$\frac{d}{dx}\left({\sin }^{-1}\left(\frac{2}{x^{-1}+x}\right)\right)$ is equal to 

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

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

$$\begin{gathered} put x= \tan \theta \\ ={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)={\sin }^{-1}\left(\sin 2\theta \right)=2\theta \end{gathered}$$

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

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