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CGPET 2018 Solved Paper

Section: Mathematics

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Question : 141 of 150

Marks: +1, -0

If

y=\sin^{-1}\left(\frac{1-x}{1+x}\right)

, then differential coefficient W.r.t.

\sqrt{x}

is

$-\frac{2}{1+x}$
$\sqrt{x}$
$\frac{2}{\sqrt{x}}$
1

Solution:

We have,

y=\sin^{-1}\left(\frac{1-x}{1+x}\right)

y=\frac{\pi}{2}-\cos^{-1}\left(\frac{1-x}{1+x}\right)

y=\frac{\pi}{2}-\cos^{-1}\left[\frac{1-(\sqrt{x})^{2}}{1+(\sqrt{x})^{2}}\right]

y=\frac{\pi}{2}-2\tan^{-1}\sqrt{x}

........(i)

On differentiating Eq. (i) w.r.t

x

, we get

\frac{dy}{dx}=0-\frac{2}{1+x}\cdot\frac{1}{2\sqrt{x}}

=-\frac{1}{(1+x)\sqrt{x}}

........(ii)

Now, let

u=\sqrt{x}

On differentiating w.r.t.

x

, we have

\frac{du}{dx}=\frac{1}{2\sqrt{x}}

........(iii)

Now,

\frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}=\frac{-\frac{1}{(1+x)\sqrt{x}}}{\frac{1}{2\sqrt{x}}}

=-\frac{2}{1+x}

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