sin x\bigr).\]Second term, by the chain rule with $\displaystyle \dfrac{d}{dt}\sin^{-1}t=\dfrac{1}{\sqrt{1-t^{2}}}$ and $\displaystyle t=\sqrt{x}$, $\displaystyle \dfrac{dt}{dx}=\dfrac{1}{2\sqrt{x}}$: \[\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}\sqrt{1-x}}=\frac{1}{2\sqrt{x-x^{2}}}.\] Squaring $\displaystyle \sqrt{x}$ to $\displaystyle x$ inside the root is the step to get right.Therefore \[\frac{dy}{dx}=(\sin x)^{x}\bigl(x\cot x+\log\sin x\bigr)+\frac{1}{2\sqrt{x-x^{2}}},\qquad 0<x<1.\]
Exercise 9
$\displaystyle x^{\sin x}+(\sin x)^{\cos x}$
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NCERT’s answer
$\displaystyle x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]+(\sin x)^{\cos x}[\cos x \cot x-\sin x \log \sin x]$
Both terms have a variable base and a variable exponent, so each is handled by taking logarithms separately. Let $\displaystyle y=u+v$ with $\displaystyle u=x^{\sin x}$, $\displaystyle v=(\sin x)^{\cos x}$, on a set where $\displaystyle x>0$ and $\displaystyle \sin x>0$.First term. $\displaystyle \log u=\sin x\cdot\log x$, so \[\frac{1}{u}\frac{du}{dx}=\cos x\log x+\sin x\cdot\frac{1}{x},\qquad \frac{du}{dx}=x^{\sin x}\left(\cos x\,\log x+\frac{\sin x}{x}\right).\]Second term. $\displaystyle \log v=\cos x\cdot\log\sin x$, so \[\frac{1}{v}\frac{dv}{dx}=-\sin x\,\log\sin x+\cos x\cdot\frac{\cos x}{\sin x}=\cos x\cot x-\sin x\,\log\sin x,\] \[\frac{dv}{dx}=(\sin x)^{\cos x}\bigl(\cos x\cot x-\sin x\,\log\sin x\bigr).\]Hence \[\frac{dy}{dx}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\,\log x\right)+(\sin x)^{\cos x}\bigl(\cos x\cot x-\sin x\,\log\sin x\bigr).\]
Exercise 10
$\displaystyle x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
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NCERT’s answer
$\displaystyle x^{x \cos x}[\cos x .(1+\log x)-x \sin x \log x]-\frac{4 x}{\left(x^{2}-1\right)^{2}}$
Let $\displaystyle y=u+v$ with $\displaystyle u=x^{x\cos x}$ and $\displaystyle v=\dfrac{x^{2}+1}{x^{2}-1}$.First term, by logarithmic differentiation ($\displaystyle x>0$). $\displaystyle \log u=x\cos x\cdot\log x$, a product of three functions, so \[\frac{1}{u}\frac{du}{dx}=(1)\cos x\log x+x(-\sin x)\log x+x\cos x\cdot\frac{1}{x}\] \[=\cos x\log x-x\sin x\log x+\cos x.\] Hence \[\frac{du}{dx}=x^{x\cos x}\bigl[\cos x\,(1+\log x)-x\sin x\,\log x\bigr].\]Second term, by the quotient rule: \[\frac{dv}{dx}=\frac{(2x)(x^{2}-1)-(x^{2}+1)(2x)}{(x^{2}-1)^{2}}=\frac{2x\bigl[(x^{2}-1)-(x^{2}+1)\bigr]}{(x^{2}-1)^{2}}=\frac{-4x}{(x^{2}-1)^{2}}.\] The two $\displaystyle x^{2}$ terms cancel; keeping the minus sign on the second bracket is essential.Therefore \[\frac{dy}{dx}=x^{x\cos x}\bigl[\cos x\,(1+\log x)-x\sin x\,\log x\bigr]-\frac{4x}{(x^{2}-1)^{2}},\qquad x>0,\ x\neq 1.\]

EXERCISE 5.5 11–18 (part 10 of 15)

Differentiate the functions given in Exercises $\displaystyle 1$ to $\displaystyle 11$ w.r.t. $\displaystyle x$.
Exercise 11
$\displaystyle (x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
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NCERT’s answer
$\displaystyle (x \cos x)^{x}[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}} \frac{x \cot x+1-\log (x \sin x)}{x^{2}}$
Let $\displaystyle y=u+v$ with $\displaystyle u=(x\cos x)^{x}$ and $\displaystyle v=(x\sin x)^{\frac{1}{x}}$, on a set where $\displaystyle x\cos x>0$ and $\displaystyle x\sin x>0$ so both bases are positive.First term. $\displaystyle \log u=x\log(x\cos x)=x\bigl[\log x+\log\cos x\bigr]$. Splitting the log of the product first makes the differentiation routine: \[\frac{1}{u}\frac{du}{dx}=\bigl[\log x+\log\cos x\bigr]+x\left[\frac{1}{x}-\tan x\right]=\log(x\cos x)+1-x\tan x.\] Hence \[\frac{du}{dx}=(x\cos x)^{x}\bigl[1-x\tan x+\log(x\cos x)\bigr].\]Second term. $\displaystyle \log v=\dfrac{1}{x}\log(x\sin x)=\dfrac{1}{x}\bigl[\log x+\log\sin x\bigr]$. By the product rule, remembering $\displaystyle \dfrac{d}{dx}\dfrac{1}{x}=-\dfrac{1}{x^{2}}$, \[\frac{1}{v}\frac{dv}{dx}=-\frac{1}{x^{2}}\log(x\sin x)+\frac{1}{x}\left[\frac{1}{x}+\cot x\right]=\frac{1}{x^{2}}\bigl[1+x\cot x-\log(x\sin x)\bigr].\] Hence \[\frac{dv}{dx}=(x\sin x)^{\frac{1}{x}}\cdot\frac{1}{x^{2}}\bigl[1+x\cot x-\log(x\sin x)\bigr].\]Therefore \[\frac{dy}{dx}=(x\cos x)^{x}\bigl[1-x\tan x+\log(x\cos x)\bigr]+(x\sin x)^{\frac{1}{x}}\left[\frac{1+x\cot x-\log(x\sin x)}{x^{2}}\right].\]
Find $\displaystyle \frac{d y}{d x}$ of the functions given in Exercises $\displaystyle 12$ to 15.
Exercise 12
$\displaystyle x^{y}+y^{x}=1$
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NCERT’s answer
$\displaystyle -\frac{y x^{y-1}+y^{x} \log y}{x^{y} \log x+x y^{x-1}}$
The relation $\displaystyle x^{y}+y^{x}=1$ defines $\displaystyle y$ implicitly. Neither term can be differentiated by the power rule, so write each as an exponential and differentiate implicitly, treating $\displaystyle y$ as a function of $\displaystyle x$.Put $\displaystyle u=x^{y}$ and $\displaystyle v=y^{x}$, so $\displaystyle u+v=1$.For $\displaystyle u$: $\displaystyle \log u=y\log x$, hence \[\frac{1}{u}\frac{du}{dx}=\frac{dy}{dx}\log x+\frac{y}{x},\qquad \frac{du}{dx}=x^{y}\left(\log x\,\frac{dy}{dx}+\frac{y}{x}\right).\]For $\displaystyle v$: $\displaystyle \log v=x\log y$, hence \[\frac{1}{v}\frac{dv}{dx}=\log y+\frac{x}{y}\frac{dy}{dx},\qquad \frac{dv}{dx}=y^{x}\left(\log y+\frac{x}{y}\frac{dy}{dx}\right).\]Because the right side of the given equation is the constant $\displaystyle 1$, $\displaystyle \dfrac{du}{dx}+\dfrac{dv}{dx}=0$: \[x^{y}\log x\,\frac{dy}{dx}+y\,x^{y-1}+y^{x}\log y+x\,y^{x-1}\frac{dy}{dx}=0,\] using $\displaystyle x^{y}\cdot\dfrac{y}{x}=y\,x^{y-1}$ and $\displaystyle y^{x}\cdot\dfrac{x}{y}=x\,y^{x-1}$.Collecting the $\displaystyle \dfrac{dy}{dx}$ terms, \[\frac{dy}{dx}\bigl(x^{y}\log x+x\,y^{x-1}\bigr)=-\bigl(y\,x^{y-1}+y^{x}\log y\bigr),\] so \[\frac{dy}{dx}=-\,\frac{y\,x^{y-1}+y^{x}\log y}{x^{y}\log x+x\,y^{x-1}}.\]
Exercise 13
$\displaystyle y^{x}=x^{y}$
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NCERT’s answer
$\displaystyle \frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}\right)$
Given $\displaystyle y^{x}=x^{y}$. Take logarithms of both sides first — this removes both variable exponents in one step: \[x\log y=y\log x.\]Now differentiate implicitly with respect to $\displaystyle x$, applying the product rule on each side and remembering $\displaystyle \dfrac{d}{dx}\log y=\dfrac{1}{y}\dfrac{dy}{dx}$: \[\log y+\frac{x}{y}\frac{dy}{dx}=\frac{dy}{dx}\log x+\frac{y}{x}.\]Collect the $\displaystyle \dfrac{dy}{dx}$ terms on one side: \[\frac{dy}{dx}\left(\frac{x}{y}-\log x\right)=\frac{y}{x}-\log y,\] \[\frac{dy}{dx}\cdot\frac{x-y\log x}{y}=\frac{y-x\log y}{x}.\]Hence \[\frac{dy}{dx}=\frac{y\,(y-x\log y)}{x\,(x-y\log x)},\] valid where $\displaystyle x>0,\ y>0$ and $\displaystyle x-y\log x\neq 0$.
Exercise 14
$\displaystyle (\cos x)^{y}=(\cos y)^{x}$
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NCERT’s answer
$\displaystyle \frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$
Given $\displaystyle (\cos x)^{y}=(\cos y)^{x}$. Both sides have a variable base and a variable exponent, so take logarithms of both sides: \[y\log\cos x=x\log\cos y.\]Differentiate implicitly with respect to $\displaystyle x$. On the left, the product rule together with $\displaystyle \dfrac{d}{dx}\log\cos x=\dfrac{-\sin x}{\cos x}=-\tan x$; on the right, the product rule together with $\displaystyle \dfrac{d}{dx}\log\cos y=-\tan y\,\dfrac{dy}{dx}$: \[\frac{dy}{dx}\log\cos x+y(-\tan x)=\log\cos y+x(-\tan y)\frac{dy}{dx}.\] The chain-rule factor $\displaystyle \dfrac{dy}{dx}$ attached to $\displaystyle \log\cos y$ is the step most often missed.Collecting the $\displaystyle \dfrac{dy}{dx}$ terms, \[\frac{dy}{dx}\bigl(\log\cos x+x\tan y\bigr)=\log\cos y+y\tan x,\] so \[\frac{dy}{dx}=\frac{\log\cos y+y\tan x}{\log\cos x+x\tan y},\] wherever $\displaystyle \cos x>0$, $\displaystyle \cos y>0$ and the denominator is non-zero.
Exercise 15
$\displaystyle x y=e^{(x-y)}$
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NCERT’s answer
$\displaystyle \frac{y(x-1)}{x(y+1)}$
Given $\displaystyle xy=e^{(x-y)}$, with $\displaystyle x>0$ and $\displaystyle y>0$ (the left side must be positive, since the right side is). Taking logarithms of both sides converts the product to a sum and strips the exponential: \[\log x+\log y=x-y.\]Differentiate implicitly with respect to $\displaystyle x$: \[\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}.\]Collect the $\displaystyle \dfrac{dy}{dx}$ terms: \[\frac{dy}{dx}\left(\frac{1}{y}+1\right)=1-\frac{1}{x},\qquad \frac{dy}{dx}\cdot\frac{1+y}{y}=\frac{x-1}{x}.\]Hence \[\frac{dy}{dx}=\frac{y\,(x-1)}{x\,(1+y)}.\]
Exercise 16
Find the derivative of the function given by $\displaystyle f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$ and hence find $\displaystyle f^{\prime}(1)$.
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NCERT’s answer
$\displaystyle (1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}\right] ; f^{\prime}(1)=120$
A product of four factors, so use logarithmic differentiation.Let $\displaystyle f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$. Each factor is positive for $\displaystyle x>0$; more generally take logarithms of the modulus: \[\log|f(x)|=\log|1+x|+\log(1+x^{2})+\log(1+x^{4})+\log(1+x^{8}).\]Differentiating, each term by the chain rule, \[\frac{f'(x)}{f(x)}=\frac{1}{1+x}+\frac{2x}{1+x^{2}}+\frac{4x^{3}}{1+x^{4}}+\frac{8x^{7}}{1+x^{8}}.\]Hence \[f'(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\left[\frac{1}{1+x}+\frac{2x}{1+x^{2}}+\frac{4x^{3}}{1+x^{4}}+\frac{8x^{7}}{1+x^{8}}\right].\]Now put $\displaystyle x=1$. Substitute into the product and the bracket separately: \[f(1)=2\cdot 2\cdot 2\cdot 2=16,\] \[\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]=\frac{1+2+4+8}{2}=\frac{15}{2}.\]Therefore \[f'(1)=16\times\frac{15}{2}=120.\]
Exercise 17
Differentiate $\displaystyle \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$ in three ways mentioned below:
(i)
by using product rule
(ii)
by expanding the product to obtain a single polynomial.
(iii)
by logarithmic differentiation. Do they all give the same answer?
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NCERT’s answer
$\displaystyle 5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$
Let $\displaystyle y=(x^{2}-5x+8)(x^{3}+7x+9)$.
(i)
Product rule. With $\displaystyle u=x^{2}-5x+8$, $\displaystyle v=x^{3}+7x+9$, so $\displaystyle u'=2x-5$ and $\displaystyle v'=3x^{2}+7$,
\[\frac{dy}{dx}=u'v+uv'=(2x-5)(x^{3}+7x+9)+(x^{2}-5x+8)(3x^{2}+7).\]
Expanding,
\[(2x-5)(x^{3}+7x+9)=2x^{4}-5x^{3}+14x^{2}-35x+18x-45,\]
\[(x^{2}-5x+8)(3x^{2}+7)=3x^{4}-15x^{3}+24x^{2}+7x^{2}-35x+56,\]
and adding,
\[\frac{dy}{dx}=5x^{4}-20x^{3}+45x^{2}-52x+11.\]
(ii)
Expanding first.
\[y=x^{5}+7x^{3}+9x^{2}-5x^{4}-35x^{2}-45x+8x^{3}+56x+72\]
\[=x^{5}-5x^{4}+15x^{3}-26x^{2}+11x+72.\]
Differentiating term by term with the power rule (the constant $\displaystyle 72$ dies),
\[\frac{dy}{dx}=5x^{4}-20x^{3}+45x^{2}-52x+11.\]
(iii)
Logarithmic differentiation. Taking logarithms of the modulus,
\[\log|y|=\log|x^{2}-5x+8|+\log|x^{3}+7x+9|,\]
\[\frac{1}{y}\frac{dy}{dx}=\frac{2x-5}{x^{2}-5x+8}+\frac{3x^{2}+7}{x^{3}+7x+9}.\]
Multiplying by $\displaystyle y=(x^{2}-5x+8)(x^{3}+7x+9)$, each fraction cancels one factor:
\[\frac{dy}{dx}=(2x-5)(x^{3}+7x+9)+(x^{2}-5x+8)(3x^{2}+7)=5x^{4}-20x^{3}+45x^{2}-52x+11.\]
Yes — all three methods give the same answer, $\displaystyle 5x^{4}-20x^{3}+45x^{2}-52x+11$. They must: the three routes differentiate one and the same function, and the derivative of a function is unique. (Method (iii) formally needs $\displaystyle y\neq 0$, but the resulting polynomial is continuous, so it holds everywhere.)
Exercise 18
If $\displaystyle u, v$ and $\displaystyle w$ are functions of $\displaystyle x$, then show that \[\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \frac{d w}{d x} \] in two ways - first by repeated application of product rule, second by logarithmic differentiation.
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Let $\displaystyle u,v,w$ be differentiable functions of $\displaystyle x$ and $\displaystyle y=u\cdot v\cdot w$.First method: repeated use of the product rule. Group the three factors as $\displaystyle y=(uv)\cdot w$ and apply the product rule $\displaystyle (fg)'=f'g+fg'$ once, with $\displaystyle f=uv$ and $\displaystyle g=w$: \[\frac{dy}{dx}=\frac{d(uv)}{dx}\cdot w+(uv)\cdot\frac{dw}{dx}.\] Now apply the product rule again to $\displaystyle \dfrac{d(uv)}{dx}=\dfrac{du}{dx}v+u\dfrac{dv}{dx}$: \[\frac{dy}{dx}=\left(\frac{du}{dx}v+u\frac{dv}{dx}\right)w+uv\frac{dw}{dx}=\frac{du}{dx}\,v\,w+u\,\frac{dv}{dx}\,w+u\,v\,\frac{dw}{dx}.\]Second method: logarithmic differentiation. Assume in addition that $\displaystyle u,v,w$ are non-zero at the point considered, so that $\displaystyle y\neq 0$ and the logarithms exist. Then \[\log|y|=\log|u|+\log|v|+\log|w|.\] Differentiating both sides with respect to $\displaystyle x$, using $\displaystyle \dfrac{d}{dx}\log|f|=\dfrac{f'}{f}$, \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.\] Multiplying through by $\displaystyle y=uvw$, each fraction cancels its own factor: \[\frac{dy}{dx}=uvw\left(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right)=\frac{du}{dx}\,v\,w+u\,\frac{dv}{dx}\,w+u\,v\,\frac{dw}{dx}.\]Both routes give \[\frac{d}{dx}(u\cdot v\cdot w)=\frac{du}{dx}\,v\cdot w+u\cdot\frac{dv}{dx}\cdot w+u\cdot v\,\frac{dw}{dx},\] as required. Note the first proof needs no extra hypothesis, whereas the second is valid only where $\displaystyle u,v,w$ are all non-zero; since both sides of the identity are continuous, the restriction is immaterial.

EXERCISE 5.6 1–11 (part 11 of 15)

If $\displaystyle x$ and $\displaystyle y$ are connected parametrically by the equations given in Exercises $\displaystyle 1$ to $\displaystyle 10$, without eliminating the parameter, Find $\displaystyle \frac{d y}{d x}$.
Exercise 1
$\displaystyle x=2 a t^{2}, y=a t^{4}$
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NCERT’s answer
$\displaystyle t^{2}$
Parametric differentiation: when $\displaystyle x$ and $\displaystyle y$ are both functions of a parameter $\displaystyle t$, the chain rule gives \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \text{provided } \frac{dx}{dt}\neq 0 . \]Differentiate each equation with respect to $\displaystyle t$ by the power rule: \[x=2at^{2}\ \Rightarrow\ \frac{dx}{dt}=4at,\qquad y=at^{4}\ \Rightarrow\ \frac{dy}{dt}=4at^{3}. \]Hence \[\frac{dy}{dx}=\frac{4at^{3}}{4at}=t^{2}. \]$\displaystyle \dfrac{dy}{dx}=t^{2}$, valid for $\displaystyle a\neq0$ and $\displaystyle t\neq0$; at $\displaystyle t=0$ we have $\displaystyle dx/dt=0$ and the formula does not apply.
Exercise 2
$\displaystyle x=a \cos \theta, y=b \cos \theta$
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NCERT’s answer
$\displaystyle \frac{b}{a}$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$, provided $\displaystyle \dfrac{dx}{d\theta}\neq0$.Differentiate each equation with respect to $\displaystyle \theta$: \[x=a\cos\theta\ \Rightarrow\ \frac{dx}{d\theta}=-a\sin\theta,\qquad y=b\cos\theta\ \Rightarrow\ \frac{dy}{d\theta}=-b\sin\theta . \]Therefore \[\frac{dy}{dx}=\frac{-b\sin\theta}{-a\sin\theta}=\frac{b}{a}. \]The cancellation of $\displaystyle \sin\theta$ is legitimate only when $\displaystyle \sin\theta\neq0$; at $\displaystyle \theta=n\pi$, $\displaystyle dx/d\theta=0$.$\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}$, a constant — which is what one should expect, since the point $\displaystyle (a\cos\theta,\;b\cos\theta)$ always satisfies $\displaystyle ay=bx$ and so moves along a straight line.
Exercise 3
$\displaystyle x=\sin t, y=\cos 2 t$
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NCERT’s answer
$\displaystyle -4 \sin t$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, provided $\displaystyle \dfrac{dx}{dt}\neq0$.\[x=\sin t\ \Rightarrow\ \frac{dx}{dt}=\cos t . \]For $\displaystyle y=\cos 2t$ the chain rule (inner function $\displaystyle 2t$) gives \[\frac{dy}{dt}=-2\sin 2t . \] Now use $\displaystyle \sin 2t=2\sin t\cos t$, so that the $\displaystyle \cos t$ coming from $\displaystyle dx/dt$ can be cancelled: \[\frac{dy}{dt}=-4\sin t\cos t . \]Therefore \[\frac{dy}{dx}=\frac{-4\sin t\cos t}{\cos t}=-4\sin t . \]$\displaystyle \dfrac{dy}{dx}=-4\sin t\;(=-4x)$, for $\displaystyle \cos t\neq0$, i.e. $\displaystyle t\neq(2n+1)\dfrac{\pi}{2}$.
Exercise 4
$\displaystyle x=4 t, y=\frac{4}{t}$
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NCERT’s answer
$\displaystyle -\frac{1}{t^{2}}$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.Write $\displaystyle y=\dfrac{4}{t}=4t^{-1}$ before differentiating, then apply the power rule: \[\frac{dx}{dt}=4,\qquad \frac{dy}{dt}=4\cdot(-1)t^{-2}=-\frac{4}{t^{2}} . \]Hence \[\frac{dy}{dx}=\frac{-4/t^{2}}{4}=-\frac{1}{t^{2}} . \]Check: here $\displaystyle xy=4t\cdot\dfrac4t=16$, so $\displaystyle y=\dfrac{16}{x}$ and $\displaystyle \dfrac{dy}{dx}=-\dfrac{16}{x^{2}}=-\dfrac{16}{16t^{2}}=-\dfrac{1}{t^{2}}$, as found.$\displaystyle \dfrac{dy}{dx}=-\dfrac{1}{t^{2}}$, $\displaystyle t\neq0$.
Exercise 5
$\displaystyle x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta$
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NCERT’s answer
$\displaystyle \frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$. Differentiate each coordinate term by term, using the chain rule on the $\displaystyle 2\theta$ terms (the inner factor $\displaystyle 2$ is the step most often dropped):\[x=\cos\theta-\cos 2\theta\ \Rightarrow\ \frac{dx}{d\theta}=-\sin\theta+2\sin 2\theta , \] \[y=\sin\theta-\sin 2\theta\ \Rightarrow\ \frac{dy}{d\theta}=\cos\theta-2\cos 2\theta . \]Therefore \[\frac{dy}{dx}=\frac{\cos\theta-2\cos 2\theta}{-\sin\theta+2\sin 2\theta}=\frac{\cos\theta-2\cos 2\theta}{2\sin 2\theta-\sin\theta}. \]$\displaystyle \dfrac{dy}{dx}=\dfrac{\cos\theta-2\cos 2\theta}{2\sin 2\theta-\sin\theta}$, valid wherever $\displaystyle 2\sin 2\theta\neq\sin\theta$.
Exercise 6
$\displaystyle x=a(\theta-\sin \theta), y=a(1+\cos \theta)$
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NCERT’s answer
$\displaystyle -\cot \frac{\theta}{2}$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$. (This is the cycloid.)\[x=a(\theta-\sin\theta)\ \Rightarrow\ \frac{dx}{d\theta}=a(1-\cos\theta),\qquad y=a(1+\cos\theta)\ \Rightarrow\ \frac{dy}{d\theta}=-a\sin\theta . \]So \[\frac{dy}{dx}=\frac{-a\sin\theta}{a(1-\cos\theta)}=\frac{-\sin\theta}{1-\cos\theta}. \]Simplify with the half-angle identities $\displaystyle \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $\displaystyle 1-\cos\theta=2\sin^{2}\frac{\theta}{2}$: \[\frac{dy}{dx}=-\frac{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}{2\sin^{2}\dfrac{\theta}{2}}=-\frac{\cos\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}}=-\cot\frac{\theta}{2}. \]$\displaystyle \dfrac{dy}{dx}=-\cot\dfrac{\theta}{2}$, valid for $\displaystyle \theta\neq 2n\pi$ (there $\displaystyle 1-\cos\theta=0$, the cusps of the cycloid, where the tangent is vertical).
Exercise 7
$\displaystyle x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$
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NCERT’s answer
$\displaystyle -\cot 3 t$
Write the two coordinates as products, $\displaystyle x=\sin^{3}t\,(\cos 2t)^{-1/2}$ and $\displaystyle y=\cos^{3}t\,(\cos 2t)^{-1/2}$, and differentiate by the product and chain rules. First the common factor: \[\frac{d}{dt}(\cos 2t)^{-1/2}=-\tfrac12(\cos 2t)^{-3/2}\cdot(-2\sin 2t)=\frac{\sin 2t}{(\cos 2t)^{3/2}} . \]Then \[\frac{dx}{dt}=\frac{3\sin^{2}t\cos t}{(\cos 2t)^{1/2}}+\sin^{3}t\cdot\frac{\sin 2t}{(\cos 2t)^{3/2}} =\frac{3\sin^{2}t\cos t\cos 2t+\sin^{3}t\sin 2t}{(\cos 2t)^{3/2}} . \] Put $\displaystyle \sin 2t=2\sin t\cos t$ and take out $\displaystyle \sin^{2}t\cos t$: \[\frac{dx}{dt}=\frac{\sin^{2}t\cos t\left(3\cos 2t+2\sin^{2}t\right)}{(\cos 2t)^{3/2}} =\frac{\sin^{2}t\cos t\left(1+2\cos 2t\right)}{(\cos 2t)^{3/2}}, \] since $\displaystyle 2\sin^{2}t=1-\cos 2t$.In the same way \[\frac{dy}{dt}=\frac{-3\cos^{2}t\sin t}{(\cos 2t)^{1/2}}+\cos^{3}t\cdot\frac{\sin 2t}{(\cos 2t)^{3/2}} =\frac{\cos^{2}t\sin t\left(-3\cos 2t+2\cos^{2}t\right)}{(\cos 2t)^{3/2}} =\frac{\cos^{2}t\sin t\left(1-2\cos 2t\right)}{(\cos 2t)^{3/2}}, \] since $\displaystyle 2\cos^{2}t=1+\cos 2t$.Dividing, the awkward power $\displaystyle (\cos 2t)^{3/2}$ cancels: \[\frac{dy}{dx}=\frac{\cos^{2}t\sin t\,(1-2\cos 2t)}{\sin^{2}t\cos t\,(1+2\cos 2t)} =\frac{\cos t\,(1-2\cos 2t)}{\sin t\,(1+2\cos 2t)} . \]Now use the triple-angle forms written with $\displaystyle \cos 2t$: \[\sin 3t=\sin t\,(3-4\sin^{2}t)=\sin t\,(1+2\cos 2t),\qquad \cos 3t=\cos t\,(4\cos^{2}t-3)=\cos t\,(2\cos 2t-1). \] Hence the numerator is $\displaystyle -\cos 3t$ and the denominator is $\displaystyle \sin 3t$: \[\frac{dy}{dx}=-\frac{\cos 3t}{\sin 3t}=-\cot 3t . \]$\displaystyle \dfrac{dy}{dx}=-\cot 3t$. The parametrisation needs $\displaystyle \cos 2t>0$, which forces $\displaystyle \cos^{2}t>\tfrac12$ so $\displaystyle \cos t\neq0$ automatically; the one extra requirement is $\displaystyle \sin t\neq0$, where $\displaystyle dx/dt=0$.
Exercise 8
$\displaystyle x=a\left(\cos t+\log \tan \frac{t}{2}\right) y=a \sin t$
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NCERT’s answer
$\displaystyle \tan t$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.The easy one first: \[y=a\sin t\ \Rightarrow\ \frac{dy}{dt}=a\cos t . \]For $\displaystyle x$ the term to handle carefully is $\displaystyle \log\tan\frac{t}{2}$; by the chain rule (derivative of $\displaystyle \log u$ is $\displaystyle u'/u$, and the inner $\displaystyle \frac{t}{2}$ contributes $\displaystyle \frac12$): \[\frac{d}{dt}\log\tan\frac{t}{2}=\frac{1}{\tan\dfrac{t}{2}}\cdot\sec^{2}\frac{t}{2}\cdot\frac12 =\frac{\cos\dfrac{t}{2}}{\sin\dfrac{t}{2}}\cdot\frac{1}{2\cos^{2}\dfrac{t}{2}} =\frac{1}{2\sin\dfrac{t}{2}\cos\dfrac{t}{2}}=\frac{1}{\sin t}=\mathrm{cosec}\,t . \]Therefore \[\frac{dx}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)=a\cdot\frac{1-\sin^{2}t}{\sin t}=\frac{a\cos^{2}t}{\sin t}. \]Dividing, \[\frac{dy}{dx}=\frac{a\cos t}{\dfrac{a\cos^{2}t}{\sin t}}=\frac{a\cos t\sin t}{a\cos^{2}t}=\frac{\sin t}{\cos t}=\tan t . \]$\displaystyle \dfrac{dy}{dx}=\tan t$, valid where $\displaystyle \tan\dfrac{t}{2}>0$ (so the logarithm exists) and $\displaystyle \cos t\neq0$ (so $\displaystyle dx/dt\neq0$).
Exercise 9
$\displaystyle x=a \sec \theta, y=b \tan \theta$
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NCERT’s answer
$\displaystyle \frac{b}{a} \operatorname{cosec} \theta$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$. Use the standard derivatives $\displaystyle \dfrac{d}{d\theta}\sec\theta=\sec\theta\tan\theta$ and $\displaystyle \dfrac{d}{d\theta}\tan\theta=\sec^{2}\theta$: \[x=a\sec\theta\ \Rightarrow\ \frac{dx}{d\theta}=a\sec\theta\tan\theta,\qquad y=b\tan\theta\ \Rightarrow\ \frac{dy}{d\theta}=b\sec^{2}\theta . \]Hence \[\frac{dy}{dx}=\frac{b\sec^{2}\theta}{a\sec\theta\tan\theta}=\frac{b\sec\theta}{a\tan\theta} =\frac{b}{a}\cdot\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta} =\frac{b}{a}\cdot\frac{1}{\sin\theta}. \]$\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}\,\mathrm{cosec}\,\theta$, valid for $\displaystyle \cos\theta\neq0$ (needed for $\displaystyle \sec\theta,\tan\theta$) and $\displaystyle \sin\theta\neq0$ (where $\displaystyle dx/d\theta=0$). The curve is the hyperbola $\displaystyle \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$.
Exercise 10
$\displaystyle x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$
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NCERT’s answer
$\displaystyle \tan \theta$
Parametric differentiation, with the product rule on $\displaystyle \theta\sin\theta$ and $\displaystyle \theta\cos\theta$.\[\frac{dx}{d\theta}=a\Big(-\sin\theta+\big(1\cdot\sin\theta+\theta\cos\theta\big)\Big)=a\big(-\sin\theta+\sin\theta+\theta\cos\theta\big)=a\theta\cos\theta , \] \[\frac{dy}{d\theta}=a\Big(\cos\theta-\big(1\cdot\cos\theta-\theta\sin\theta\big)\Big)=a\big(\cos\theta-\cos\theta+\theta\sin\theta\big)=a\theta\sin\theta . \] The whole point of the question is this cancellation: the product rule supplies exactly the term that kills the first one, and getting the sign in $\displaystyle \dfrac{d}{d\theta}(\theta\cos\theta)=\cos\theta-\theta\sin\theta$ right is what makes it happen.Therefore \[\frac{dy}{dx}=\frac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta . \]$\displaystyle \dfrac{dy}{dx}=\tan\theta$, valid for $\displaystyle \theta\neq0$ and $\displaystyle \cos\theta\neq0$ (this is the involute of a circle).
Exercise 11
If $\displaystyle x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}$, show that $\displaystyle \frac{d y}{d x}=-\frac{y}{x}$
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Write both coordinates as single powers of $\displaystyle a$ (take $\displaystyle a>0$, so the powers are defined): \[x=\sqrt{a^{\sin^{-1}t}}=a^{\frac12\sin^{-1}t},\qquad y=\sqrt{a^{\cos^{-1}t}}=a^{\frac12\cos^{-1}t}. \]Use logarithmic differentiation, since the variable sits in the exponent. Taking natural logarithms, \[\log x=\tfrac12\big(\sin^{-1}t\big)\log a,\qquad \log y=\tfrac12\big(\cos^{-1}t\big)\log a . \]Differentiate each with respect to $\displaystyle t$, using $\displaystyle \dfrac{d}{dt}\sin^{-1}t=\dfrac{1}{\sqrt{1-t^{2}}}$ and $\displaystyle \dfrac{d}{dt}\cos^{-1}t=-\dfrac{1}{\sqrt{1-t^{2}}}$ — that minus sign is the entire content of the result: \[\frac{1}{x}\frac{dx}{dt}=\frac{\log a}{2\sqrt{1-t^{2}}}\ \Rightarrow\ \frac{dx}{dt}=\frac{x\log a}{2\sqrt{1-t^{2}}}, \] \[\frac{1}{y}\frac{dy}{dt}=-\frac{\log a}{2\sqrt{1-t^{2}}}\ \Rightarrow\ \frac{dy}{dt}=-\frac{y\log a}{2\sqrt{1-t^{2}}} . \]Hence, by parametric differentiation, \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt} =\frac{-\dfrac{y\log a}{2\sqrt{1-t^{2}}}}{\dfrac{x\log a}{2\sqrt{1-t^{2}}}} =-\frac{y}{x}, \] which is what was to be shown.Check, independently: $\displaystyle xy=a^{\frac12(\sin^{-1}t+\cos^{-1}t)}=a^{\frac{\pi}{4}}$, a constant, because $\displaystyle \sin^{-1}t+\cos^{-1}t=\dfrac{\pi}{2}$. Differentiating $\displaystyle xy=$ constant implicitly gives $\displaystyle x\dfrac{dy}{dx}+y=0$, i.e. $\displaystyle \dfrac{dy}{dx}=-\dfrac{y}{x}$.Valid for $\displaystyle t\in(-1,1)$ and $\displaystyle a>0,\ a\neq1$ (if $\displaystyle a=1$ then $\displaystyle x=y=1$ and nothing varies).

EXERCISE 5.7 1–10 (part 12 of 15)

Find the second order derivatives of the functions given in Exercises $\displaystyle 1$ to $\displaystyle 10$ .
Exercise 1
$\displaystyle x^{2}+3 x+2$
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NCERT’s answer
$\displaystyle 2$
Let $\displaystyle y=x^{2}+3x+2$. The second order derivative is obtained by differentiating $\displaystyle \frac{dy}{dx}$ once more.By the power rule $\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$, applied term by term, \[\frac{dy}{dx}=2x+3 .\] Differentiating this, \[\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(2x+3)=2 .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=2\ }\] for every real $\displaystyle x$ (the second derivative of a quadratic is the constant $\displaystyle 2a$).
Exercise 2
$\displaystyle x^{20}$
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NCERT’s answer
$\displaystyle 380 x^{18}$
Let $\displaystyle y=x^{20}$. Apply the power rule $\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ twice.First derivative: \[\frac{dy}{dx}=20x^{19} .\] Differentiating $\displaystyle 20x^{19}$ by the same rule (the constant $\displaystyle 20$ rides along): \[\frac{d^{2}y}{dx^{2}}=20\cdot 19\,x^{18}=380x^{18} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=380x^{18}\ }\]
Exercise 3
$\displaystyle x \cdot \cos x$
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NCERT’s answer
$\displaystyle -x \cos x-2 \sin x$
Let $\displaystyle y=x\cos x$. Use the product rule $\displaystyle (uv)'=u'v+uv'$.With $\displaystyle u=x,\ v=\cos x$: \[\frac{dy}{dx}=1\cdot\cos x+x(-\sin x)=\cos x-x\sin x .\] Now differentiate again. The term $\displaystyle x\sin x$ needs the product rule a second time — this is the step usually dropped: \[\frac{d^{2}y}{dx^{2}}=-\sin x-\big(1\cdot\sin x+x\cos x\big)=-\sin x-\sin x-x\cos x .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-(x\cos x+2\sin x)\ }\]
Exercise 4
$\displaystyle \log x$
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NCERT’s answer
$\displaystyle -\frac{1}{x^{2}}$
Let $\displaystyle y=\log x$, which is defined only for $\displaystyle x>0$.Standard derivative: \[\frac{dy}{dx}=\frac{1}{x}=x^{-1} .\] Differentiate again by the power rule with $\displaystyle n=-1$: \[\frac{d^{2}y}{dx^{2}}=(-1)x^{-2}=-\frac{1}{x^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{1}{x^{2}},\qquad x>0\ }\]
Exercise 5
$\displaystyle x^{3} \log x$
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NCERT’s answer
$\displaystyle x(5+6 \log x)$
Let $\displaystyle y=x^{3}\log x$, defined for $\displaystyle x>0$. Use the product rule $\displaystyle (uv)'=u'v+uv'$.With $\displaystyle u=x^{3},\ v=\log x$: \[\frac{dy}{dx}=3x^{2}\log x+x^{3}\cdot\frac{1}{x}=3x^{2}\log x+x^{2} .\] Differentiate again; the first term needs the product rule once more: \[\frac{d^{2}y}{dx^{2}}=\Big(6x\log x+3x^{2}\cdot\frac{1}{x}\Big)+2x=6x\log x+3x+2x .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=6x\log x+5x=x\,(6\log x+5),\qquad x>0\ }\]
Exercise 6
$\displaystyle e^{x} \sin 5 x$
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NCERT’s answer
$\displaystyle 2 e^{x}(5 \cos 5 x-12 \sin 5 x)$
Let $\displaystyle y=e^{x}\sin 5x$. Use the product rule, and the chain rule on $\displaystyle \sin 5x$ — the inner factor $\displaystyle 5$ is what students most often lose.\[\frac{dy}{dx}=e^{x}\sin 5x+e^{x}\cdot 5\cos 5x=e^{x}\big(\sin 5x+5\cos 5x\big).\] Differentiate again by the product rule, differentiating the bracket with the chain rule: \[\frac{d^{2}y}{dx^{2}}=e^{x}\big(\sin 5x+5\cos 5x\big)+e^{x}\big(5\cos 5x-25\sin 5x\big).\] Collecting like terms, \[\frac{d^{2}y}{dx^{2}}=e^{x}\big(10\cos 5x-24\sin 5x\big).\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=e^{x}(10\cos 5x-24\sin 5x)=2e^{x}(5\cos 5x-12\sin 5x)\ }\]
Exercise 7
$\displaystyle e^{6 x} \cos 3 x$
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NCERT’s answer
$\displaystyle 9 e^{6 x}(3 \cos 3 x-4 \sin 3 x)$
Let $\displaystyle y=e^{6x}\cos 3x$. Product rule, with the chain rule supplying the inner factors $\displaystyle 6$ and $\displaystyle 3$.\[\frac{dy}{dx}=6e^{6x}\cos 3x+e^{6x}(-3\sin 3x)=3e^{6x}\big(2\cos 3x-\sin 3x\big).\] Differentiate again, keeping the factor $\displaystyle 3$ outside: \[\frac{d^{2}y}{dx^{2}}=3\Big[6e^{6x}\big(2\cos 3x-\sin 3x\big)+e^{6x}\big(-6\sin 3x-3\cos 3x\big)\Big],\] since $\displaystyle \frac{d}{dx}(2\cos 3x-\sin 3x)=-6\sin 3x-3\cos 3x$. Hence \[\frac{d^{2}y}{dx^{2}}=3e^{6x}\big(12\cos 3x-6\sin 3x-6\sin 3x-3\cos 3x\big)=3e^{6x}\big(9\cos 3x-12\sin 3x\big).\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=9e^{6x}(3\cos 3x-4\sin 3x)\ }\]
Exercise 8
$\displaystyle \tan ^{-1} x$
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NCERT’s answer
$\displaystyle -\frac{2 x}{\left(1+x^{2}\right)^{2}}$
Let $\displaystyle y=\tan^{-1}x$. Use the standard derivative $\displaystyle \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}$.\[\frac{dy}{dx}=\frac{1}{1+x^{2}}=(1+x^{2})^{-1} .\] Differentiate by the chain rule (outer power $\displaystyle -1$, inner derivative $\displaystyle 2x$): \[\frac{d^{2}y}{dx^{2}}=(-1)(1+x^{2})^{-2}\cdot 2x=\frac{-2x}{(1+x^{2})^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=\frac{-2x}{(1+x^{2})^{2}}\ }\] valid for all real $\displaystyle x$, since $\displaystyle 1+x^{2}$ never vanishes.
Exercise 9
$\displaystyle \log (\log x)$
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NCERT’s answer
$\displaystyle -\frac{(1+\log x)}{(x \log x)^{2}}$
Let $\displaystyle y=\log(\log x)$. For this to be defined we need $\displaystyle \log x>0$, i.e. $\displaystyle x>1$; the whole answer carries that restriction.Chain rule (outer $\displaystyle \log$, inner $\displaystyle \log x$): \[\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x\log x}=(x\log x)^{-1} .\] Differentiate again by the chain rule, and note that the inner function needs the product rule: \[\frac{d}{dx}(x\log x)=1\cdot\log x+x\cdot\frac{1}{x}=\log x+1 .\] Therefore \[\frac{d^{2}y}{dx^{2}}=-(x\log x)^{-2}\,(1+\log x)=-\frac{1+\log x}{(x\log x)^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{(1+\log x)}{(x\log x)^{2}},\qquad x>1\ }\]
Exercise 10
$\displaystyle \sin (\log x)$
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NCERT’s answer
$\displaystyle -\frac{\sin (\log x)+\cos (\log x)}{x^{2}}$
Let $\displaystyle y=\sin(\log x)$, defined for $\displaystyle x>0$.Chain rule: \[\frac{dy}{dx}=\cos(\log x)\cdot\frac{1}{x}=\frac{\cos(\log x)}{x} .\] Now the quotient rule $\displaystyle \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ with $\displaystyle u=\cos(\log x),\ v=x$; the numerator's derivative again needs the chain rule, $\displaystyle \frac{d}{dx}\cos(\log x)=-\frac{\sin(\log x)}{x}$: \[\frac{d^{2}y}{dx^{2}}=\frac{x\left(-\dfrac{\sin(\log x)}{x}\right)-\cos(\log x)}{x^{2}}=\frac{-\sin(\log x)-\cos(\log x)}{x^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{\sin(\log x)+\cos(\log x)}{x^{2}},\qquad x>0\ }\]

EXERCISE 5.7 11–17 (part 13 of 15)

Exercise 11
If $\displaystyle y=5 \cos x-3 \sin x$, prove that $\displaystyle \frac{d^{2} y}{d x^{2}}+y=0$
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Given $\displaystyle y=5\cos x-3\sin x$. Differentiate twice using $\displaystyle \frac{d}{dx}\cos x=-\sin x$ and $\displaystyle \frac{d}{dx}\sin x=\cos x$.First derivative: \[\frac{dy}{dx}=5(-\sin x)-3(\cos x)=-5\sin x-3\cos x .\] Second derivative: \[\frac{d^{2}y}{dx^{2}}=-5\cos x-3(-\sin x)=-5\cos x+3\sin x .\] The key step is to recognise the bracket that reappears: \[\frac{d^{2}y}{dx^{2}}=-\big(5\cos x-3\sin x\big)=-y .\] Hence \[\frac{d^{2}y}{dx^{2}}+y=-y+y=0 .\]So $\displaystyle \frac{d^{2}y}{dx^{2}}+y=0$, as required. $\displaystyle \blacksquare$
Exercise 12
If $\displaystyle y=\cos ^{-1} x$, Find $\displaystyle \frac{d^{2} y}{d x^{2}}$ in terms of $\displaystyle y$ alone.
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NCERT’s answer
$\displaystyle -\cot y \operatorname{cosec}^{2} y$
Given $\displaystyle y=\cos^{-1}x$. Its domain is $\displaystyle -1<x<1$ for differentiability, and then the principal value gives $\displaystyle y\in(0,\pi)$ — remember this, because the sign of $\displaystyle \sin y$ depends on it.Standard derivative: \[\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}}=-(1-x^{2})^{-1/2} .\] Differentiate by the chain rule: \[\frac{d^{2}y}{dx^{2}}=-\left[\left(-\frac{1}{2}\right)(1-x^{2})^{-3/2}\cdot(-2x)\right]=-\frac{x}{(1-x^{2})^{3/2}} .\]Now convert to $\displaystyle y$. From $\displaystyle y=\cos^{-1}x$ we have $\displaystyle x=\cos y$ and $\displaystyle 1-x^{2}=1-\cos^{2}y=\sin^{2}y$. Since $\displaystyle y\in(0,\pi)$, $\displaystyle \sin y>0$, so \[(1-x^{2})^{3/2}=\big(\sin^{2}y\big)^{3/2}=\sin^{3}y \quad(\text{not }-\sin^{3}y).\] Therefore \[\frac{d^{2}y}{dx^{2}}=-\frac{\cos y}{\sin^{3}y}=-\frac{\cos y}{\sin y}\cdot\frac{1}{\sin^{2}y} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\cot y\,\mathrm{cosec}^{2}y,\qquad y\in(0,\pi)\ }\]
Exercise 13
If $\displaystyle y=3 \cos (\log x)+4 \sin (\log x)$, show that $\displaystyle x^{2} y_{2}+x y_{1}+y=0$
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Given $\displaystyle y=3\cos(\log x)+4\sin(\log x)$, $\displaystyle x>0$. Write $\displaystyle y_{1}=\frac{dy}{dx}$, $\displaystyle y_{2}=\frac{d^{2}y}{dx^{2}}$.Differentiate by the chain rule (inner derivative $\displaystyle \frac{1}{x}$): \[y_{1}=-3\sin(\log x)\cdot\frac{1}{x}+4\cos(\log x)\cdot\frac{1}{x}=\frac{-3\sin(\log x)+4\cos(\log x)}{x} .\] The efficient move is to clear the denominator before differentiating again: \[x\,y_{1}=-3\sin(\log x)+4\cos(\log x).\] Differentiate both sides with respect to $\displaystyle x$, using the product rule on the left: \[y_{1}+x\,y_{2}=-3\cos(\log x)\cdot\frac{1}{x}-4\sin(\log x)\cdot\frac{1}{x}=\frac{-\big(3\cos(\log x)+4\sin(\log x)\big)}{x}.\] Multiply through by $\displaystyle x$: \[x\,y_{1}+x^{2}y_{2}=-\big(3\cos(\log x)+4\sin(\log x)\big)=-y ,\] the last equality being the original definition of $\displaystyle y$. Hence \[x^{2}y_{2}+x\,y_{1}+y=0 . \qquad\blacksquare\]
Exercise 14
If $\displaystyle y=\mathrm{A} e^{m x}+\mathrm{B} e^{n x}$, show that $\displaystyle \frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$
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Given $\displaystyle y=\mathrm{A}e^{mx}+\mathrm{B}e^{nx}$, with $\displaystyle \mathrm{A},\mathrm{B},m,n$ constants. Use $\displaystyle \frac{d}{dx}e^{kx}=k\,e^{kx}$.\[\frac{dy}{dx}=\mathrm{A}m\,e^{mx}+\mathrm{B}n\,e^{nx},\qquad \frac{d^{2}y}{dx^{2}}=\mathrm{A}m^{2}e^{mx}+\mathrm{B}n^{2}e^{nx}.\] Substitute into the left side of the required identity and group the $\displaystyle e^{mx}$ and $\displaystyle e^{nx}$ terms separately: \[\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx}+mny =\mathrm{A}e^{mx}\big[m^{2}-(m+n)m+mn\big]+\mathrm{B}e^{nx}\big[n^{2}-(m+n)n+mn\big].\] Each bracket vanishes: \[m^{2}-m^{2}-mn+mn=0,\qquad n^{2}-mn-n^{2}+mn=0 .\] Therefore \[\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx}+mny=0 . \qquad\blacksquare\]
Exercise 15
If $\displaystyle y=500 e^{7 x}+600 e^{-7 x}$, show that $\displaystyle \frac{d^{2} y}{d x^{2}}=49 y$
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Given $\displaystyle y=500e^{7x}+600e^{-7x}$. Use $\displaystyle \frac{d}{dx}e^{kx}=k\,e^{kx}$, taking care with the sign of $\displaystyle k=-7$.First derivative: \[\frac{dy}{dx}=500\cdot 7e^{7x}+600\cdot(-7)e^{-7x}=3500e^{7x}-4200e^{-7x}.\] Second derivative — the second $\displaystyle -7$ makes the term positive again: \[\frac{d^{2}y}{dx^{2}}=3500\cdot 7e^{7x}-4200\cdot(-7)e^{-7x}=24500e^{7x}+29400e^{-7x}.\] Factor out $\displaystyle 49$: \[\frac{d^{2}y}{dx^{2}}=49\big(500e^{7x}+600e^{-7x}\big)=49y .\]Hence $\displaystyle \frac{d^{2}y}{dx^{2}}=49y$. $\displaystyle \blacksquare$
Exercise 16
If $\displaystyle e^{y}(x+1)=1$, show that $\displaystyle \frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$
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Given $\displaystyle e^{y}(x+1)=1$. Since $\displaystyle e^{y}>0$ always, $\displaystyle x+1=e^{-y}>0$, so the relation lives on $\displaystyle x>-1$.Differentiate both sides with respect to $\displaystyle x$, using the product rule on the left and the chain rule on $\displaystyle e^{y}$ (implicit differentiation: $\displaystyle \frac{d}{dx}e^{y}=e^{y}\frac{dy}{dx}$): \[e^{y}\frac{dy}{dx}(x+1)+e^{y}\cdot 1=0 .\] Factor $\displaystyle e^{y}$, which is never zero, and divide it out: \[(x+1)\frac{dy}{dx}+1=0\ \Longrightarrow\ \frac{dy}{dx}=-\frac{1}{x+1}.\] Differentiate once more (power rule on $\displaystyle (x+1)^{-1}$): \[\frac{d^{2}y}{dx^{2}}=-\Big[-(x+1)^{-2}\Big]=\frac{1}{(x+1)^{2}} .\] But $\displaystyle \left(\frac{dy}{dx}\right)^{2}=\left(-\frac{1}{x+1}\right)^{2}=\frac{1}{(x+1)^{2}}$ as well.Therefore $\displaystyle \frac{d^{2}y}{dx^{2}}=\left(\frac{dy}{dx}\right)^{2}$. $\displaystyle \blacksquare$
Exercise 17
If $\displaystyle y=\left(\tan ^{-1} x\right)^{2}$, show that $\displaystyle \left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}=2$
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Given $\displaystyle y=\left(\tan^{-1}x\right)^{2}$. Write $\displaystyle y_{1}=\frac{dy}{dx}$, $\displaystyle y_{2}=\frac{d^{2}y}{dx^{2}}$.Chain rule (outer square, inner $\displaystyle \tan^{-1}x$ with derivative $\displaystyle \frac{1}{1+x^{2}}$): \[y_{1}=2\tan^{-1}x\cdot\frac{1}{1+x^{2}} .\] Rather than differentiate this quotient directly, clear the denominator first: \[(1+x^{2})\,y_{1}=2\tan^{-1}x .\] Now differentiate both sides with respect to $\displaystyle x$, product rule on the left: \[(1+x^{2})\,y_{2}+2x\,y_{1}=\frac{2}{1+x^{2}} .\] Finally multiply the whole equation by $\displaystyle (1+x^{2})$ to remove the remaining fraction: \[(1+x^{2})^{2}y_{2}+2x(1+x^{2})\,y_{1}=2 .\]That is $\displaystyle \left(x^{2}+1\right)^{2}y_{2}+2x\left(x^{2}+1\right)y_{1}=2$, as required. $\displaystyle \blacksquare$

Miscellaneous Exercise 1–10 (part 14 of 15)

Differentiate w.r.t. $\displaystyle x$ the function in Exercises $\displaystyle 1$ to 11.
Exercise 1
$\displaystyle \left(3 x^{2}-9 x+5\right)^{9}$
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NCERT’s answer
$\displaystyle 27\left(3 x^{2}-9 x+5\right)^{8}(2 x-3)$
Chain rule for a power of a function: if $\displaystyle y=u^{9}$ with $\displaystyle u=3x^{2}-9x+5$, then $\displaystyle \dfrac{dy}{dx}=9u^{8}\dfrac{du}{dx}$.Here $\displaystyle \dfrac{du}{dx}=6x-9$, so \[\frac{dy}{dx}=9\left(3x^{2}-9x+5\right)^{8}(6x-9)=27\left(3x^{2}-9x+5\right)^{8}(2x-3). \]
Differentiate w.r.t. x the function in Exercises $\displaystyle 1$ to 11.
Exercise 2
$\displaystyle \sin^{3} x+\cos^{6} x$
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NCERT’s answer
$\displaystyle 3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$
Differentiate term by term, chain rule on each power.\[\frac{d}{dx}\left(\sin^{3}x\right)=3\sin^{2}x\cdot\frac{d}{dx}(\sin x)=3\sin^{2}x\cos x \] \[\frac{d}{dx}\left(\cos^{6}x\right)=6\cos^{5}x\cdot\frac{d}{dx}(\cos x)=-6\cos^{5}x\sin x \]Adding the two, \[\frac{dy}{dx}=3\sin^{2}x\cos x-6\sin x\cos^{5}x=3\sin x\cos x\left(\sin x-2\cos^{4}x\right). \]
Differentiate w.r.t. $\displaystyle x$ the function in Exercises $\displaystyle 1$ to 11.
Exercise 3
$\displaystyle (5 x)^{3 \cos 2 x}$
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NCERT’s answer
$\displaystyle (5 x)^{3 \cos 2 x}\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
The base $\displaystyle 5x$ and the exponent $\displaystyle 3\cos 2x$ both contain $\displaystyle x$, so neither the power rule nor the exponential rule applies on its own $\displaystyle -$ take logarithms first (logarithmic differentiation), for $\displaystyle x>0$.Let $\displaystyle y=(5x)^{3\cos 2x}$. Then \[\log y=3\cos 2x\,\log (5x). \]Differentiate both sides w.r.t. $\displaystyle x$, product rule on the right and chain rule on $\displaystyle \cos 2x$: \[\frac{1}{y}\frac{dy}{dx}=3(-2\sin 2x)\log (5x)+3\cos 2x\cdot\frac{1}{5x}\cdot 5=-6\sin 2x\,\log (5x)+\frac{3\cos 2x}{x}. \]Hence \[\frac{dy}{dx}=(5x)^{3\cos 2x}\left[\frac{3\cos 2x}{x}-6\sin 2x\,\log (5x)\right]. \]
Differentiate w.r.t. x the function in Exercises $\displaystyle 1$ to 11.
Exercise 4
$\displaystyle \sin^{-1}(x \sqrt{x}), 0 \leq x \leq 1$
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NCERT’s answer
$\displaystyle \frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}$
Write $\displaystyle x\sqrt{x}=x^{3/2}$, so $\displaystyle y=\sin^{-1}\left(x^{3/2}\right)$; this is defined because $\displaystyle 0\le x^{3/2}\le 1$ on $\displaystyle 0\le x\le 1$.Chain rule with $\displaystyle \dfrac{d}{du}\sin^{-1}u=\dfrac{1}{\sqrt{1-u^{2}}}$, taking $\displaystyle u=x^{3/2}$, so that $\displaystyle u^{2}=x^{3}$ and $\displaystyle \dfrac{du}{dx}=\dfrac{3}{2}x^{1/2}$: \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^{3}}}\cdot\frac{3}{2}\sqrt{x}=\frac{3\sqrt{x}}{2\sqrt{1-x^{3}}}. \]This holds for $\displaystyle 0\le x<1$; at $\displaystyle x=1$ the denominator vanishes and the graph has a vertical tangent, so the function is not differentiable at the right endpoint.
Differentiate w.r.t. $\displaystyle x$ the function in Exercises $\displaystyle 1$ to 11.
Exercise 5
$\displaystyle \frac{\cos ^{-1} \dfrac{x}{2}}{\sqrt{2 x+7}},-2<x<2$
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NCERT’s answer
$\displaystyle -\left[\frac{1}{\sqrt{4-x^{2}} \sqrt{2 x+7}}+\frac{\cos ^{-1} \dfrac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]$
Quotient rule with $\displaystyle u=\cos^{-1}\dfrac{x}{2}$ and $\displaystyle v=\sqrt{2x+7}$.\[\frac{du}{dx}=-\frac{1}{\sqrt{1-\dfrac{x^{2}}{4}}}\cdot\frac{1}{2}=-\frac{1}{\sqrt{4-x^{2}}},\qquad \frac{dv}{dx}=\frac{2}{2\sqrt{2x+7}}=\frac{1}{\sqrt{2x+7}} \]The factor $\displaystyle \tfrac12$ in $\displaystyle \dfrac{du}{dx}$ is the derivative of the inner function $\displaystyle x/2$ $\displaystyle -$ the step most often dropped; note also that $\displaystyle \sqrt{4-\frac{4x^{2}}{4}}$ simplifies to $\displaystyle \sqrt{4-x^{2}}$ only because $\displaystyle 2\sqrt{1-\frac{x^{2}}{4}}=\sqrt{4-x^{2}}$.Then \[\frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}=\frac{-\dfrac{\sqrt{2x+7}}{\sqrt{4-x^{2}}}-\dfrac{\cos^{-1}\dfrac{x}{2}}{\sqrt{2x+7}}}{2x+7}. \]Hence, for $\displaystyle -2<x<2$, \[\frac{dy}{dx}=-\frac{1}{\sqrt{4-x^{2}}\,\sqrt{2x+7}}-\frac{\cos^{-1}\dfrac{x}{2}}{(2x+7)^{3/2}}. \]
Exercise 6
$\displaystyle \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$
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NCERT’s answer
$\displaystyle \frac{1}{2}$
Simplify the argument before differentiating, using the half-angle identities \[1+\sin x=\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2},\qquad 1-\sin x=\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}. \]For $\displaystyle 0<x<\frac{\pi}{2}$ we have $\displaystyle 0<\frac{x}{2}<\frac{\pi}{4}$, so $\displaystyle \cos\frac{x}{2}>\sin\frac{x}{2}>0$ and both square roots are taken positive: \[\sqrt{1+\sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\qquad \sqrt{1-\sin x}=\cos\frac{x}{2}-\sin\frac{x}{2}. \] This sign decision is the whole point of the given interval; writing $\displaystyle \sqrt{A^{2}}=A$ without checking the quadrant is the usual error.Therefore \[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2\cos\dfrac{x}{2}}{2\sin\dfrac{x}{2}}=\cot\frac{x}{2}, \] and since $\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{4}\right)$ lies inside the principal branch $\displaystyle (0,\pi)$ of $\displaystyle \cot^{-1}$, \[y=\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}. \]Hence \[\frac{dy}{dx}=\frac{1}{2}. \]
Exercise 7
$\displaystyle (\log x)^{\log x}, x>1$
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Base and exponent both involve $\displaystyle x$, so use logarithmic differentiation. For $\displaystyle x>1$ we have $\displaystyle \log x>0$, which is exactly what makes $\displaystyle \log(\log x)$ defined.Let $\displaystyle y=(\log x)^{\log x}$. Then \[\log y=\log x\cdot\log(\log x). \]Differentiate w.r.t. $\displaystyle x$, product rule on the right and chain rule on $\displaystyle \log(\log x)$: \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\log(\log x)+\log x\cdot\frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x}\left[\log(\log x)+1\right]. \]Hence \[\frac{dy}{dx}=\frac{(\log x)^{\log x}}{x}\left[1+\log(\log x)\right],\qquad x>1. \]
Exercise 8
$\displaystyle \cos (a \cos x+b \sin x)$, for some constant $\displaystyle a$ and $\displaystyle b$.
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NCERT’s answer
$\displaystyle (a \sin x-b \cos x) \sin (a \cos x+b \sin x)$
Chain rule with inner function $\displaystyle u=a\cos x+b\sin x$, where $\displaystyle a$ and $\displaystyle b$ are constants.\[\frac{du}{dx}=-a\sin x+b\cos x \]\[\frac{dy}{dx}=-\sin u\cdot\frac{du}{dx}=-\sin\left(a\cos x+b\sin x\right)\left(b\cos x-a\sin x\right) \]That is, \[\frac{dy}{dx}=\left(a\sin x-b\cos x\right)\sin\left(a\cos x+b\sin x\right). \]
Exercise 9
$\displaystyle (\sin x-\cos x)^{(\sin x-\cos x)}, \frac{\pi}{4}<x<\frac{3 \pi}{4}$
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NCERT’s answer
$\displaystyle (\sin x-\cos x)^{\sin x-\cos x}(\cos x+\sin x)(1+\log (\sin x-\cos x)), \sin x>\cos x$
Put $\displaystyle u=\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)$. On $\displaystyle \frac{\pi}{4}<x<\frac{3\pi}{4}$ we have $\displaystyle 0<x-\frac{\pi}{4}<\frac{\pi}{2}$, so $\displaystyle u>0$ and $\displaystyle \log u$ exists $\displaystyle -$ that is why the interval is given.With $\displaystyle y=u^{u}$, logarithmic differentiation gives $\displaystyle \log y=u\log u$, so \[\frac{1}{y}\frac{dy}{dx}=\left(\log u+u\cdot\frac{1}{u}\right)\frac{du}{dx}=(1+\log u)\frac{du}{dx}, \] and $\displaystyle \dfrac{du}{dx}=\cos x+\sin x$.Hence \[\frac{dy}{dx}=(\sin x-\cos x)^{(\sin x-\cos x)}\left(\cos x+\sin x\right)\left[1+\log\left(\sin x-\cos x\right)\right]. \]
Exercise 10
$\displaystyle x^{x}+x^{a}+a^{x}+a^{a}$, for some fixed $\displaystyle a>0$ and $\displaystyle x>0$
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NCERT’s answer
$\displaystyle x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$
Four terms of four different types, each needing its own rule.$\displaystyle x^{x}$: base and exponent both variable, so log-differentiate. With $\displaystyle u=x^{x}$, $\displaystyle \log u=x\log x$, hence $\displaystyle \dfrac{1}{u}\dfrac{du}{dx}=\log x+x\cdot\dfrac{1}{x}=1+\log x$ and \[\frac{d}{dx}\left(x^{x}\right)=x^{x}(1+\log x). \]$\displaystyle x^{a}$: variable base, constant exponent, so the power rule, $\displaystyle \dfrac{d}{dx}x^{a}=ax^{a-1}$.$\displaystyle a^{x}$: constant base, variable exponent, so the exponential rule, $\displaystyle \dfrac{d}{dx}a^{x}=a^{x}\log a$.$\displaystyle a^{a}$: a pure constant, derivative $\displaystyle 0$.Adding, \[\frac{dy}{dx}=x^{x}(1+\log x)+ax^{a-1}+a^{x}\log a. \]

Miscellaneous Exercise 11–22 (part 15 of 15)

Differentiate w.r.t. $\displaystyle x$ the function in Exercises $\displaystyle 1$ to 11.
Exercise 11
$\displaystyle x^{x^{2}-3}+(x-3)^{x^{2}}$, for $\displaystyle x>3$
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Split as $\displaystyle y=u+v$ with $\displaystyle u=x^{x^{2}-3}$ and $\displaystyle v=(x-3)^{x^{2}}$, and take logarithms of each piece separately $\displaystyle -$ a sum cannot be logged as a whole. For $\displaystyle x>3$ both $\displaystyle \log x$ and $\displaystyle \log(x-3)$ are defined.For $\displaystyle u$: $\displaystyle \log u=(x^{2}-3)\log x$, so by the product rule \[\frac{1}{u}\frac{du}{dx}=2x\log x+\left(x^{2}-3\right)\cdot\frac{1}{x},\qquad \frac{du}{dx}=x^{x^{2}-3}\left[2x\log x+\frac{x^{2}-3}{x}\right]. \]For $\displaystyle v$: $\displaystyle \log v=x^{2}\log(x-3)$, so \[\frac{1}{v}\frac{dv}{dx}=2x\log(x-3)+x^{2}\cdot\frac{1}{x-3},\qquad \frac{dv}{dx}=(x-3)^{x^{2}}\left[2x\log(x-3)+\frac{x^{2}}{x-3}\right]. \]Hence, for $\displaystyle x>3$, \[\frac{dy}{dx}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2x\log x\right]+(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2x\log(x-3)\right]. \]
Exercise 12
Find $\displaystyle \frac{d y}{d x}$, if $\displaystyle y=12(1-\cos t), x=10(t-\sin t),-\frac{\pi}{2}<t<\frac{\pi}{2}$
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NCERT’s answer
$\displaystyle \frac{6}{5} \cot \frac{t}{2}$
Parametric differentiation: $\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, valid wherever $\displaystyle \dfrac{dx}{dt}\neq 0$.\[\frac{dy}{dt}=12\sin t,\qquad \frac{dx}{dt}=10(1-\cos t) \]Using $\displaystyle 1-\cos t=2\sin^{2}\dfrac{t}{2}$ and $\displaystyle \sin t=2\sin\dfrac{t}{2}\cos\dfrac{t}{2}$, \[\frac{dy}{dx}=\frac{12\sin t}{10(1-\cos t)}=\frac{12\cdot 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}}{10\cdot 2\sin^{2}\dfrac{t}{2}}=\frac{6}{5}\cot\frac{t}{2}. \]So $\displaystyle \dfrac{dy}{dx}=\dfrac{6}{5}\cot\dfrac{t}{2}$ for $\displaystyle -\dfrac{\pi}{2}<t<\dfrac{\pi}{2}$, $\displaystyle t\neq 0$ (at $\displaystyle t=0$ both $\displaystyle \dfrac{dx}{dt}$ and $\displaystyle \dfrac{dy}{dt}$ vanish and the formula does not apply).
Exercise 13
Find $\displaystyle \frac{d y}{d x}$, if $\displaystyle y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}, 0<x<1$
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NCERT’s answer
$\displaystyle 0$
Simplify before differentiating $\displaystyle -$ the derivative is immediate once the expression is recognised.Substitute $\displaystyle x=\sin\theta$. Since $\displaystyle 0<x<1$, $\displaystyle \theta=\sin^{-1}x\in\left(0,\frac{\pi}{2}\right)$, so $\displaystyle \cos\theta>0$ and \[\sqrt{1-x^{2}}=\sqrt{\cos^{2}\theta}=\cos\theta. \]Also \[\sin^{-1}(\cos\theta)=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-\theta\right)\right)=\frac{\pi}{2}-\theta, \] because $\displaystyle \frac{\pi}{2}-\theta\in\left(0,\frac{\pi}{2}\right)$ lies in the principal branch $\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ of $\displaystyle \sin^{-1}$.Hence \[y=\theta+\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}, \] a constant on $\displaystyle 0<x<1$, and therefore \[\frac{dy}{dx}=0,\qquad 0<x<1. \]
Exercise 14
If $\displaystyle x \sqrt{1+y}+y \sqrt{1+x}=0$, for , $\displaystyle -1<x<1$, prove that \[\frac{d y}{d x}=-\frac{1}{(1+x)^{2}} \]
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First turn the implicit relation into an explicit one. From $\displaystyle x\sqrt{1+y}+y\sqrt{1+x}=0$, \[x\sqrt{1+y}=-y\sqrt{1+x}. \]Squaring both sides (legitimate, but it can create extra roots, which are discarded below): \[x^{2}(1+y)=y^{2}(1+x)\;\Longrightarrow\;x^{2}-y^{2}+x^{2}y-xy^{2}=0 \] \[(x-y)(x+y)+xy(x-y)=0\;\Longrightarrow\;(x-y)\left(x+y+xy\right)=0. \]The factor $\displaystyle x-y=0$ must be rejected: putting $\displaystyle y=x$ back into the original equation gives $\displaystyle 2x\sqrt{1+x}=0$, i.e. only the single point $\displaystyle x=0$, not a relation valid on $\displaystyle -1<x<1$. Discarding it is the step the proof turns on. So \[x+y+xy=0\;\Longrightarrow\;y(1+x)=-x\;\Longrightarrow\;y=-\frac{x}{1+x}\quad(1+x>0). \]Differentiating by the quotient rule, \[\frac{dy}{dx}=-\frac{(1+x)\cdot 1-x\cdot 1}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}, \] which is what was to be proved.
Exercise 15
If $\displaystyle (x-a)^{2}+(y-b)^{2}=c^{2}$, for some $\displaystyle c>0$, prove that \[\frac{\left[1+\left(\dfrac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\dfrac{d^{2} y}{d x^{2}}} \] is a constant independent of $\displaystyle a$ and $\displaystyle b$.
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Differentiate $\displaystyle (x-a)^{2}+(y-b)^{2}=c^{2}$ implicitly w.r.t. $\displaystyle x$: \[2(x-a)+2(y-b)\frac{dy}{dx}=0\;\Longrightarrow\;\frac{dy}{dx}=-\frac{x-a}{y-b}\quad(y\neq b). \]Differentiate $\displaystyle (x-a)+(y-b)\dfrac{dy}{dx}=0$ once more, product rule on the second term: \[1+\left(\frac{dy}{dx}\right)^{2}+(y-b)\frac{d^{2}y}{dx^{2}}=0\;\Longrightarrow\;\frac{d^{2}y}{dx^{2}}=-\frac{1+\left(\dfrac{dy}{dx}\right)^{2}}{y-b}. \]Now use the given equation to remove $\displaystyle a$ and $\displaystyle b$: \[1+\left(\frac{dy}{dx}\right)^{2}=1+\frac{(x-a)^{2}}{(y-b)^{2}}=\frac{(x-a)^{2}+(y-b)^{2}}{(y-b)^{2}}=\frac{c^{2}}{(y-b)^{2}}. \]Therefore \[\frac{d^{2}y}{dx^{2}}=-\frac{c^{2}}{(y-b)^{3}},\qquad \left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{\frac{3}{2}}=\frac{c^{3}}{\left|y-b\right|^{3}}, \] the modulus appearing because a $\displaystyle 3/2$ power is a positive square root.Dividing, \[\frac{\left[1+\left(\dfrac{dy}{dx}\right)^{2}\right]^{\frac{3}{2}}}{\dfrac{d^{2}y}{dx^{2}}}=\frac{c^{3}/\left|y-b\right|^{3}}{-c^{2}/(y-b)^{3}}=-c\cdot\frac{(y-b)^{3}}{\left|y-b\right|^{3}}, \] which equals $\displaystyle -c$ on the upper semicircle $\displaystyle y>b$ and $\displaystyle +c$ on the lower one.Either way the quotient is $\displaystyle \pm c$: it depends only on the radius $\displaystyle c$ and not on $\displaystyle x$, $\displaystyle y$, $\displaystyle a$ or $\displaystyle b$. Hence it is a constant independent of $\displaystyle a$ and $\displaystyle b$.
Exercise 16
If $\displaystyle \cos y=x \cos (a+y)$, with $\displaystyle \cos a \neq \pm 1$, prove that $\displaystyle \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$.
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Since $\displaystyle x$ stands alone in the relation, it is cleaner to differentiate $\displaystyle x$ with respect to $\displaystyle y$ and then invert.Because $\displaystyle \cos a\neq\pm 1$, $\displaystyle \sin a\neq 0$. From $\displaystyle \cos y=x\cos(a+y)$, \[x=\frac{\cos y}{\cos (a+y)}. \]Quotient rule w.r.t. $\displaystyle y$, remembering $\displaystyle \dfrac{d}{dy}\cos(a+y)=-\sin(a+y)$: \[\frac{dx}{dy}=\frac{-\sin y\cos (a+y)-\cos y\left(-\sin (a+y)\right)}{\cos^{2}(a+y)}=\frac{\sin (a+y)\cos y-\cos (a+y)\sin y}{\cos^{2}(a+y)}. \]The numerator is $\displaystyle \sin\big((a+y)-y\big)=\sin a$, so \[\frac{dx}{dy}=\frac{\sin a}{\cos^{2}(a+y)}. \]This is non-zero (that is exactly what $\displaystyle \cos a\neq\pm 1$ guarantees), so it may be inverted: \[\frac{dy}{dx}=\frac{\cos^{2}(a+y)}{\sin a}, \] as required.
Exercise 17
If $\displaystyle x=a(\cos t+t \sin t)$ and $\displaystyle y=a(\sin t-t \cos t)$, find $\displaystyle \frac{d^{2} y}{d x^{2}}$.
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NCERT’s answer
$\displaystyle \frac{\sec ^{3} t}{a t}, 0<t<\frac{\pi}{2}$
Parametric second derivative. The rule is \[\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}, \] and NOT $\displaystyle \dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$ $\displaystyle -$ that is the standard trap here.First derivatives, product rule on $\displaystyle t\sin t$ and $\displaystyle t\cos t$: \[\frac{dx}{dt}=a\left(-\sin t+\sin t+t\cos t\right)=a\,t\cos t \] \[\frac{dy}{dt}=a\left(\cos t-\cos t+t\sin t\right)=a\,t\sin t \]Hence, for $\displaystyle t\neq 0$, \[\frac{dy}{dx}=\frac{a\,t\sin t}{a\,t\cos t}=\tan t. \]Then \[\frac{d^{2}y}{dx^{2}}=\frac{d}{dt}(\tan t)\cdot\frac{1}{dx/dt}=\sec^{2}t\cdot\frac{1}{a\,t\cos t}=\frac{\sec^{3}t}{a\,t}. \]So $\displaystyle \dfrac{d^{2}y}{dx^{2}}=\dfrac{\sec^{3}t}{a\,t}$, valid where $\displaystyle t\neq 0$ and $\displaystyle \cos t\neq 0$ (for instance $\displaystyle 0<t<\frac{\pi}{2}$).
Exercise 18
If $\displaystyle f(x)=|x|^{3}$, show that $\displaystyle f^{\prime \prime}(x)$ exists for all real $\displaystyle x$ and find it.
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Write the modulus piecewise: \[f(x)=|x|^{3}=\begin{cases}x^{3}, & x\ge 0\\ -x^{3}, & x<0\end{cases} \]First derivative. For $\displaystyle x>0$, $\displaystyle f'(x)=3x^{2}$; for $\displaystyle x<0$, $\displaystyle f'(x)=-3x^{2}$. At the corner $\displaystyle x=0$ the pieces do not settle the matter, so use the definition: \[f'(0)=\lim_{h\to 0}\frac{|h|^{3}-0}{h}=\lim_{h\to 0}\frac{|h|^{3}}{h}=\lim_{h\to 0}\left(\pm h^{2}\right)=0. \] All three cases are covered by the single formula $\displaystyle f'(x)=3x|x|$.Second derivative. For $\displaystyle x>0$, $\displaystyle f''(x)=6x$; for $\displaystyle x<0$, $\displaystyle f''(x)=-6x$. At $\displaystyle x=0$, again from the definition applied to $\displaystyle f'$: \[f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h}=\lim_{h\to 0}\frac{3h|h|-0}{h}=\lim_{h\to 0}3|h|=0, \] the left- and right-hand limits both being $\displaystyle 0$. So $\displaystyle f''(0)$ exists $\displaystyle -$ this check at the corner is the point of the question.Hence $\displaystyle f''(x)$ exists for every real $\displaystyle x$, and \[f''(x)=6|x|=\begin{cases}6x, & x\ge 0\\ -6x, & x<0\end{cases} \]
Exercise 19
Using the fact that $\displaystyle \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$ and the differentiation, obtain the sum formula for cosines.
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Treat $\displaystyle \mathrm{B}$ as a constant and differentiate the identity \[\sin(\mathrm{A}+\mathrm{B})=\sin \mathrm{A}\cos \mathrm{B}+\cos \mathrm{A}\sin \mathrm{B} \] with respect to $\displaystyle \mathrm{A}$. This is allowed because the relation is an identity: the two sides are the same function of $\displaystyle \mathrm{A}$, so their derivatives agree.Left side, chain rule with $\displaystyle \dfrac{d}{d\mathrm{A}}(\mathrm{A}+\mathrm{B})=1$: \[\frac{d}{d\mathrm{A}}\sin(\mathrm{A}+\mathrm{B})=\cos(\mathrm{A}+\mathrm{B}). \]Right side, with $\displaystyle \cos \mathrm{B}$ and $\displaystyle \sin \mathrm{B}$ constant multipliers: \[\frac{d}{d\mathrm{A}}\left(\sin \mathrm{A}\cos \mathrm{B}+\cos \mathrm{A}\sin \mathrm{B}\right)=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\sin \mathrm{B}. \]Equating the two, \[\cos(\mathrm{A}+\mathrm{B})=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\sin \mathrm{B}, \] which is the sum formula for cosines.
Exercise 20
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
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Yes, such a function exists. Take \[f(x)=|x|+|x-1|. \]Continuity: $\displaystyle x\mapsto|x|$ and $\displaystyle x\mapsto|x-1|$ are continuous on all of $\displaystyle \mathbf{R}$, and the sum of two continuous functions is continuous, so $\displaystyle f$ is continuous everywhere.Removing the moduli, \[f(x)=\begin{cases}1-2x, & x<0\\ 1, & 0\le x<1\\ 2x-1, & x\ge 1\end{cases} \]Differentiability: on each of $\displaystyle (-\infty,0)$, $\displaystyle (0,1)$ and $\displaystyle (1,\infty)$ the function is a polynomial, hence differentiable there. Only the two joins need checking, and there the one-sided derivatives disagree: \[\text{at }x=0:\quad \text{LHD}=-2,\quad \text{RHD}=0 \] \[\text{at }x=1:\quad \text{LHD}=0,\quad \text{RHD}=2 \]So $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$ and fails to be differentiable at exactly the two points $\displaystyle x=0$ and $\displaystyle x=1$. The answer is therefore yes; more generally $\displaystyle |x-p|+|x-q|$ with $\displaystyle p\neq q$ does the same job at any prescribed pair of points.
Exercise 21
If $\displaystyle y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right|$, prove that $\displaystyle \frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right|$
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Expand the determinant along the first row $\displaystyle -$ the only row containing $\displaystyle x$, since $\displaystyle l,m,n,a,b,c$ are constants: \[y=f(x)(mc-nb)-g(x)(lc-na)+h(x)(lb-ma). \]Each bracket is a constant, so differentiating term by term (the derivative of a constant multiple is the constant times the derivative): \[\frac{dy}{dx}=f'(x)(mc-nb)-g'(x)(lc-na)+h'(x)(lb-ma). \]This right-hand side is precisely the first-row expansion of the determinant in which only the first row has been differentiated, the other two rows being unchanged. Hence \[\frac{dy}{dx}=\left|\begin{array}{ccc}f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c\end{array}\right|, \] as required.
Exercise 22
If $\displaystyle y=e^{a \cos ^{-1} x},-1 \leq x \leq 1$, show that $\displaystyle \left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$.
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This solution has not been cross-checked against the answer printed in NCERT.
Differentiate $\displaystyle y=e^{a\cos^{-1}x}$ by the chain rule, using $\displaystyle \dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$: \[\frac{dy}{dx}=e^{a\cos^{-1}x}\left(-\frac{a}{\sqrt{1-x^{2}}}\right)=-\frac{ay}{\sqrt{1-x^{2}}},\qquad -1<x<1. \]Clear the radical before differentiating again $\displaystyle -$ this is the step that keeps the second derivative manageable: \[\sqrt{1-x^{2}}\,\frac{dy}{dx}=-ay. \]Differentiate w.r.t. $\displaystyle x$, product rule on the left: \[\sqrt{1-x^{2}}\,\frac{d^{2}y}{dx^{2}}+\frac{-x}{\sqrt{1-x^{2}}}\frac{dy}{dx}=-a\frac{dy}{dx}. \]Now replace the right-hand side using the first derivative: \[-a\frac{dy}{dx}=-a\left(-\frac{ay}{\sqrt{1-x^{2}}}\right)=\frac{a^{2}y}{\sqrt{1-x^{2}}}. \]So \[\sqrt{1-x^{2}}\,\frac{d^{2}y}{dx^{2}}-\frac{x}{\sqrt{1-x^{2}}}\frac{dy}{dx}=\frac{a^{2}y}{\sqrt{1-x^{2}}}. \]Multiplying throughout by $\displaystyle \sqrt{1-x^{2}}$, which is non-zero for $\displaystyle -1<x<1$, \[\left(1-x^{2}\right)\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}-a^{2}y=0, \] as required (the endpoints $\displaystyle x=\pm 1$ are excluded, since $\displaystyle \dfrac{dy}{dx}$ is not finite there).

EXERCISE 5.1 1–10 (part 1 of 15)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10