KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 अवकलन

Question 1

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=acosθ , y=asinθ
Sol :
Differentiate w.r.t θ

$\frac{d x}{d \theta}=-a \sin \theta$..(i)

$\frac{d y}{d \theta}=a \cos \theta$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \cos \theta}{- a \sin \theta}$

=-cotθ

(ii) x=acosθ , y=bcosθ
Sol :
Differentiate w.r.t θ

$\frac{d x}{d \theta}=-a \sin \theta$..(i)

$\frac{d y}{d \theta}=-b \sin \theta$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-b \sin \theta}{- a \sin \theta}$

$\frac{d y}{d x}=\frac{b}{a}$

Question 2

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=at²,y=2at
Sol :
Differentiate w.r.t t

$\frac{d x}{d t}=2 a t$..(i)

$\frac{d y}{d t}=2 a$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2a}{2at}$

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

(ii) x=4t , $y=\frac{4}{t}$
Sol :
Differentiate w.r.t t

$\frac{d x}{d t}=4$..(i)

$\frac{d y}{d t}=-\frac{4}{t^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\dfrac{\frac{d y}{d t}}{\frac{dx}{dt}}=\dfrac{\frac{-4}{t^2}}{4}$

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

Question 3

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=sint,y=cos2t
Sol :
Differentiate w.r.t t

$\frac{d x}{d t}=\cos \theta$..(i)

$\frac{d y}{d t}=-\sin 2 t \times 2$

$\frac{dy}{d t}=-2 .\sin 2t$

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-2 \cdot \cos t}{\cos t}$

$\frac{d y}{d x}=\frac{-2 \times 2\sin t \cos t}{\cos t}$

⇒-4sint

(ii) x=asecθ,y=btanθ
Sol :

Differentiating w.r.t θ

$\frac{d x}{d \theta}=a \sec \theta \tan \theta$..(i)

$\frac{d y}{d \theta}=b \sec ^{2} \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}$

$\frac{d y}{d x}=\frac{b}{a}\times \frac{1}{\cos \theta}$

$=\frac{b}{a} \operatorname{cosec} \theta$

Question 4

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=a(θ-sinθ), y=a(1-cosθ)
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=a(1-\cos \theta)$

$\frac{d x}{d \theta}=a \cdot 2 \sin ^{2} \frac{\theta}{2}$

$\frac{d x}{d \theta}=2 a \sin ^{2} \frac{\theta}{2}$..(i)

अब , y=(1-cosθ)

Differentiating w.r.t θ

$\frac{d y}{d \theta}=a \sin \theta=a \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$

$\frac{d y}{d \theta}=2 a \sin \frac{\theta}{2} \cos \frac{\theta}{2}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 a \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 a \sin^2 \frac{\theta}{2}}$

$\frac{d y}{d x}=\cot \frac{\theta}{2}$

(ii) x=a(θ-sinθ),y=a(1+cosθ)
Sol :

Question 5

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=sinθ,y=θ+cosθ
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=\cos \theta$..(i)

$\frac{d y}{d \theta}=1-\sin \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{dx}{d\theta}}=\frac{1-\sin \theta}{\cos \theta}$

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

(ii) x=10(t-sint),y=12(1-cost),$-\frac{\pi}{2}<t \leq \frac{\pi}{2}$
Sol :
Differentiating w.r.t t

$\frac{d x}{d t}=10(1-\cos t)$

$\frac{d x}{d t}=10 \times 2 \sin ^{2} \frac{t}{2}$

$\frac{d x}{d t}=20 \sin^2 \frac{t}{2}$

अब , y=12(1-cost)

Differentiating w.r.t t

$\frac{d y}{d t}=12 \sin t=12 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}$

$\frac{d y}{d t}=24 \sin \frac{t}{2} \cos \frac{t}{2}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{24 \sin \frac{t}{2} \cos \frac{t}{2}}{20 \sin ^{2} \frac{t}{2}}$

$\frac{dy}{d t}=\frac{6}{5} \cos \frac{t}{2}$

Question 6

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=3cosθ-cos³θ,y=3sinθ-sin³θ
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=-3 \sin \theta-3 \cos ^{2} \theta(-\sin \theta)$

$\frac{d x}{d \theta}=-3 \sin \theta+3 \cos ^{2} \theta \sin \theta$

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

$\frac{d x}{d t}=-3 \sin^{3} \theta$..(i)

अब, y=3sinθ-sin³θ

Differentiating w.r.t θ

$\frac{d y}{d \theta}=3 \cos \theta-3 \sin ^{2} \theta \cdot \cos \theta$

=3cosθ(1-sin²θ)

$\frac{d y}{d \theta}=3 \cos ^{3} \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \cos ^{3} \theta}{-3 \sin^3 \theta}$

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

(ii) x=cosθ-cos2θ,y=sinθ-sin³θ
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=-\sin \theta+\sin 2 \theta \cdot 2$

$\frac{d x}{d \theta}=2 \sin 2 \theta-\sin \theta$..(i)

अब , y=sinθ–sin³θ

Differentiating w.r.t θ

$\frac{d y}{d t}=\cos \theta-3 \sin ^{2} \theta \cdot \cos \theta$

$\frac{d y}{d \theta}=\cos \theta\left(1-3 \sin ^{2} \theta\right)$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

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

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

Question 7

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) $x=\sqrt{1+t^{2}}, y=\sqrt{1-t^{2}}$ , $x=t+\frac{1}{t}, y=t-\frac{1}{t}$
Sol :
$x=\sqrt{1+t^{2}}$

Differentiating w.r.t t

$\frac{d x}{d t}=\frac{1}{2 \sqrt{1+t^2}} \times 2t$

$\frac{d x}{d t}=\frac{t}{\sqrt{1+t^{2}}}$..(i)

अब, $y=\sqrt{1-t^{2}}$

Differentiating w.r.t t

$\frac{d y}{d t}=\frac{1}{2 \sqrt{1-t^{2}}} \times(-2 t)$

$\frac{d y}{d t}=\frac{-t}{\sqrt{1-t^{2}}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

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

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

(ii) $x=t+\frac{1}{t}, y=t-\frac{1}{t}$
Sol :
$x=t+\frac{1}{t}$

Differentiating w.r.t t

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

$\frac{d x}{d t}=\frac{t^{2}-1}{t^{2}}$..(i)

अब ,$y=t-\frac{1}{t}$

Differentiating w.r.t t

$\frac{dy}{d t}=1+\frac{1}{t^{2}}$

$\frac{d y}{d t}=\frac{t^{2}+1}{t^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

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

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

Question 8

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=e^t+sint,t=logt
Sol :
Differentiating w.r.t t

$\frac{d x}{d t}=e^{t}+\cos t$..(i)

$\frac{d y}{d t}=\frac{1}{t}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{t}}{e^{t}+\cos t}$

$\frac{d_{y}}{d x}=\frac{1}{t\left(e^{t}+\cos t\right)}$

(ii) $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$
Sol :
Differentiating w.r.t t

$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\tan \frac{t}{2}} \times \sec ^{2} \frac{t}{2} \times \frac{1}{2}\right)$

$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\frac{\sin t / 2}{\cos t / 2}} \times \frac{1}{\cos ^{2} \frac{t}{2}} \times \frac{1}{2}\right)$

$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right)$

$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\sin t}\right)$

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

$\frac{d x}{d t}=a \frac{\cos ^{2} t}{\sin t}$..(i)

अब,y=asint

Differentiating w.r.t t

$\frac{d y}{d t}=a \cos t$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \cos t}{\frac{a \cos^{2} t}{sin t}}$

$\frac{d y}{d x}=\tan t$

Question 9

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

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

Differentiating w.r.t t

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

$\frac{d x}{d t}=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}$

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

$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$..(i)

अब, $y=\frac{1-t^{2}}{1+t^{2}}$

Differentiating w.r.t t

$\frac{d y}{d t}=\frac{-2 t\left(1+t^{2}\right)-\left(1-t^{2}\right) \cdot 2 t}{\left(1+t^{2}\right)^{2}}$

$=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}}$

$\frac{d y}{d t}=-\frac{4 t}{\left(1+t^{2}\right)^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{-4 t}{\left(1+t^{2}\right)^{2}}}{\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$

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

Question 10

यदि(If) $x=\frac{1-t^{2}}{1+t^{2}}, y=\frac{2 t}{1+t^{2}}$ , सिद्ध करे कि (prove that) $\frac{d y}{d x}+\frac{x}{y}=0$
Sol :
$x=\frac{1-t^{2}}{1+t^{2}}$

Differentiating w.r.t t

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

$=\frac{-2 t-2 t^{3}-2 t+2t^{3}}{\left(1+t^{2}\right)^{2}}$

$\frac{d x}{d t}=\frac{-4 t}{\left(1+t^{2}\right)^{2}}$..(i)

अब , $y=\frac{2 t}{1+t^{2}}$

Differentiating w.r.t t

$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)-2 t \cdot 2 t}{\left(1+t^{2}\right)^{2}}$

$=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}$

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

$\frac{d y}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d{y}}{d t}}{\frac{dx}{d t}}=\dfrac{\frac{2(1-t^2)}{(1+t^2)^2}}{\frac{-4t}{(1+t^2)^2}}$

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

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

$\frac{dy}{d x}+\frac{x}{y}=0$

Question 11

यदि(If) y=acos²θ,x=bsin²θ , सिद्ध करे कि (prove that) $\frac{d y}{d x}+\frac{a}{b}=0$
Sol :
y=acos²θ

Differentiating w.r.t θ

$\frac{d y}{d \theta}=2 a \cos \theta \cdot(-\sin \theta)$

$\frac{d y}{d \theta}=-a \sin 2 \theta$..(i)

अब , x=bsin²θ

Differentiating w.r.t θ

$\frac{d x}{d \theta}=2b \sin \theta \cdot \cos \theta$

$\frac{d x}{d \theta}=b\sin2 \theta$

समीकरण (i) मे (ii) से भाग देने पर ,

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

$\frac{dy}{dx}+\frac{a}{b}=0$

Question 12

$t=\frac{\pi}{3}$ पर $\frac{d y}{d x}$ निकले जब (Find $\frac{d y}{d x}$ at $t=\frac{\pi}{3}$ when) x=a(cost+tsint) and y=a(sint-tcost)
Sol :
x=a(cost+tsint)

Differentiating w.r.t t

$\frac{d x}{d t}=a(-\sin t+1.\sin t+t \cos t)$

$\frac{d x}{d t}=a t \cos t$..(i)

अब,

y=a(sint-tcost)

Differentiating w.r.t t

$\frac{d y}{d t}=a[\cos t-1 \cdot \cos t-t(-\sin t)]$

$\frac{d y}{d t}=a t \sin t$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{dy}{d t}}{\frac{d x}{dt}}=\frac{at\sin t}{a t \cos t}$

$\frac{d y}{d x}=\tan t$

At $t=\frac{\pi}{3}$

$\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{3}}=\tan \frac{\pi}{3}=\sqrt{3}$

Question 13

sin²x का e^cosx के सापेक्ष अवकलज निकाले ।
[Find the derivative of sin²x w.r.t e^cosx ]
Sol :
माना u=sin²x , v=e^cosx

Differentiating w.r.t x

$\frac{d u}{d x}=2 \sin x \cos x$..(i)

$\frac{d v}{d x}=e^{\cos x}(-\sin x)$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{2 \sin x \cos x}{e^{\cos x}(-\sin x)}$

$\frac{d u}{d v}=-\frac{2 \cos x}{e^{\cos x}}$

Question 14

tan x का cot x के सापेक्ष अवकलज निकाले ।
[Find the derivative of  tan x w.r.t cot x]
Sol :
माना u=tanc, v=cotx

Differentiating w.r.t x

$\frac{du}{d{x}}=\sec ^{2} x$..(i)

$\frac{d v}{d x}=-\text{cosec}^2x$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{du}{dx}}{\frac{d v}{d x}}=\frac{\sec ^{2} x}{-cosec^2x}$

$\frac{d u}{d v}=-\tan ^{2} x$

Question 15

cosx का x³ के सापेक्ष अवकलन ज्ञात करें ।
[Find the derivative of cosx w.r.t x³]
Sol :
माना u=cosx , v=x³

Differentiating w.r.t x

$\frac{d u}{d x}=-\sin x$..(i)

$\frac{d v}{d x}=3 x^{2}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{-\sin x}{3 x^{2}}$

$\frac{d u}{d v}=-\frac{\sin x}{3 x^{2}}$

Question 16

$\frac{x}{\sin x}$ का sinx के सापेक्ष अवकलन गुणांक निकाले ।

[Find the derivative of $\frac{x}{\sin x}$ w.r.t sin x]

Sol :
माना $u=\frac{x}{\sin x}$ , v=sinx

Differentiating w.r.t x

$\frac{d u}{d x}=\frac{1 \sin x-x \cdot \cos x}{\sin ^{2} x}$

$\frac{d u}{d x}=\frac{\sin x-x \cos x}{\sin ^{2} x}$..(i)

v=sinx

Differentiating w.r.t x

$\frac{d v}{d x}=\cos x$..(i)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d y}{d x}}{\frac{dv}{dx}}=\frac{\frac{\sin x-x \cos x}{\sin ^{2} x}}{\cos x}$

$\frac{du}{d v}=\frac{\sin x-x \cos x}{\sin ^{2} x \cos x}$

Question 17

$\sqrt{\sin \left(1+x^{2}\right)^{2}}$ का 1+x² के सापेक्ष अवकलन गुणांक ज्ञात करे ।

[Find the d.c. of $\sqrt{\sin \left(1+x^{2}\right)^{2}}$ w.r.t $1+x^2$]

Sol :
माना $u=\sqrt{\sin \left(1+x^{2}\right)^{2}}$ ,v=1+x²

Differentiating w.r.t x

$\frac{d u}{d x}=\frac{1}{2 \sqrt{\sin \left(1+x^{2}\right)^{2}}} \times \cos \left(1+x^{2}\right)^{2} \times 2\left(1+x^{2}\right) \cdot 2 x$

$\frac{d u}{d x}=\frac{2 x\left(1+x^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$..(i)

अब,

v=1+x²

Differentiating w.r.t x

$\frac{d v}{d x}=2 x$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d u}{d v}}{\frac{dv}{dx}}=\frac{\frac{2x\left(1+x^{2}\right) \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}}{2x}$

$\frac{d u}{d v}=\frac{\left(1+t^{2}\right) \cdot \cos \left(1+x^{2}\right)^{2}}{\sqrt{\sin \left(1+x^{2}\right)^{2}}}$

Question 18

$\tan ^{-1} \frac{2 x}{1-x^{2}}$ का $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ के सापेक्ष अवकलन गुणांक निकालें जहाँ |x|<1

[Find the d.c. of $\tan ^{-1} \frac{2 x}{1-x^{2}}$ w.r.t $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ where |x|<1]

Sol :
माना $u=\tan ^{-1} \frac{2 x}{1-x^{2}}$ $v=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$

u=2tan^-1x,v=2tan^-1x

Differentiating w.r.t x

$\frac{d u}{d x}=\frac{2}{1+x^{2}}$..(i)

$\frac{d v}{dx}=\frac{2}{1+t^{2}}$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d u}{d x}}{\frac{d v}{dx}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}$

$\frac{d u}{d v}=1$

Question 19

$\tan ^{-1} \sqrt{\frac{1-x^{2}}{1+x^{2}}}$ का cos^-1x² के सापेक्ष अवकलन गुणांक निकालें ।

[Find the d.c. of $\tan ^{-1} \sqrt{\frac{1-x^{2}}{1+x^{2}}}$ w.r.t cos^-1x²]

Sol :
माना $u=\tan ^{-1} \sqrt{\frac{1-x^{2}}{1+x^{2}}}$ , v=cos^-1x²

Differentiating w.r.t x

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

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

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

$\frac{d u}{d x}=\frac{-x}{{1+x^{2}}{}} \times \sqrt{\frac{1+x^{2}}{1-x^{2}}}$

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

$\frac{d u}{d x}=\frac{-x}{\sqrt{1-x^{4}}}$..(i)

अब, v=cos^-1x²

Differentiating w.r.t x

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

$\frac{d v}{d x}=\frac{-2 x}{\sqrt{1-x^{4}}}$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-x}{\sqrt{1-x^4}}}{\frac{-2 x}{\sqrt{1-x^{4}}}}$

$\frac{d y}{d v}=\frac{1}{2}$

Question 20

$\tan ^{-1} \frac{2 x}{1-x^{2}}$ का $\sin ^{-1} \frac{2 x}{1+x^{2}}$ के सापेक्ष अवकलन गुणांक निकालें ।

[Find the d.c. of $\tan ^{-1} \frac{2 x}{1-x^{2}}$ w.r.t $\sin ^{-1} \frac{2 x}{1+x^{2}}$ ]
Sol :
माना $\tan ^{-1} \frac{2 x}{1-x^{2}}$ , $\sin ^{-1} \frac{2 x}{1+x^{2}}$

u=2tan^-1x,v=2tan^-1x

Differentiating w.r.t x

$\frac{d u}{d x}=\frac{2}{1+x^{2}}$..(i)

$\frac{d u}{d x}=\frac{2}{1+x^{2}}$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}$

$\frac{d u}{d v}=1$

Question 21

यदि $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$ , दिखाएँ कि $t=\frac{\pi}{6}$ पर $\frac{d y}{d x}=0$

[If $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$ show that $\frac{dy}{dx}=0$ at $t=\frac{\pi}{6}$]

Sol :
Differentiating w.r.t x

$\frac{d x}{d t}=\frac{\frac{3 \sin ^{2} t \cdot \cos t \sqrt{\cos 2 t}}{1}-\sin ^{3} t \cdot \frac{1}{2 \sqrt{\cos ^{2} t}} \times(-\sin 2 t) \cdot 2}{(\sqrt{\cos ^{2} t})^{2}}$

$=\frac{\frac{3 \sin^{2}+\cos t-\cos^2 t+\sin^{3}+\sin 2 t}{\sqrt{\cos^ 2 t}}}{\frac{\cos 2 t}{1}}$

$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t\left(1-2 \sin ^{2} t\right)+\sin ^{3} t \cdot 2 \sin t \cos \theta}{\cos ^{2} t \sqrt{\cos 2 t}}$

$\frac{d x}{d t}=\frac{3 \sin ^{2}t\cos t-6 \sin ^{2} t \cos t+2 \sin ^{4} t \cos t}{\cos 2 t \sqrt{\cos 2 t}}$

$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t-4 \sin ^{4} t \cos t}{\cos 2 t \sqrt{\cos 2 t}}$

$\frac{d{x}}{d t}=\frac{\sin t\cos t\left(3 \sin t-4\sin^{3} t\right)}{\cos 2+\sqrt{\cos 2t}}$

$\frac{d x}{d t}=\frac{\sin t \cos t\sin 3t}{\cos 2 t \sqrt{\cos 2 t}}$..(i)

अब,

$y=\frac{\cos 3 t}{\sqrt{\cos 2 t}}$

Differentiating w.r.t x

$\frac{d y}{d t}=\frac{3 \cos ^{2} t \cdot(-\sin t) \cdot \sqrt{\cos 2t}-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \times\left(-\sin 2 t\right).2}{(\sqrt{\cos 2t})^{2}}$

$=\frac{\frac{-3 \cos ^{2} t \sin t \cdot \cos 2 t+\cos ^{3} t \sin 2 t}{\sqrt{\cos 2 t}}}{\cos 2 t}$

$\frac{d y}{d t}=\frac{-3 \cos ^{2} t \sin t-\left(2 \cos ^{2} t-1\right)+\cos ^{3} t \cdot 2 \sin t \cos t}{\cos 2 t \sqrt{\cos 2 t}}$

$\frac{d y}{d t}=\frac{-6 \cos ^{4} t \sin t+3 \cos ^{2} t \sin t+2 \cos ^{4} t \operatorname{sin} t}{\cos 2t \sqrt{\cos 2t}}$

$\frac{d{y}}{d t}=\frac{-4 \cos ^{4} t {\sin t}+3 \cos^{2} t{\sin } t}{\cos 2 t \sqrt{\cos 2t}}$

$\frac{d y}{d t}=\frac{-\sin t \cos \left(4 \cos ^{3} t-3 \cos t\right)}{\cos 2 t \sqrt{\cos 2t}}$

$\frac{d y}{d t}=-\frac{\sin t \cos t \cdot \cos 3 t}{\cos 2 t \sqrt{\cos 2t}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-\sin t \cos t \cdot \cos 3 t}{\cos 2t \sqrt{\cos 2t}}}{\frac{\sin t+\cos t \cdot \sin 3 t}{\cos {2} t \sqrt{\cos 2 t}}}$

$\frac{d y}{d x}=-\cot 3 t$

$t=\frac{\pi}{6}$ पर ,

$\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{6}}=-\cot 3 \left( \frac{\pi}{6}\right)$

$=-\cot \frac{\pi}{2}$

=0