physicscatalyst.com logo






Continuity and Differentiability NCERT Solutions for Class 12 Maths Exercise 5.2





In this page we have Continuity and Differentiability NCERT Solutions for Class 12 Maths Exercise 5.3. Hope you like them and do not forget to like , social share and comment at the end of the page.


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}}}$
 

 


Class 12 Maths Home page Class 12 Physics Home page