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

Last updated date: 05th Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 - FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 5 Ex 5.3 by Vedantu, is crucial for understanding how functions behave, how they change, and how they can be manipulated using calculus. Exercise 5.3 focuses on the concepts of differentiability and provides a variety of problems that help reinforce your understanding of this topic. By working through these solutions, you’ll gain confidence in tackling different differentiation problems, which is essential for success in board exams and competitive exams like JEE. Vedantu’s step-by-step solutions are designed to make learning easy and effective.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 5 Exercise 5.3 Class 12 | Vedantu
3. Topics Covered in Class 12 Maths Chapter 5 Exercise 5.3
4. Access PDF for Maths NCERT Chapter 5 Continuity and Differentiability Exercise 5.3 Class 12
5. Class 12 Maths Chapter 5: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 5 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 5 Exercise 5.3 Class 12 | Vedantu

• Product and Quotient rules help to differentiate a function that is multiplied or divided by each other.

• The chain rule is used to break complex functions into simple parts to find their derivatives.

• Higher order derivatives are the second, third, or further derivatives of a function. In simple words, it is differentiating a function multiple times results in higher order derivatives.

• Derivatives of implicit functions is the process of differentiating an implicit equation with respect to the desired variable x while treating the other variables as unspecified functions of x.

• Derivatives of inverse trigonometric functions are the process of finding angles for the given trigonometric value.

• Ex 5.3 Class 12 has 15 fully solved questions for Chapter 5 Continuity and Differentiability.

## Topics Covered in Class 12 Maths Chapter 5 Exercise 5.3

1. Derivatives of Implicit functions

2. Derivatives of Inverse trigonometric functions

Competitive Exams after 12th Science

## Access PDF for Maths NCERT Chapter 5 Continuity and Differentiability Exercise 5.3 Class 12

1. Find  $\frac{dy}{dx}:2x+3y=\sin x$

Ans: The given relationship is  $2x+3y=\sin x$

Differentiating the above relationship with respect to $x$ ,

We get

$\Rightarrow \text{ }$  $\frac{d}{dy}(2x+3y)=\frac{d}{dx}(\sin x)$

$\Rightarrow \frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\cos x$

$\Rightarrow 2+3\frac{dy}{dx}=\cos x$

$\Rightarrow 3\frac{dy}{dx}=\cos x-2$

$\therefore \frac{dx}{dy}=\frac{\cos x-2}{3}$

2. Find  $\frac{dy}{dx}:2x+3y=\sin y$

Ans: The given relationship is  $2x+3y=\sin y$

Differentiating the above relationship with respect to $x$ , We obtain

$\Rightarrow \text{ }$   $\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin y)$

$\Rightarrow 2+3\frac{dy}{dx}=\cos y\frac{dy}{dx}\quad [$  By using chain rule]

$\Rightarrow 2=(cosy$  $-3)\frac{dy}{dx}$

$\therefore \frac{dy}{dx}=\frac{2}{\cos y-3}$

3. Find  $\frac{dy}{dx}:ax+b{{y}^{2}}=\cos y$

Ans: The given relationship is $ax+b{{y}^{2}}=\cos y$ .

Differentiating the above relationship with respect to $x$ , we obtain

$\Rightarrow \text{ }$  $\frac{d}{dx}(\alpha x)+\frac{d}{dx}\left( b{{y}^{2}} \right)=\frac{d}{dx}(\cos y)$

$\Rightarrow a+b\frac{d}{dx}\left( {{y}^{2}} \right)=\frac{d}{dx}(\cos y)$

$\frac{d}{dx}\left( {{y}^{2}} \right)=2y\frac{dy}{dx}$  and  $\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$

Using the chain rule,

We get,

$\Rightarrow \text{ }$  $a+b\times 2y\frac{dy}{dx}=-\sin y\frac{dy}{dx}$

$\Rightarrow (2by+\sin y)\frac{dy}{dx}=-a$

$\therefore \frac{dy}{dx}=\frac{-a}{2by+\sin y}$

4. Find  $\frac{dy}{dx}:xy+{{y}^{2}}=\tan x+y$

Ans:  The given relationship is  $xy+{{y}^{2}}=\tan x+y$

Differentiating the above relationship with respect to $x$ , We obtain

$\Rightarrow \text{ }$  $\frac{d}{dx}\left( xy+{{y}^{2}} \right)=\frac{d}{dx}(\tan x+y)$

$\Rightarrow \frac{d}{dx}(xy)+\frac{d}{dx}\left( {{y}^{2}} \right)=\frac{d}{dx}(\tan x)+\frac{dy}{dx}$

$\Rightarrow \left[ y\cdot \frac{d}{dx}(x)+x\cdot \frac{dy}{dx} \right]+2y\frac{dy}{dx}={{\sec }^{2}}x+\frac{dy}{dx}\quad$  [Using product rule and chain rule]

$\Rightarrow y\cdot 1+x\frac{dy}{dx}+2y\frac{dy}{dx}={{\sec }^{2}}x+\frac{dy}{dx}$

$\Rightarrow (x+2y-1)\frac{dy}{dx}={{\sec }^{2}}x-y$

$\therefore \frac{dy}{dx}=\frac{{{\sec }^{2}}x-y}{(x+2y-1)}$

5. Find  $\frac{dy}{dx}:{{x}^{2}}+xy+{{y}^{2}}=100$

Ans:  The given relationship is ${{x}^{2}}+xy+{{y}^{2}}=100$ .

Differentiating the above relationship with respect to $x$ , We obtain

$\Rightarrow \text{ }$   $\frac{d}{dx}\left( {{x}^{2}}+xy+{{y}^{2}} \right)=\frac{d}{dx}(100)$                   Derivative of the constant function is  0

$\Rightarrow \frac{d}{dx}\left( {{x}^{2}} \right)+\frac{d}{dx}(xy)+\frac{d}{dx}\left( {{y}^{2}} \right)=0$

$\Rightarrow 2x+\left[ y\cdot \frac{d}{dx}(x)+x\cdot \frac{dy}{dx} \right]+2y\frac{dy}{dx}=0$   Using product rule and chain rule

$\Rightarrow 2x+y.1+x\cdot \frac{dy}{dx}+2y\frac{dy}{dx}=0$

$\Rightarrow 2x+y+(x+2y)\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}=-\frac{2x+y}{x+2y}$

6. Find  $\frac{dy}{dx}:{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}=81$

Ans: The given relationship is  ${{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}=81$  .

Differentiating the above relationship with respect to  $x$ ,

We get,

$\Rightarrow \text{ }$   $\frac{d}{dx}\left( {{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}} \right)=\frac{d}{dx}(81)$

$\Rightarrow \frac{d}{dx}\left( {{x}^{3}} \right)+\frac{d}{dx}\left( {{x}^{2}}y \right)+\frac{d}{dx}{x{y}^{2}}+\frac{d}{dx}\left( {{y}^{3}} \right)=0$

$\Rightarrow 3{{x}^{2}}+\left[ y\frac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}\frac{dy}{dx} \right]+\left[ {{y}^{2}}\frac{d}{dx}(x)+x\frac{d}{dx}\left( {{y}^{2}} \right) \right]+3{{y}^{2}}\frac{dy}{dx}=0$

$\Rightarrow 3{{x}^{2}}+\left[ y\cdot 2x+{{x}^{2}}\frac{dy}{dx} \right]+\left[ {{y}^{2}}\cdot 1+x\cdot 2y\cdot \frac{dy}{dx} \right]+3{{y}^{2}}\frac{dy}{dx}=0$

$\Rightarrow \left( {{x}^{2}}+2xy+3{{y}^{2}} \right)\frac{dy}{dx}+\left( 3{{x}^{2}}+2xy+{{y}^{2}} \right)=0$

$\therefore \frac{dy}{dx}=\frac{-\left( 3{{x}^{2}}+2xy+{{y}^{2}} \right)}{\left( {{x}^{2}}+2xy+3{{y}^{2}} \right)}$

7. Find  $\frac{dx}{dy}:{{\sin }^{2}}y+\cos xy=k$

Ans:   The given relationship is  ${{\sin }^{2}}y+\cos xy=k$

Differentiating the above relationship with respect to $x$ ,

We get,

$\Rightarrow \text{ }$   $\frac{d}{dx}\left( {{\sin }^{2}}y+\cos xy \right)=\frac{d}{dx}(k )$

$\Rightarrow \frac{d}{dx}\left( {{\sin }^{2}}y \right)+\frac{d}{dx}(\cos xy)=0$

Using the chain rule,

We get  $\frac{d}{dx}\left( {{\sin }^{2}}y \right)=2\sin y\frac{d}{dx}(\sin y)=2\sin y\cos y\frac{dy}{dx}$

$\frac{d}{dx}(\cos xy)=-\sin xy\frac{d}{dx}(xy)=-\sin xy\left[ y\frac{d}{dx}(x)+x\frac{dy}{dx} \right]$

$=-\sin xy\left[ y\cdot 1+x\frac{dy}{dx} \right]=-y\sin xy-x\sin xy\frac{dy}{dx}$

From the above equations we get  $2\sin y\cos y\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx}=0$

$\Rightarrow (2\sin y\cos y-x\sin xy)\frac{dy}{dx}=y\sin xy$

$\Rightarrow (\sin 2y-x\sin xy)\frac{dx}{dy}=y\sin xy$

$\frac{dx}{dy}=\frac{y\sin xy}{\sin 2y-x\sin xy}$

8. Find  $\frac{dy}{dx}={{\sin }^{2}}x+{{\cos }^{2}}y=1$

Ans:  The given relationship is  ${{\sin }^{2}}x+{{\cos }^{2}}y=1$

Differentiating the above relationship with respect to  $x$ ,

We get  $\frac{dy}{dx}\left( {{\sin }^{2}}x+{{\cos }^{2}}y \right)=\frac{d}{dx}(1)$

$\Rightarrow \frac{d}{dx}\left( {{\sin }^{2}}x \right)+\frac{d}{dx}\left( {{\cos }^{2}}y \right)=0$

$\Rightarrow 2\sin x\cdot \frac{d}{dx}(\sin x)+2\cos y\cdot \frac{d}{dx}(\cos y)=0$

$\Rightarrow 2\sin x\cos x+2\cos y(-\sin y)\cdot \frac{dy}{dx}=0$

$\sin 2x-\sin 2y\frac{dy}{dx}=0$

$\frac{dx}{dy}=\frac{\sin 2x}{\sin 2y}$

9. Find  $\frac{dy}{dx}=y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)$

Ans:  The given relationship is  $y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)$

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

$\Rightarrow \sin y=\frac{2x}{1+{{x}^{2}}}$

Differentiating the above relationship with respect to  $x$ ,

We get,

$\Rightarrow \text{ }$  $\frac{d}{dx}(\sin y)=\frac{d}{dx}\left( \frac{2x}{1+{{x}^{2}}} \right)$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{d}{dx}\left( \frac{2x}{1+{{x}^{2}}} \right)$

The function $\frac{2x}{1+{{x}^{2}}}$  is of the form of $\frac{u}{v}$ . Therefore, by quotient rule, we get  $\frac{d}{dx}\left( \frac{2x}{1+{{x}^{2}}} \right)=\frac{\left( 1+{{x}^{2}} \right)\frac{d}{dx}(2x)-2x\cdot \frac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$=\frac{\left( 1+{{x}^{2}} \right)\cdot 2-2x[0+2x]}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\frac{2+2{{x}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\frac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

Also,  $\sin y=\frac{2x}{1+{{x}^{2}}}$

$\Rightarrow \cos y=\sqrt{1-{{\sin }^{2}}y}=\sqrt{1-{{\left( \frac{2x}{1+{{x}^{2}}} \right)}^{2}}}=\sqrt{\frac{{{\left( 1+{{x}^{2}} \right)}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$

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

From above equations , we get

$\Rightarrow \text{ }$   $\frac{1-{{x}^{2}}}{1+{{x}^{2}}}\times \frac{dy}{dx}=\frac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$\Rightarrow \frac{dy}{dx}=\frac{2}{1+{{x}^{2}}}$

10. Find $\dfrac{d y}{d x} \text { in, } y=\tan ^{-1}\left(\dfrac{3 x-x^{3}}{1-3 x^{2}}\right),-\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}$

Ans: $y=\tan ^{-1}\left(\dfrac{3 x-x^{3}}{1-3 x^{2}}\right)$

Putting $\mathrm{x}=\tan \theta$

$y=\tan ^{-1}\left(\dfrac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \\$

$y=\tan ^{-1}(\tan 3 \theta) \quad\left(\tan 3 \theta=\dfrac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \\$

$y=3 \theta$

Differentiating both sides w.r.t. $x$.

$\dfrac{d(y)}{d x}=\dfrac{d 3\left(\tan ^{-1} x\right)}{d x} \\$

$\dfrac{d y}{d x}=3 \dfrac{d\left(\tan ^{-1} x\right)}{d x} \\$

$\dfrac{d y}{d x}=3\left(\dfrac{1}{1+x^{2}}\right) \quad\left(\left(\tan ^{-1} x\right)^{\prime}=\dfrac{1}{1+x^{2}}\right) \\$

$\dfrac{d y}{d x}=\dfrac{3}{1+x^{2}}$

11. Find  $\frac{dy}{dx}:y={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right),0 < x < 1$

Ans: The given relationship is,

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

$\Rightarrow \cos y=\frac{1-{{x}^{2}}}{1+{{x}^{2}}}$

$\Rightarrow \frac{1-{{\tan }^{2}}\frac{y}{2}}{1+{{\tan }^{2}}\frac{y}{2}}=\frac{1-{{x}^{2}}}{1+{{x}^{2}}}$

On comparing L.H.S. and R.H.S. of the above relationship,

We get

$\Rightarrow \text{ }$  $\tan \frac{y}{2}=x$

$\Rightarrow {\text{\; }}y = 2\left( {{{\tan }^{ - 1}}x} \right)$

Differentiating the above relationship with respect to

x,

We get

$\Rightarrow {\text{ }}$   $\frac{{dy}}{{dx}} = \frac{{d\left( {2{{\tan }^{ - 1}}x} \right)}}{{dx}}$

$\Rightarrow \frac{{dy}}{{dx}} = 2\frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}}$

$\Rightarrow \frac{{dy}}{{dx}} = 2\left( {\frac{1}{{1 + {x^2}}}} \right)$

$\therefore \frac{{dy}}{{dx}} = \frac{2}{{1 + {x^2}}}$

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

Ans: The given relationship is  $y={{\sin }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$

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

$\Rightarrow \sin y=\frac{1-{{x}^{2}}}{1+{{x}^{2}}}$

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

$\Rightarrow (1+\sin y){{x}^{2}}=1-\sin y$

$\Rightarrow {{x}^{2}}=\frac{1-\sin y}{1+\sin y}$

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

$\Rightarrow x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}$

$\Rightarrow x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}$

$\Rightarrow x=\tan \left( \frac{\pi }{4}-\frac{\pi }{2} \right)$

Differentiating the above relationship with respect to $x$ ,

We get,

$\Rightarrow \text{ }$   $\frac{d}{dx}(x)=\frac{d}{dx}\left[ \tan \left( \frac{\pi }{4}-\frac{y}{2} \right) \right]$

$\Rightarrow 1={{\sec }^{2}}\left( \frac{\pi }{4}-\frac{y}{2} \right)\cdot \frac{d}{dt}\left( \frac{\pi }{4}-\frac{y}{2} \right)$

$\Rightarrow 1=\left[ 1+{{\tan }^{2}}\left( \frac{\pi }{4}-\frac{y}{2} \right)\left( -\frac{1}{2}\frac{dy}{dx} \right) \right.$

$\Rightarrow 1=\left( 1+{{x}^{2}} \right)\left( -\frac{1}{2}\frac{dy}{dx} \right)$

$\Rightarrow \frac{dy}{dx}=\frac{-2}{1+{{x}^{2}}}$

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

Ans: The given relationship is

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

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

$\Rightarrow \cos y=\frac{2x}{1+{{x}^{2}}}$

Differentiating the above relationship with respect to $\text{x}$ ,

We get,

$\Rightarrow \text{ }$  $\frac{d}{dx}(\cos y)=\frac{d}{dx}\left( \frac{2x}{1+{{x}^{2}}} \right)$

$\Rightarrow -\sin y\frac{dy}{dx}=\frac{\left( 1+{{x}^{2}} \right)\cdot \frac{d}{dx}(2x)-2x\cdot \frac{d}{dt}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$\Rightarrow -\sqrt{1-{{\cos }^{2}}y}\frac{dy}{dx}=\frac{\left( 1+{{x}^{2}} \right)\times 2- 2x\cdot 2x}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$\Rightarrow \left[ \sqrt{1-{{\left( \frac{2x}{1+{{x}^{2}}} \right)}^{2}}} \right]\frac{dy}{dx}=-\left[ \frac{2{{(1-{{x}^{2}})}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right]$

$\Rightarrow \sqrt{\frac{{{\left( 1+{{x}^{2}} \right)}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}\frac{dy}{dx}=\frac{-2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$=\sqrt{\frac{{{\left( 1-{{x}^{2}} \right)}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}\frac{dy}{dx}=\frac{-2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$\Rightarrow \frac{1-{{x}^{2}}}{1+{{x}^{2}}}\cdot \frac{dy}{dx}=\frac{-2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$

$\Rightarrow \frac{dy}{dx}=\frac{-2}{1+{{x}^{2}}}$

14. Find  $\frac{dy}{dx}:y={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right),-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Ans:  Relationship is  $y={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$

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

$\Rightarrow \sin y=2x\sqrt{1-{{x}^{2}}}$

Differentiating the above relationship with respect to  $x$ , we get  $\cos y \frac{dy}{dx}=2\left[ x\frac{d}{dx}\left( \sqrt{1-{{x}^{2}}} \right)+\sqrt{1-{{x}^{2}}}\frac{dx}{dx} \right]$

$=\sqrt{1-{{\sin }^{2}}y}\frac{dy}{dx}=2\left[ \frac{x}{2}\cdot \frac{-2x}{\sqrt{1-{{x}^{2}}}}+\sqrt{1-{{x}^{2}}} \right]$

$=\sqrt{1-{\left( 2x\sqrt{1-{{x}^{2}}}\right)}^{2}}\frac{dy}{dx}=2\left[ \frac{-{{x}^{2}}+1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}} \right]$

$=\sqrt{1-4{{x}^{2}}\left( 1-{{x}^{2}} \right)}\frac{dy}{dx}=2\left[ \frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}} \right]$

$=\sqrt{{{(1-2{x}^{2})}^{2}}}\frac{dy}{dx}=2\left[ \frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}} \right]$

$=\left( 1-2{{x}^{2}} \right)\frac{dy}{dx}=2\left[ \frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}} \right]$

$\Rightarrow \frac{dy}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}}$

15. Find  $\frac{dy}{dx}:y={{\sec }^{-1}}\left( \frac{1}{2{{x}^{2}}-1} \right),0 < x < \frac{1}{\sqrt{2}}$

Ans:  The given relationship is  $y={{\sec }^{-1}}\left( \frac{1}{2{{x}^{2}}-1} \right)$

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

$\Rightarrow \sec y=\frac{1}{2{{x}^{2}}-1}$

$\Rightarrow \cos y=2{{x}^{2}}-1$

$\Rightarrow 2{{x}^{2}}=1+\cos y$

$\Rightarrow 2{{x}^{2}}=2{{\cos }^{2}}\frac{y}{2}$

$\Rightarrow x=\cos \frac{y}{2}$

Differentiating the above relationship with respect to $x$ ,

We get,

$\Rightarrow \text{ }$  $\frac{d}{dx}(x)=\frac{d}{dx}\left( \cos \frac{y}{2} \right)$

$\Rightarrow 1=-\sin \frac{y}{2}\cdot \frac{d}{dx}\left( \frac{y}{2} \right)$

$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-{{\cos }^{2}}\frac{y}{2}}}$

$\Rightarrow \frac{dy}{dx}=\frac{-2}{\sqrt{1-{{x}^{2}}}}$

## Conclusion

NCERT Solutions for Class 12 Maths Ex 5.3 Continuity and Differentiability by Vedantu are essential for mastering the concepts of differentiability. This exercise emphasizes understanding and applying various differentiation techniques such as the product rule, quotient rule, chain rule, implicit differentiation, and higher-order derivatives. These skills are crucial for solving complex problems in board exams and competitive exams like JEE. In previous year's question papers, there were around 3-4 questions based on these topics, highlighting their importance in exams. Paying close attention to the detailed solutions provided by Vedantu will help you gain confidence and proficiency in handling a wide range of differentiation problems.

## Class 12 Maths Chapter 5: Exercises Breakdown

 Exercises Number of Questions Exercise 5.1 10 Short Questions and 24 Long Questions Exercise 5.2 2 Short Questions 8 Long Questions Exercise 5.4 5 Short Questions and 5 Long Questions Exercise 5.5 4 Short Questions and 14 Long Questions Exercise 5.6 11 Questions and Solutions Exercise 5.7 10 Short Questions and 7 Long Questions Miscellaneous Exercise 22 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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

1. What topics are covered in Class 12 Ex 5.3 Maths Chapter 5?

Exercise 5.3 Class 12 Maths covers various differentiation techniques including the product rule, quotient rule, chain rule, implicit differentiation, and higher-order derivatives.

2. How do you apply the quotient rule in differentiation?

The quotient rule is applied to differentiate the ratio of two functions. u(x) and v(x) are two functions, their ratio is $\frac{u\left (x \right )}{v\left ( x \right )}$ is differentiated as

$\frac{{u}'\left ( x \right )v\left ( x \right )-u\left ( x \right ){v}'\left ( x \right )}{\left [ v\left ( x \right ) \right ]^{2}}$