Contents
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
$1 \cdot \cos y+x(-\sin y) \cdot \frac{d y}{d x}+\frac{d y}{d x} \cdot \sin x+y \cdot \cos x=\sec ^{2}(x+y)\left[1+\frac{d y}{d}\right]$
$\cos-x \sin y \frac{dy}{dx}+\sin \frac{d y}{d{x}}+y \cos x=\sec^2(x+y)+\sec^2(x+y)\frac{dy}{dx}$
$\left[\sin x-x \sin y-\sec ^{2}(x+y)\right] \frac{d y}{d x}=\sec^{2}(x+y)-\cos y-y\cos x$
$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)-\cos y-y \cos x}{\sin x-x \sin y-\sec ^{2}(x+y)}$
Question 4
$\frac{d y}{d x}$ ज्ञात करें यदि
[Find $\frac{dy}{dx}$ , if]
(i) $x^{m} y^{n}=(x-y)^{m+n}$
Sol :
Differentiating with respect to x
$m x^{m-1} \cdot y^{n}+x^{m} \cdot n y^{n-1} \cdot \frac{d y}{d x}=(m+n)(x+y)^{m+n-1} \cdot\left[1-\frac{d y}{d x}\right]$
$x^{m} y^{n}\left[\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}\right]=\frac{(m+n)(x-y)^{m+n}}{x-y}\left[1-\frac{dy}{dx}\right]$
$\left[\because x^{m} y^{n}=(x-y)^{m+n}\right]$
$\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x-y}-\frac{m+n}{x-y} \cdot \frac{d y}{dx}$
$\frac{n}{y} \frac{d y}{d x}+\frac{m+n}{x-y} \frac{d y}{d x}=\frac{m+n}{x-y}-\frac{m}{x}$
$\left[\frac{n}{y}+\frac{m+n}{x-y}\right] \frac{d y}{d x}=\frac{m+n}{x-y}-\frac{m}{x}$
$\left[\frac{n x-n y+m y+n y}{y(x-y)}\right] \frac{d y}{d x}=\frac{m x+n x-m x+m y}{x(x-y)}$
$\left[\frac{n x+m y}{y}\right] \frac{d y}{dx}=\frac{n x+m y}{x}$
$\frac{d y}{d x}=\frac{y}{x}$
(ii) $x^{3} y^{4}=(x-y)^{7}$
Sol :
Differentiating with respect to x
$3 x^{2} y^{4}+x^{3} \cdot 4 y^{3} \cdot \frac{d y}{dx}=7(x-y)^{7-1} \cdot\left[1-\frac{d y}{d x}\right]$
$x^{3} y^{4}\left[\frac{3}{x}+\frac{4}{y} \frac{d y}{d x}\right]=\frac{7(x-y)^{7}}{x-y}\left[1-\frac{dy}{d x}\right]$
$\left[\because x^{3} y^{4}=(x-y)^{7}\right]$
$\frac{3}{x}+\frac{4}{y} \frac{d}{d}=\frac{7}{x-y}-\frac{7}{x-y} \cdot \frac{dy}{dx}$
$\frac{4}{y} \frac{d y}{d x}+\frac{7}{x-y} \frac{d y}{d x}=\frac{7}{x-y}-\frac{3}{x}$
$\left[\frac{4}{y}+\frac{7}{x-y}\right] \frac{dy}{d x}=\frac{7}{x-y}-\frac{3}{x}$
$\left[\frac{4 x-4 y+7 y}{y(x-y)}\right] \frac{d y}{d x}=\frac{7 x-3 x+3 y}{x(x-y)}$
$\left[\frac{4 x+3 y}{y(x-y)}\right] \frac{d y}{d x}=\frac{4 x+3 y}{x(x-y)}$
$\frac{d y}{d x}=\frac{y}{x}$
(iii) $x^{2} y^{2}=(x-y)^{4}$
Sol :
Differentiating with respect to x
$2 x \cdot y^{2}+x^{2} \cdot 2 y \cdot \frac{d y}{d x}=4(x-y)^{4-1} \cdot\left[1-\frac{d y}{d x}\right]$
$x^{2} y^{2}\left[\frac{2}{x}+\frac{2}{y} \frac{d y}{d x}\right]=4 \frac{(x-y)^{4}}{x-y}\left[1-\frac{dy}{d x}\right]$
$\left[\because x^{2} y^{2}=(x-y)^{4}\right]$
$\frac{2}{x}+\frac{2}{y} \frac{d y}{d x}=\frac{4}{x-y}-\frac{4}{x-y} \frac{dy}{dx}$
$\frac{2}{y} \frac{d y}{d x}+\frac{4}{x-y} \frac{d y}{dx}=\frac{4}{x-y}-\frac{2}{x}$
$\left[\frac{2}{y}+\frac{4}{x-y}\right] \frac{d y}{dx}=\frac{4}{x-y}-\frac{2}{x}$
$\left[\frac{2 x-2 y+4 y}{y(x-y)]}\right] \frac{d y}{d x}=\frac{4 x-2 x+2 y}{x(x-y)}$
$\left[\frac{2 x+2 y}{y(x-y)}\right] \frac{d y}{d x}=\frac{2 x+2 y}{x(x-y)}$
$\frac{d y}{d x}=\frac{y}{x}$
(iv) $x^{2} y=(2 x+3 y)^{3}$
Sol :
Differentiating with respect to x
$2 x \cdot y+x^{2} \cdot \frac{d y}{dx}=3(2 x+3 y)^{3-1} \cdot\left(2+\frac{3 dy}{dx}\right)$
$x^{2} y\left[\frac{2}{x}+\frac{1}{y} \cdot \frac{d y}{d x}\right]=3\left(\frac{2 x+3 y}{2 x+3 y}\right)^{3}\left[2+3 \frac{d y}{d x}\right]$
$\left[\because x^{2} y=(2 x+3 y)^{3}\right]$
$\frac{2}{x}+\frac{1}{y} \frac{dy}{dx}=\frac{6}{2 x+3 y}+\frac{9}{2 x+3y}- \frac{dy}{dx}$
$\left[\frac{1}{y}-\frac{9}{2 x+3 y}\right] \frac{d y}{d x}=\frac{6}{2 x+3 y}-\frac{2}{x}$
$\left[\frac{2 x+3 y-9 y}{y(2 x+3 y)}\right] \frac{d y}{d x}=\frac{6 x-4 x-6 y}{x(2 x+3y)}$
$\left[\frac{2 x-6 y}{y(2 x+3 y)}\right] \frac{d y}{d x}=\frac{2 x-6 y}{x(2 x+3 y)}$
$\frac{d{y}}{d x}=\frac{y}{x}$
(v) $x^3y^4=(x+y)^7$
Sol :
Question 5
यदि (If ) cos y=x cos(a+y) with cosa≠ ±1 सिद्ध करे कि(prove that) $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$
Sol :
cos y=x cos(a+y)
Differentiating with respect to x
$-\sin y \cdot \frac{d y}{dx}=1 \cdot \cos (a+y)+x \cdot[-\sin (a+y)]\left[0+\frac{dy}{d x}\right]$
$-\sin y \frac{d y}{dx}=\cos(a+y)-x \sin (a+y) \frac{d y}{d x}$
$x \sin (a+y) \frac{d y}{d x}-\sin \frac{d y}{dx}=\cos (a+b)$
$[x \sin (a+y)-\sin y] \frac{d y}{dx}=\cos (a+y)$
$\left[\frac{\cos y \sin (a+y)}{\cos (a+y)}-\sin y\right] \frac{d y}{d x}=\cos (a+y)$
$\left[\frac{\sin (a+y) \cdot \cos y-\cos (a+y) \sin y}{\cos (a+y)}\right] \frac{dy}{d x}=\cos (a+y)$
$\frac{\sin (a+y-y)}{\cos (a+y)} \frac{d y}{dx}=\cos (a+y)$
$\frac{d y}{dx}=\frac{\cos ^{2}(a+y)}{\sin a}$
Question 6
यदि (If) $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots . \text { to } \infty}}}$ सिद्ध करें कि (prove that) $(2 y-1) \frac{d y}{d x}+\sin x=0$
Sol :
$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots . \text { to } \infty}}}$
$y=\sqrt{\cos x+y}$
Squaring both sides
$y^{2}=\cos x+y$
Differentiating with respect to x
$2 y \cdot \frac{d y}{d x}=-\sin x+\frac{d y}{d x}$
$2 y \frac{d y}{d x}-\frac{d y}{d x}+\sin x=0$
$(2 y-1) \frac{d y}{d x}+\sin x=0$
Question 7
यदि (If) $y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$ सिद्ध करें कि( prove that) $\left(x^{2}-y^{2}+3\right) \frac{d y}{d x}=1$
Sol :
$y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$
$y=x+\frac{1}{y}$
Differentiating with respect to x
$\frac{d y}{d x}=1-\frac{1}{y^{2}} \cdot \frac{d y}{d x}$
$\frac{d y}{dx}+\frac{1}{y^{2}} \frac{d y}{d x}=1$
$\left[1+\frac{1}{y^{2}}\right] \frac{d y}{d x}=1$
$\left[y^{2}-2 \cdot y \cdot \frac{1}{y}+\frac{1}{y^{2}}-y^{2}+3\right] \frac{d y}{d x}=1$
$\left[\left(y-\frac{1}{y}\right)^{2}-y^{2}+3\right] \frac{d y}{d x}=1$
$\left[x^{2}-y^{2}+3\right] \frac{d y}{d x}=1$
Question 8
यदि (If) $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$ सिद्ध करे कि (prove that) $\frac{d y}{d x}=\frac{1}{2 y-1}$
Sol :
$y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$
$y=\sqrt{x+y}$
Squaring both sides
$y^{2}=x+y$
Differentiating with respect to x
$2 y.\frac{d y}{dx}=1+\frac{d y}{d x}$
$2 y \frac{d y}{d x}-\frac{d y}{d x}=1$
$(2 y-1) \frac{d y}{d x}=1$
$\frac{d y}{d x}=\frac{1}{2 y-1}$
Question 9
यदि (If) $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Sol :
$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$
Differentiating with respect to x
$\frac{1}{2 \sqrt{1-x^{2}}} \times(-2 x)+\frac{1}{2 \sqrt{1-y^{2}}} \times(-2 y) \cdot \frac{d y}{d x}=a\left[1-\frac{dy}{d x}\right]$
$\frac{-x}{\sqrt{1-x^{2}}}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a-\frac{a d y}{d x}$
$a \frac{d y}{d x}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$
$\left[a-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}+\frac{x}{\sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}+1-y^{2}-x y+y^{2}}{(x-y) \sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{1-x^{2}+\sqrt{1-x^{2}} \sqrt{1-y^{2}}+x^{2}-xy}{(x-y) \sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-x^{2}}}$
$\frac{dy}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Question 10
यदि (If) $y=x+\frac{1}{x}$ सिद्ध करें कि (prove that) $x \frac{d y}{d x}+y=2 x$ $x \frac{d y}{d x}+y=2 x$
Sol :
$y=x+\frac{1}{x}$
$y=\frac{x^{2}+1}{x}$
$x y=x^{2}+1$
Differentiating with respect to x
$1 \cdot y+x \frac{d y}{d x}=2 x+0$
$x \frac{d y}{d x}+y=2 x$