# NCERT Solutions for Exercise 5.3 Class 12 Maths Chapter 5 - Continuity and Differentiability

In this article, you will get NCERT solutions for Class 12 Maths chapter 5 exercise 5.3 consists of questions related to finding derivatives of implicit functions. In chapter 2 of Class 12th Math NCERT syllabus, you have already learned about inverse trigonometric functions, domain, and range of the inverse trigonometric functions. In exercise 5.3 Class 12 Maths, you will get questions related to finding derivatives of inverse trigonometric functions.

The basic knowledge of domain and range of inverse trigonometric functions is always beneficial to understand Class 12 Maths ch 5 ex 5.3. You might have to remember some derivatives for inverse trigonometric functions like you remembered for trigonometric functions. Solving more problems from Class 12th Maths chapter 5 exercise 5.3 will help you to memorize these formulas at your fingertips. To check NCERT Solutions for other classes, click on the given link.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.2
• Continuity and Differentiability Exercise 5.4
• Continuity and Differentiability Exercise 5.5
• Continuity and Differentiability Exercise 5.6
• Continuity and Differentiability Exercise 5.7
• Continuity and Differentiability Exercise 5.8
• Continuity and Differentiability Miscellaneous Exercise

## Continuity and Differentiability Exercise: 5.3

Question:1. Find dy/dx in the following:

$2 x + 3 y = \sin x$

Given function is
$2 x + 3 y = \sin x$
We can rewrite it as
$3y = \sin x - 2x$
Now, differentiation w.r.t. x is
$3\frac{dy}{dx} = \frac{d(\sin x - 2x)}{dx} = \cos x - 2$
$\frac{dy}{dx} = \frac{\cos x-2}{3}$
Therefore, the answer is $\frac{\cos x-2}{3}$

Question:2. Find dy/dx in the following: $2 x + 3y = \sin y$

Given function is
$2 x + 3 y = \sin y$
We can rewrite it as
$\sin y - 3y = 2x$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(\sin y - 3y) = \frac{d( 2x)}{dx}$

$(\cos y\frac{dy}{dx} - 3\frac{dy}{dx}) = 2$
$\frac{dy}{dx} = \frac{2}{\cos y -3}$
Therefore, the answer is $\frac{2}{\cos y -3}$

Question:3. Find dy/dx in the following: $ax + by ^2 = \cos y$

Given function is
$ax + by ^2 = \cos y$
We can rewrite it as
$by^2-\cos y = -ax$
Now, differentiation w.r.t. x is
$\frac{dy}{dx}(2by - (-\sin y)) = \frac{d( -ax)}{dx} = -a$
$\frac{dy}{dx} = \frac{-a}{2b y +\sin y}$
Therefore, the answer is $\frac{-a}{2b y +\sin y}$

Question:4. Find dy/dx in the following:

$xy + y^2 = \tan x + y$

Given function is
$xy + y^2 = \tan x + y$
We can rewrite it as
$xy+y^2-y= \tan x$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y-1) = \frac{d( \tan x)}{dx} = \sec^2 x$
$\frac{dy}{dx} = \frac{\sec^2 x- y}{x+2y-1}$
Therefore, the answer is $\frac{\sec^2 x- y}{x+2y-1}$

Question:5. Find dy/dx in the following: $x^2 + xy + y^2 = 100$

Given function is
$x^2 + xy + y^2 = 100$
We can rewrite it as
$xy + y^2 = 100 - x^2$
Now, differentiation w.r.t. x is
$y+\frac{dy}{dx}(x+2y) = \frac{d( 100-x^2)}{dx} = -2x$
$\frac{dy}{dx} = \frac{-2 x- y}{x+2y}$
Therefore, the answer is $\frac{-2 x- y}{x+2y}$

Question:6 Find dy/dx in the following:

$x ^3 + x^2 y + xy^2 + y^3 = 81$

Given function is
$x ^3 + x^2 y + xy^2 + y^3 = 81$
We can rewrite it as
$x^2 y + xy^2 + y^3 = 81 - x^3$
Now, differentiation w.r.t. x is
$\frac{d(x^2 y + xy^2 + y^3)}{dx} = \frac{d(81 - x^3)}{dx}$
$2xy+y^2+\frac{dy}{dx}(x^2+2xy+3y^2) = -3x^2\\ \frac{dy}{dx}=\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$
Therefore, the answer is $\frac{-(3x^2+2xy+y^2)}{(x^2+2xy+3y^2}$

Question:7. Find dy/dx in the following: $\sin ^ 2 y + \cos xy = k$

Given function is
$\sin ^ 2 y + \cos xy = k$
Now, differentiation w.r.t. x is
$\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}$
$2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0\\ \frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx} = \frac{y\sin xy}{2\sin y \cos y-x\sin xy} = \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)$
Therefore, the answer is $\frac{y\sin xy}{\sin 2y -x\sin xy}$

Question:8. Find dy/dx in the following:

$\sin ^2 x + \cos ^ 2 y = 1$

Given function is
$\sin ^2 x + \cos ^ 2 y = 1$
We can rewrite it as
$\cos ^ 2 y = 1-\sin^2x$
Now, differentiation w.r.t. x is
$\frac{d(\cos^2y)}{dx} = \frac{d(1-\sin^2x)}{dx}$
$2\cos y (-\sin y)\frac{dy}{dx} = -2\sin x \cos x\\ \frac{dy}{dx} = \frac{2\sin x\cos x}{2\sin y \cos y} = \frac{\sin 2x }{\sin 2y} \ \ \ \ \ \ (\because2\sin a \cos a = \sin 2a)$
Therefore, the answer is $\frac{\sin 2x}{\sin 2y }$

Question:9 Find dy/dx in the following:

$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$

Given function is
$y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})$
Our equation reduces to
$y = \sin^{-1}(\sin 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question:10. Find dy/dx in the following:
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }$

Given function is
$y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )$
Lets consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )$
Our equation reduces to
$y = \tan^{-1}(\tan 3t)$
$y = 3t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}$
Therefore, the answer is $\frac{3}{1+x^2}$

Question:11. Find dy/dx in the following:

$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right ) , 0 < x < 1$

Given function is
$y = \cos ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$y = \cos^{-1}(\cos 2t)$
$y = 2t$
Now, differentiation w.r.t. x is
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}$
Therefore, the answer is $\frac{2}{1+x^2}$

Question:12. Find dy/dx in the following: $y = \sin ^{-1 } \left ( \frac{1- x ^2 }{1+ x^2} \right ) , 0< x < 1$

Given function is
$y = \sin ^{-1} \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
We can rewrite it as
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2t }{1+\tan^2t} = \cos 2t \ \ \ \ \ \ (\because \cos 2x = \frac{1-\tan^2x }{1+\tan^2x} )$
Our equation reduces to
$\sin y = \cos 2t$
Now, differentiation w.r.t. x is
$\frac{d(\sin y)}{dx} = \frac{d(\cos2t)}{dt}.\frac{dt}{dx}$
$\cos y\frac{dy}{dx} = 2(-\sin 2t).\frac{1}{1+x^2} = \frac{-2\sin2t}{1+x^2}$$= \frac{-2.\frac{2\tan t}{1+\tan^2t}}{1+x^2} =\frac{-2.\frac{2x}{1+x^2}}{1+x^2} =\frac{-4x}{(1+x^2)^2}$
$(\because \sin 2x = \frac{2\tan x}{1+\tan^2x} \ and \ x = \tan t)$
$\sin y = \ \left ( \frac{1 - x^2 }{1+ x^2 } \right )\Rightarrow \cos y = \frac{2x}{1+x^2}$
$\frac{2x}{1+x^2}\frac{dy}{dx} = \frac{-4x}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question:13. Find dy/dx in the following:

$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1$

Given function is
$y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )$
We can rewrite it as
$\cos y = \left ( \frac{2x}{1+ x^2 } \right )$
Let's consider $x = \tan t$
Then,
$\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)$
$1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)$
Now,
$\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )$
Our equation reduces to
$\cos y = \sin 2t$
Now, differentiation w.r.t. x is
$\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}$
$(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2}$$= \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}$
$(\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)$
$\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\Rightarrow \sin y = \frac{1-x^2}{1+x^2}$
$-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}$
$\frac{dy}{dx} = \frac{-2}{(1+x^2)}$
Therefore, the answer is $\frac{-2}{1+x^2}$

Question:14. Find dy/dx in the following:

$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }$

Given function is
$y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )$
Lets take $x = \sin t$
Then,
$\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =\cos t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}$
$(\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )$
And
$2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t} = 2\sin t\cos t =\sin 2t$
$(\because \cos x = \sqrt{1-\sin^2x} \ and \ 2\sin x\cos x = \sin2x )$
Now, our equation reduces to
$y = \sin ^ { -1 } ( \sin 2t )$
$y = 2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{2}{\sqrt{1-x^2}}$

Question:15. Find dy/dx in the following:

$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2$

Given function is
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )$
Let's take $x = \cos t$
Then,
$\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )$
And
$\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t$
$(\because \cos 2x = \sqrt{2\cos^2x-1} \ and \ \cos x = \frac{1}{\sec x} )$

Now, our equation reduces to
$y = \sec ^{-1} \left ( \sec 2t \right )$
$y = 2t$
Now, differentiation w.r.t. x
$\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}$
$\frac{dy}{dx} = 2.\frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}$
Therefore, the answer is $\frac{-2}{\sqrt{1-x^2}}$

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3:-

In Class 12 Maths chapter 5 exercise 5.3 solutions, there are 15 long answer types questions out of which seven questions are related to finding derivatives of inverse trigonometric functions. There are two solved examples related to derivatives of implicit functions and two examples are related to inverse trigonometric functions given before the Class 12th Maths chapter 5 exercise 5.3. You can try to solve these examples and exercise questions by yourself. This will require a good knowledge of trigonometric functions and inverse trigonometric functions.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Also see-

• NCERT Solutions for Class 12 Maths Chapter 5

• NCERT Exemplar Solutions Class 12 Maths Chapter 5

## NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT Solutions for Class 12 Biology

## Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3:-

• NCERT Solutions for Class 12 Maths chapter 5 exercise 5.3 are important for the students to get conceptual clarity and perform well in the board exams.
• In NCERT book Class 12 Maths chapter 5 exercise 5.3 solutions you will get important inverse trigonometric functions derivatives formulas also.
• Class 12th Maths chapter 5 exercise 5.3 is provided at one place which you can download.

Happy learning!!!