NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

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 \(\displaystyle x\in\mathbf{R}\).

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 \(\displaystyle x\in\mathbf{R}\).

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 \(\displaystyle x\in\mathbf{R}\).

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.

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 \(\displaystyle \cos(cx+d)\neq 0\).

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 \(\displaystyle 5x^{4}\cos\left(x^{3}\right)\sin\left(2x^{5}\right)-3x^{2}\sin\left(x^{3}\right)\sin^{2}\left(x^{5}\right)\).

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 \(\displaystyle \cot\left(x^{2}\right)>0\).

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 \(\displaystyle x\to 0^{+}\).

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 \(\displaystyle (1,0)\).)

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 \(\displaystyle x=2\).