NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

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: \[3\frac{dy}{dx}=\cos x-2 \]\[\boxed{\ \frac{dy}{dx}=\frac{\cos x-2}{3}\ } \]

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.\[\boxed{\ \frac{dy}{dx}=\frac{2}{\cos y-3}\ } \]

\(\displaystyle a\) 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: \[2by\frac{dy}{dx}+\sin y\,\frac{dy}{dx}=-a \] \[(2by+\sin y)\frac{dy}{dx}=-a \]\[\boxed{\ \frac{dy}{dx}=\frac{-a}{2by+\sin y}\ },\qquad 2by+\sin y\ne 0. \]

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: \[x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=\sec^{2}x-y \] \[(x+2y-1)\frac{dy}{dx}=\sec^{2}x-y \]\[\boxed{\ \frac{dy}{dx}=\frac{\sec^{2}x-y}{x+2y-1}\ },\qquad x+2y\ne 1. \]

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: \[(x+2y)\frac{dy}{dx}=-(2x+y) \]\[\boxed{\ \frac{dy}{dx}=-\frac{2x+y}{x+2y}\ },\qquad x+2y\ne 0. \]

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 \(\displaystyle \frac{dy}{dx}\): \[\left(x^{2}+2xy+3y^{2}\right)\frac{dy}{dx}=-\left(3x^{2}+2xy+y^{2}\right) \]\[\boxed{\ \frac{dy}{dx}=-\frac{3x^{2}+2xy+y^{2}}{x^{2}+2xy+3y^{2}}\ },\qquad x^{2}+2xy+3y^{2}\ne 0. \]

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 \(\displaystyle \frac{dy}{dx}\): \[\left(\sin 2y-x\sin (xy)\right)\frac{dy}{dx}=y\sin (xy) \]\[\boxed{\ \frac{dy}{dx}=\frac{y\sin (xy)}{\sin 2y-x\sin (xy)}\ },\qquad \sin 2y\ne x\sin (xy). \]

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 \(\displaystyle \sin 2y=\sin(\pm 2x+2n\pi)=\pm\sin 2x\).

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.\[\boxed{\ \frac{dy}{dx}=\frac{2}{1+x^{2}}\ \text{for }|x|<1,\qquad \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ \text{for }|x|>1;\ \text{not differentiable at }x=\pm1.\ } \]

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 \(\displaystyle \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}\): \[\boxed{\ \frac{dy}{dx}=\frac{3}{1+x^{2}}\ } \]