NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

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 \(\displaystyle x>3\), \[\frac{dy}{dx}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2x\log x\right]+(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2x\log(x-3)\right]. \]

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).

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 \[\frac{dy}{dx}=0,\qquad 0<x<1. \]

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.

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 \(\displaystyle b\).

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.

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 \(\displaystyle 0<t<\frac{\pi}{2}\)).

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 \[f''(x)=6|x|=\begin{cases}6x, & x\ge 0\\ -6x, & x<0\end{cases} \]

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.

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.

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.

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).