Find :
Question:

Find $\frac{d y}{d x}$ :

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

Solution:

We have,

$y=\sin ^{-1}\left[\frac{2 x}{1+x^{2}}\right]$

put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

Now,

$y=\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]$

$\Rightarrow y=\sin ^{-1}(\sin 2 \theta), \quad\left(\right.$ as $\left.\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$

$\Rightarrow y=2 \theta, \quad\left(\right.$ as $\left.\sin ^{-1}(\sin x)=x\right)$

$\Rightarrow y=2 \tan ^{-1} x$

$\Rightarrow \frac{d y}{d x}=2 \times \frac{1}{1+x^{2}},\left\{\right.$ because $\left.\frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}\right\}$

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

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