Question

Differentiate with respect to $x$

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

Answer 1

Given tnat,

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

$letx=tan\theta$

$dx={\sec }^2\theta d\theta$

$dx=1+x^{2}d\theta ⟹\frac{d\theta }{dx}=\frac{1}{1+x^2}$

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

$y={\sin }^{-1}\left(\sin 2\theta \right)$

$\frac{dy}{dx}=\frac{d\theta }{dx}=\frac{2}{1+x^2}$