Question
Find $\frac{d y}{d x}$ if y = sin–1$\left(\frac{2 x}{1+x^{2}}\right)$

Answer

Here,  y = $sin^{-1}(\frac{2x}{1+x^2})$
Let x = tan A
then, A = tan-1x
$\Rightarrow \frac{d A}{d x}=\frac{1}{1+x^{2}}$
y = $\sin ^{-1}\left(\frac{2 \tan A}{1+\tan ^{2} A}\right)$
Also, we know $\left[\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}\right]$
$\Rightarrow$ y = sin-1 (sin2A)
$\Rightarrow$ y = 2A
$\Rightarrow$ $\frac{d y}{d x}=\frac{dy}{dA} \times\frac{dA}{dx}=2 \frac{d A}{d x}$    ...[By chain rule]
$\Rightarrow$ $\frac{d y}{d x}=\frac{2}{1+x^{2}}$

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