Class 12 ncert solution math exercise 5.3

12/10/202323/01/2023 by Team Gmath

The Class 12 NCERT solution math exercise 5.3 prepared by expert Mathematics teacher at gmath.in as per CBSE guidelines. See our Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your Board and School exams.

Exercise 5.3

Find $\frac{dy}{dx}$ in the following(Class 12 ncert solution math exercise 5.3)

Question 1: 2x + 3y = sin x

Solution: 2x + 3y = sin x

Differentiate with respect to x

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

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

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

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

Question 2:- 2x + 3y = sin y

Solution :- 2x + 3y = sin y

Differentiate w.r.t. x
$\frac{d}{dx}(2x+3y)=\frac{d}{dx}\sin y$

\frac{}{} $\Rightarrow 2.1 + 3.\frac{dy}{dx}= \cos y\frac{dy}{dx}$

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

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

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

Question3: $ax + by^2 = cos y$

Solution: $ax + by^2 = cos y$

Differentiate with respect to x

$\frac{d}{dx}(ax+by^2)=\frac{d}{dx}\cos y$

$\Rightarrow a.1+b.2y\frac{dy}{dx}=-\sin y\frac{dy}{dx}$

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

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

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

Question 4:- $xy + y^2 = \tan x + y$

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

Differentiate with respect to x

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

$\Rightarrow x\frac{d}{dx}y+y\frac{d}{dx}x+\frac{d}{dx}y^2=\frac{d}{dx}\tan x+\frac{d}{dx}y$

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

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

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

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

Question 5: $x^2 + xy + y^2 = 100$

Solution : $x^2 + xy + y^2 = 100$

Differentiate with respect to x

$\frac{d}{dx}(x^2+xy+y^2)=\frac{d}{dx}100$

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

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

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

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

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

Question 6: $x^3 + x^2y + xy^2 + y^3 = 81$

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

Differentiate with respect to x

$\frac{d}{dx}(x^3 + x^2y + xy^2 + y^3)=\frac{d}{dx}81$

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

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+y.2x+x.2y\frac{dy}{dx}+y^2.1+3y^2\frac{dy}{dx}=0$

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

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

Question 7: $\sin^2 y + \cos (xy) =k$

Solution: $\sin^2 y + \cos (xy) =k$

Differentiate with respect to x

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

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

$\Rightarrow 2\sin y\frac{d}{dx}\sin y -\sin(xy)\frac{d}{dx}x.y=0$

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

$\Rightarrow \sin 2y\frac{dy}{dx}-\sin(xy)(x\frac{dy}{dx}+y.1)= 0$

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

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

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

Question 8: $\sin^2x+\cos^2y=1$

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

Differentiate with respect to x

$\frac{d}{dx}(\sin^2x+\cos^2y)=\frac{d}{dx}1$

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

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

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

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

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

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

Question 9: $y=\sin^{-1}\left(\frac{2x}{1+x^2}\right)$

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

Let $x = \tan\theta$

Then,

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

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

$\Rightarrow y = 2\theta$

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

Differentiate with respect to x

$\frac{dy}{dx} = \frac{d}{dx}2\tan^{-1}x$

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

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

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

Question 10: $y =\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

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

Let $x = \tan\theta$

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

$\Rightarrow y = \tan^{-1}\tan 3\theta$

$\Rightarrow y = 3\theta$

$\Rightarrow y = 3\tan^{-1}x$

Differentiate with respect to x

$\frac{dy}{dx}=3\frac{d}{dx}\tan^{-1}x$

$\Rightarrow \frac{dy}{dx} = 3\frac{1}{1+x^2}$

$\Rightarrow \frac{dy}{dx} = \frac{3}{1+x^2}$

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

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

Let $x = \tan\theta$

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

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

$\Rightarrow y = 2\theta$

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

Differentiate with respect to x

$\frac{dy}{dx} = 2\frac{d}{dx}\tan^{1}x$

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

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

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

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

Let $x = \tan\theta$

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

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

$\Rightarrow y = \sin^{-1}\sin(\frac{\pi}{2}-2\theta)$

$\Rightarrow y = \frac{\pi}{2}-2\theta$

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

Differentiate with respect to x

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

$\Rightarrow \frac{dy}{dx} = 0-2\frac{d}{dx}\tan^{-1}x$

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

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

Question 13: $y =\cos^{-1}\left(\frac{2x}{1+x^2}\right),-1<x<1$

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

Let $x= \tan\theta$

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

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

$\Rightarrow y= \cos^{-1}\cos(\frac{\pi}{2}-2\theta)$

$\Rightarrow y = \frac{\pi}{2}-2\theta$

$\Rightarrow y = \frac{\pi}{2}-2\tan^{-1}\theta$

Differentiate with respect to x

$\frac{dy}{dx}=0-2\frac{d}{dx}\tan^{-1}x$

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

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

Question 14: $y= \sin^{-1}(2x\sqrt{1-x^2}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

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

Let $x = \sin\theta$

$y = \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})$

$\Rightarrow y= \sin^{-1}(2\sin\theta\cos\theta)$

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

$\Rightarrow y = 2\theta$

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

Differentiate with respect to x

$\frac{dy}{dx} = 2\frac{d}{dx}\sin^{-1}x$

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

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

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

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

Let, $x = \cos \theta$

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

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

$\Rightarrow y = \sec^{-1}\sec2\theta$

$\Rightarrow y = 2\theta$

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

Differentiate with respect to x

$\frac{dy}{dx} = 2\frac{d}{dx}\cos^{-1}x$

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

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

NCERT solution chapter 5 continuity and differentiability

1. Ncert solution Exercise 5.1 continuity and differentiability

2. Ncert solution Exercise 5.2 continuity and differentiability

3. Ncert solution Exercise 5.4 continuity and differentiability

4. Ncert solution Exercise 5.5 continuity and differentiability

5. Ncert solution Exercise 5.6 continuity and differentiability

6. Ncert solution Exercise 5.7 continuity and differentiability

7. Ncert solution Chapter 5 Miscellaneous continuity and differentiability

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