# Continuity and Differentiability NCERT Solutions for Class 12 Maths Exercise 5.2

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### Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ in the following

Question 1
$2x+3y=\sin x$
Solution
The given relationship is $2x+3y=\sin x$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)$
$\frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)$
$2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos x$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\cos x-2}{3}$.

Question 2
$2x+3y=\sin y$
Solution

The given relationship is $2x+3y=\sin y$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(2x+3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)$
$\frac{\mathrm{d} }{\mathrm{d} x} (2x) + \frac{\mathrm{d} }{\mathrm{d} x}(3y)=\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)$
$2 + 3\frac{\mathrm{d} y}{\mathrm{d} x}=\cos y \frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\cos y -3}$.

Question 3
$ax+by^{2}=\cos y$.
Solution
The given relationship is $ax+by^{2}=\cos y$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(ax+by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)$
$\frac{\mathrm{d} }{\mathrm{d} x} (ax) + \frac{\mathrm{d} }{\mathrm{d} x}(by^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)$
$a + 2by\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin y \frac{\mathrm{d} y}{\mathrm{d} x}$ ( Using Chain rule)
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-a}{(\sin y+2by )}$.

Question 4
$xy+y^{2}=\tan x+ y$
Solution

The given relationship is $xy+y^{2}=\tan x+ y$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x + y)$
$\frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)+ \frac{\mathrm{d} y}{\mathrm{d} x}$
$y.\frac{\mathrm{d} }{\mathrm{d} x} (x)+x .\frac{\mathrm{d} y}{\mathrm{d} x} +2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}$ ( Using Chain rule and Product rule)
$y.1 + x.\frac{\mathrm{d} y}{\mathrm{d} x}+2y. \frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x+ \frac{\mathrm{d} y}{\mathrm{d} x}$
$(x+2y-1)\frac{\mathrm{d} y}{\mathrm{d} x}=\sec ^{2}x-y$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sec ^{2}x-y}{(x+2y-1 )}$.

Question 5
$x^{2}+xy+y^{2}=100$.
Solution
The given relationship is $x^{2}+xy+y^{2}=100$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}+xy+y^{2})=\frac{\mathrm{d} }{\mathrm{d} x}(100)$
$\frac{\mathrm{d} }{\mathrm{d} x} (x^{2})+\frac{\mathrm{d} }{\mathrm{d} x} (xy) + \frac{\mathrm{d} }{\mathrm{d} x}(y^{2})=0$       (derivatives of constant function is 0)
$2x+ y.1+x.\frac{\mathrm{d} y}{\mathrm{d} x} +2y.\frac{\mathrm{d} y}{\mathrm{d} x}=0$ (Using Chain rule and Product rule)
$2x+ y+(x+2y)\frac{\mathrm{d} y}{\mathrm{d} x} =0$
$\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2x+y}{x+2y}$

Question 6
$x^{3}+x^{2}y+xy^{2}+y^{3}$.
Solution
The given relationship is $x^{3}+x^{2}y+xy^{2}+y^{3}=81$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(x^{3}+x^{2}y+xy^{2}+y^{3})=\frac{\mathrm{d} }{\mathrm{d} x}(81)$
$\frac{\mathrm{d} }{\mathrm{d} x} (x^{3})+\frac{\mathrm{d} }{\mathrm{d} x} (x^{2}y) + \frac{\mathrm{d} }{\mathrm{d} x}(xy^{2})+\frac{\mathrm{d} }{\mathrm{d} x}(y^{3})=0$                        (derivatives of constant function is 0)
$3x^{2}+ y.\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}\frac{\mathrm{d} }{\mathrm{d} x}(x)+x\frac{\mathrm{d} }{\mathrm{d} x}(y^{2}+3y^{2}.\frac{\mathrm{d} y}{\mathrm{d} x} =0$
$3x^{2}+ y.2x+x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+y^{2}.1+x.2y.\frac{\mathrm{d} y}{\mathrm{d} x}+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x} =0$
$(x^{2}+2xy+3y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}+(3x^{2}+2xy+y^{2}=0$
$therefore,\; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(3x^{2}+2xy+y^{2})}{x^{2}+2xy+3y^{2}}$

Question 7
$\sin ^{2}y +\cos xy=\Pi$.
Solution
The given relationship is $\sin ^{2}y+\cos xy=\Pi$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y+\cos xy)=\frac{\mathrm{d} }{\mathrm{d} x}(\Pi)$
$\frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos xy) =0$ (derivatives of constant function is 0)
We can find these differentiation using chain rule as we get
$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}y=2 \sin y \frac{\mathrm{d} }{\mathrm{d} x}(\sin y) =2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} }{\mathrm{d} x}(\cos xy)=-\sin xy \frac{\mathrm{d} }{\mathrm{d} x}(xy)= -\sin xy (y. \frac{\mathrm{d} }{\mathrm{d} x}(x)+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -\sin xy (y. 1+x. \frac{\mathrm{d} y}{\mathrm{d} x})= -y \sin xy -x\sin xy \frac{\mathrm{d} y}{\mathrm{d} x}$
So we have
$2\sin y \cos y \frac{\mathrm{d} y}{\mathrm{d} x}- y\sin xy- x \sin xy \frac{\mathrm{d} y}{\mathrm{d} x}=0$
$(2\sin y \cos y- x \sin xy)\frac{\mathrm{d} y}{\mathrm{d} x}= y \sin xy$
$(\sin 2y-x\sin xy )\frac{\mathrm{d} y}{\mathrm{d} x}=y \sin xy$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{y \sin xy}{\sin 2y -x \sin xy}$
Question 8
$\sin ^{2}x +\cos ^{2}y=1$.
Solution
The given relationship is $\sin ^{2}x +\cos ^{2}y=1$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(\sin ^{2}x+\cos ^{2}y)=\frac{\mathrm{d} }{\mathrm{d} x}(1)$
$\frac{\mathrm{d} }{\mathrm{d} x} (\sin ^{2}y)+\frac{\mathrm{d} }{\mathrm{d} x} (\cos ^{2}y) =0$ (derivatives of constant function is 0)
$2\sin x \frac{\mathrm{d} }{\mathrm{d} x} (\sin x)+2\cos y.\frac{\mathrm{d} }{\mathrm{d} x} (\cos y) =0$
$2\sin x \cos x+2\cos y(-\sin y).\frac{\mathrm{d} y}{\mathrm{d} x} =0$
$\sin 2x-\sin 2y.\frac{\mathrm{d} y}{\mathrm{d} x} =0$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin 2x}{\sin 2y}$

Question 9
$y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$.
Solution
The given relationship is $y=\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$
$\sin y=\left ( \frac{2x}{1+x^{2}} \right )$
Differentiating the relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )$
$\cos y \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{2x}{1+x^{2}} \right )$      …
The function $\left ( \frac{2x}{1+x^{2}} \right )$ is of the form $\frac{u}{v}$
So, by quotient rule, we obtain differentiation as
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )=\frac{(1+x^{2}).\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}$
$=\frac{(1+x^{2}).2-2x.(0+2x)}{(1+x^{2})^{2}}=\frac{2+2x^{2}-4x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}$          …
Also $\sin y=\frac{2x}{1+x^{2}}$
$\cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}=\sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}$
$=\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}}$
Substituting these values ,we obtain
$=\frac{1-x^{2}}{1+x^{2}}\times \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x^{2}}$

Question 10
$y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right ),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$.
Solution
The given relationship is $y=\tan ^{-1}\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )$
$\tan y=\left ( \frac{3x-x^{3}}{1-3x^{2}} \right )$
We know that, $\tan y=\frac{3\tan \frac{y}{3}-\tan ^{3}\frac{y}{3}}{1-3\tan ^{2}\frac{y}{3}}$
Comparing both equation , we have
$x=\tan \frac{y}{3}$ Differentiating this relationship w.r.t. x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x} \left ( \tan \frac{y}{3} \right )$
$1=\sec ^{2}\frac{y}{3}.\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{y}{3} \right )$
$1=\sec ^{2}\frac{y}{3}.\frac{1}{3}\frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{\sec ^{2}\frac{y}{3}}=\frac{3}{1+\tan ^{2}\frac{y}{3}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+x^{2}}$

Question 11
$y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$.
Solution
The given relationship is $y=\cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$
$\cos y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )$
$\frac{1-\tan ^{2}\frac{y}{2}}{1+\tan ^{2}\frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}$
Comparing both sides equation
$\tan \frac{y}{2}=x$
Differentiating the above relationship with respect to x, we have
$\sec ^{2} \frac{y}{2}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )=\frac{\mathrm{d} }{\mathrm{d} x}(x)$
$\sec ^{2} \frac{y}{2}\times \frac{1}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}=1$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sec ^{2}\frac{y}{2}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+\tan ^{2}\frac{y}{2}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{1+x ^{2}}$

Question 12
$y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$.
Solution
The given relationship is $y=\sin ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right ), 0<x<1$
$\sin y= \left ( \frac{1-x^{2}}{1+x^{2}} \right )$ Differentiating the realtionship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(\sin y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )$
Using chain rule
$\cos y \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )$
Now,
$\cos y=\sqrt{1-\sin ^{2}y}=\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}$
$=\frac{2x}{1+x^{2}}$
Also $\frac{1-x^{2}}{1+x^{2}}$ can be treated as u/v,So applying quotient rule of differentiation
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1-x^{2}}{1+x^{2}} \right )=\frac{(1+x^{2}).(1-x^{2})-(1-x^{2}).(1+x^{2})}{(1+x^{2})^{2}}$
$=\frac{(1+x^{2}).(-2x)-(1-x^{2}).(2x)}{(1+x^{2})^{2}}$
$=\frac{-2x-2x^{3}-2x+2x^{3}}{(1+x^{2})^{2}}$
$=\frac{-4x}{(1+x^{2})^{2}}$
So we have
$\frac{2x}{(1+x^{2})}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-4x}{(1+x^{2})^{2}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{(1+x^{2})}$

Question 13
$y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right ), -1<x<1$.
Solution
The given relationship is $y=\cos ^{-1}\left ( \frac{2x}{1+x^{2}} \right )$
$\cos y= \left ( \frac{2x}{1+x^{2}} \right )$
Differentiating the realtionship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(\cos y)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^{2}} \right )$
$-\sin y \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{(1+x^{2}.\frac{\mathrm{d} }{\mathrm{d} x}(2x)-2x.\frac{\mathrm{d} }{\mathrm{d} x}(1+x^{2})}{(1+x^{2})^{2}}$
$-\sqrt{1-\cos^{2}y}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{(1+x^{2}.2-2x.2x)}{(1+x^{2})^{2}}$
$\sqrt{1-\left ( \frac{2x}{1+x^{2}} \right )^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$
$\sqrt{\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$
$\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}.\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [\frac{2(1-x^{2})}{(1+x^{2})^{2}} \right ]$
$\frac{(1-x^{2})}{(1+x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{1+x^{2}}$

Question 14
$y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$.
Solution
The given relationship is $y=\sin^{-1}(2x\sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
$\sin y= (2x\sqrt{1-x^{2}})$
Differentiating the equation with respect to x, we have
$\cos y \frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ x.\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{1-x^{2}} +\sqrt{1-x^{2}}\frac{\mathrm{d} x}{\mathrm{d} x}\right ]$
$\sqrt{1-\sin^{2}y} \frac{\mathrm{d}y }{\mathrm{d} x}=2\left [ \frac{x}{2}.\frac{-2x}{\sqrt{1-x^{2}}} +\sqrt{1-x^{2}}\right ]$
$\sqrt{1-(2x\sqrt{1-x^{2}})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}} \right ]$
$\sqrt{1-4x^{2}(1-x^{2})}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]$
$\sqrt{(1-2x^{2})^{2}}.\frac{\mathrm{d} y}{\mathrm{d} x}=2\left [ \frac{1-2x^{2}}{\sqrt{1-x^{2}}} \right ]$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2}{\sqrt{1-x^{2}}}$

Question 15
$y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}$.
Solution
The given relationship is $y=\sec^{-1}\left ( \frac{1}{2x^{2}-1} \right ),0<x<\frac{1}{\sqrt{2}}$
$\sec y= \left ( \frac{1}{2x^{2}-1} \right )$
$\cos y=2x^{2}-1$
$2x^{2}=1+\cos y$
$2x^{2}=2\cos ^{2}\frac{y}{2}$
$x=cos \frac{y}{2}$
Differentiating the above relationship with respect to x, we have
$\frac{\mathrm{d} }{\mathrm{d} x}(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{y}{2} \right )$
$1=-\sin \frac{y}{2}.\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{2} \right )$
$\frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos^{2}\frac{y}{2}}}$
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2}{\sqrt{1-x^{2}}}$