Class 12 ncert solution math exercise 5.3

23/01/2023 by Team Gmath

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

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

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

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

Case study relation and function 4 — Student of gade 9,planned to plant saplings along straight lines

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