KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.5 (Q19-Q21)
[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 ,
$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-x}{\sqrt{1-x^4}}}{\frac{-2 x}{\sqrt{1-x^{4}}}}$
$\frac{d y}{d v}=\frac{1}{2}$
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 ,
$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}$
$\frac{d u}{d v}=1$
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
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
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