Question

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

Answer 1

$y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
Put $x=\tan \theta$
$\Rightarrow y={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$
$={\sin }^{-1}\left(\sin 2\theta \right)=2\theta =2{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=2\cdot \frac{1}{1+x^2}=\frac{2}{1+x^2}$