KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.3 (Q1-Q3)

Exercise 11.3

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Question 1

$\frac{d y}{d x}$ ज्ञात करें यदि

[Find $\frac{d y}{d x}$ if ,]

(i) $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Sol :
Differentiating with respect to x

$\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} \cdot \frac{d y}{d x}=0$

$\frac{2}{3} y^{-\frac{1}{3}} \frac{d y}{d x}=-\frac{2}{3} x^{-\frac{1}{3}}$

$\frac{d y}{d x}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

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

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


(ii) $x^{n}+y^{n}=a^{n}$
Sol :
Differentiating with respect to x

$n \cdot x^{n-1}+n y^{n-1} \cdot \frac{d y}{d x}=0$

$\frac{d\left(x^{n}\right)}{dx}+\frac{d\left(y^{n}\right)}{d(y)} \times \frac{d y}{d x}=\frac{d\left(a^{n}\right)}{dx}$

$ny^{n-1} \cdot \frac{d y}{dx}=-n x^{n-1}$

$\frac{d y}{d x}=-\frac{x^{n-1}}{y^{n-1}}$


(iii) $x^{2}+y^{2}-x y=4$
Sol :
Differentiating with respect to x

$2 x+2 y \frac{d y}{d x}-\left(1 . y+x \cdot \frac{d y}{d}\right)=0$

$2 x+2 y \frac{d y}{dx}-y-x \frac{d y}{d x}=0$

$(2 y-x) \frac{d y}{dx}=y-2 x$

$\frac{d y}{d x}=\frac{y-2 x}{2 y-x}$


(iv) $x+y=x y^{3}$
Sol :
Differentiating with respect to x

$1+\frac{d y}{dx}=1 \cdot y^{3}+x \cdot 3 y^{2} \cdot \frac{d y}{dx}$

$\frac{d y}{dx}-3 x y^{2} \frac{d y}{d x}=y^{3}-1$

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

$\frac{d y}{d x}=\frac{y^{3}-1}{1-3x y^{2}}$

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

$=\frac{\frac{x+y-x}{x}}{1-3 x y^{2}}$

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


(v) $x \sqrt{y}+y \sqrt{x}=1$
Sol :
Differentiating with respect to x

$1 . \sqrt{y}+x \cdot \frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}+\frac{d y}{d x} \cdot \sqrt{x}+y \frac{1}{2 \sqrt{x}}=0$

$\left(\frac{x}{2\sqrt y}+\sqrt{x}\right) \frac{d y}{d}=-\left(\frac{y}{2 \sqrt{x}}+\sqrt{y}\right)$

$\left(\frac{x+2 \sqrt{x y}}{2 \sqrt{y}}\right) \frac{d y}{d x}=-\left(\frac{y+2 \sqrt{x y}}{2 \sqrt{x}}\right)$

$\frac{dy}{d x}=\frac{-\sqrt{y}(y+2 \sqrt{2 y})}{\sqrt{x}(x+2 \sqrt{x y})}$

$=-\frac{y(\sqrt{y}+\sqrt{x})}{x(\sqrt{x}+2 \sqrt{y})}$


(vi) $\left(x^{3}+y^{3}\right) x y=x^{5}-y^{5}$
Sol :
$x^{4} y+x y^{4}=x^{5}-y^{5}$

Differentiating with respect to x

$4 x^{3} \cdot y+x^{4} \frac{d y}{d x}+1 y^{4}+x \cdot 4 y^{3} \frac{d y}{d x}=5 x^{4}-5 y^{4}\frac{dy}{dx}$

$x^{4} \frac{d y}{d x}+4 x y^{3} \frac{d y}{d x}+5 y^{4} \frac{d y}{d}=5 x^{4}-4 x^{3} y -y^{4}$

$\left(x^{4}+4 x y^{3}+5 y^{4}\right) \frac{d y}{dx}=5 x^{4}-4 x^{3} y-y^{4}$

$\frac{d y}{d x}=\frac{5 x^{4}-4 x^{3} y-y^{4}}{x^{4}+4 x y^{3}+5 y^{4}}$

Question 2

$\frac{d y}{d x}$ ज्ञात करें यदि

[Find $\frac{d y}{d x}$ , if]

(i) x-y=𝜋
Sol :
Differentiating with respect to x

$1-\frac{d{y}}{d{x}}=0$

$1=\frac{d{y}}{d x}$


(ii) 2x+3y=sin x
Sol :
Differentiating with respect to x

$2+\frac{3 \cdot d y}{dx}=\cos x$

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

$\frac{d y}{d x}=\frac{1}{3}(\cos x-2)$


(iii) y+sin y =cos x
Sol :
Differentiating with respect to x

$\frac{d y}{d x}+\cos y \cdot \frac{d y}{d x}=-\sin x$

$\left(1+\cos y\right) \cdot \frac{d y}{d x}=-\sin x$

$\frac{d y}{d x}=\frac{-\sin x}{1+\cos y}$


(iv) ax+by²=cos y
Sol :
Differentiating with respect to x

$a+b \cdot 2 y \cdot \frac{d y}{d x}=-\sin y \cdot \frac{d y}{dx}$

$2 b y \frac{d y}{dx}+\sin y \frac{d y}{d x}=-a$

$(2 b y+\sin y) \frac{d y}{d x}=-a$

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


(v) x²+xy+y²=100
Sol :
Differentiating with respect to x

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

$(x+2 y) \frac{d y}{d x}=-(2 x+y)$

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


(vi) sin²y+cos(xy)=𝜋
Sol :
Differentiating with respect to x

$2 \sin y \cdot \cos y \cdot \frac{d y}{d x}+(-\sin (x y)] \cdot\left[1 \cdot y+x \cdot \frac{d y}{d x}\right]=0$

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

$[\sin 2 y-x \sin (x y)] \frac{d y}{d x}=y \sin (x y)$

$\frac{d y}{d x}=\frac{y \sin (x y)}{\sin 2 y-x \sin (x y)}$

Question 3

$\frac{d y}{d x}$ ज्ञात करें यदि

[Find $\frac{d y}{d x}$ , if]

(i) y=sin(x+y)
Sol :
Differentiating with respect to x

$\frac{d y}{d x}=\cos (x+y) \cdot\left[1+\frac{d y}{d x}\right]$

$\frac{d y}{d x}=\cos (x+y)+\cos (x+y) \frac{d y}{d x}$

$\frac{d y}{d x}-\cos (x+y) \frac{d y}{d x}=\cos (x+y)$

$[1-\cos (x+y)] \frac{d y}{d x}=\cos (x+y)$

$\frac{d y}{d x}=\frac{\cos (x+y)}{1-\cos(x+y)}$



(ii) y=sec(x+y)

Sol :
Differentiating with respect to x

$\frac{dy}{dx}=\sec (x+y) \cdot \tan (x+y) \cdot\left[1+\frac{d y}{dx}\right]$

$\frac{d{y}}{d{x}}=\sec (x+y) \tan (x+y)+\sec (x+y) \tan (x+y)\frac{dy}{dx}$

$\frac{d y}{dx}-\sec (x+y) \cdot \tan (x+y) \frac{d y}{d x}=\sec (x+y) \tan [x+y]$

$[1-\sec (x+y) \tan (x+y)] \frac{d y}{d x}=\sec (x+y) \tan (x +y)$

$\frac{d y}{d x}=\frac{\sec (x+y) \tan (x+y)}{1-\sec (x+y) \tan (x+y)}$


(iii) xy=sin(x+y)
Sol :
Differentiating with respect to x

$1 \cdot y+x \cdot \frac{d y}{d x}=\cos (x+y) \cdot\left[1+\frac{d y}{d x}\right]$

$y+x\frac{d y}{d x}=\cos (x+y)+\cos(x+y) \frac{d y}{d x}$

$x \frac{dy}{dx}-\cos (x+y) \frac{d y}{d x}=\cos (x+y)-y$

[x-\cos (x+y)] \frac{d y}{d x}=\cos (x+y)-y

$\frac{d y}{d x}=\frac{\cos (x+y)-y}{x-\cos (x+y)}$


(iv) xy=sec(x+y)
Sol :
$x=\frac{\sec (x+y)}{y}$

$y=\frac{\sec(x+y)}{x}$

Differentiating with respect to x

$1 . y+x\frac{ dy}{dx}=\sec (x+y) \cdot \tan (x+y)\left[1+\frac{d y}{d x}\right]$

$y+x\frac{dy}{d x}=\sec(x+y) \tan (x+y)+\sec (x+y)\tan(x+y)\frac{dy}{dx}$

$x \frac{d y}{d x}-\sec (x+y) \tan (x+y) \frac{d y}{d x}=\sec (x+y) \tan (x+y)-y$

$\frac{d y}{d x}=\frac{\sec (x+y) \tan (x+y)-y}{x-\sec (x+y) \cdot \tan (x+y)}$

$=\frac{y\left[\frac{\sec (x+y) \tan (x-y)}{y}-1\right]}{x\left[1-\frac{\sec (x+y) \cdot \tan (x+y)}{x}\right]}$

$=\frac{y[x \tan (x+y)-1]}{x[1-y \tan (x+y)]}$


(v) x+y=tan(xy)
Sol :
Differentiating with respect to x

$1+\frac{d y}{d x}=\sec ^{2}(x y)\left[1 \cdot y+x \cdot \frac{d y}{d x}\right]$

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

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

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

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


(vi) x=y cosec(xy)
Sol :
$x=\frac{y}{\sin \left(x{y}\right)}$

x.sin(xy)=y

Differentiating with respect to x

$1 \cdot \sin (x y)+x \cdot \cos (x y)\left[1 \cdot y+x \cdot \frac{d y}{d x}\right]=\frac{d y}{d x}$

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

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

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


(vii) x-y=sec(x+y)
Sol :
Differentiating with respect to x

$\frac{1-d y}{d x}=\sec (x+y) \tan (x+y)\left[1+\frac{d y}{d x}\right]$

$1-\frac{dy}{dx}=\sec (x+y) \tan (x+y)+\sec (x+y)\tan (x+y)\frac{dy}{d x}$

$1-\sec (x+y) \tan (x+y)=\frac{d y}{d x}+\sec (x+y)\tan(x+y)\frac{dy}{dx}$

$\frac{1-\sec (x+y) \tan (x+y)}{1+\sec(x+)) \tan (x+y)}=\frac{d y}{d x}$


(viii) $x^{2} y^{2}=\sin (x y)$
Sol :
Differentiating with respect to x

$2 x \cdot y^{2}+x^{2} \cdot 2 y \cdot \frac{d y}{d x}=\cos (x y)\left[1. y+x \cdot \frac{d y}{dx}\right]$

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

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

$x[2 x y-\cos (x y)] \frac{d y}{d x}=-y\left[2 x y-\cos\left(x{y}\right)\right]$

$\frac{d y}{dx}=\frac{-y}{x}$


(ix) xy=tan(x+y)
Sol :
Differentiating with respect to x

$1 \cdot y+x \cdot \frac{d y}{d}=\sec ^{2}(x+y)\left[1+\frac{dy}{dx}\right]$

$y+x \frac{d y}{d x}=\sec ^{2}(x+y)+\sec ^{2}(x+y) \cdot \frac{d y}{d x}$

$\left[x-\sec ^{2}(x+y)\right] \frac{d y}{dx}=\sec ^{2}(x+y)-y$

$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)-y}{x-\sec ^{2}(x+y)}$


(x) xcosy+ysinx=tan(x+y)
Sol :
Differentiating with respect to x

$1 \cdot \cos y+x(-\sin y) \cdot \frac{d y}{d x}+\frac{d y}{d x} \cdot \sin x+y \cdot \cos x=\sec ^{2}(x+y)\left[1+\frac{d y}{d}\right]$

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

$\left[\sin x-x \sin y-\sec ^{2}(x+y)\right] \frac{d y}{d x}=\sec^{2}(x+y)-\cos y-y\cos x$

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