The function f(x) given by $f(x)=arc \sin \left(\frac{2 x}{1+x^2}\right)$, is

such that $f'(x)=\left\{\begin{array}{l}\frac{2}{1+x^2},-1<x<1 \\ \frac{-2}{1+x^2},|x|>1\end{array}\right.$

We have,

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

$\Rightarrow f(x) = \begin{cases}2 \tan ^{-1} x, & \text { if }-1 \leq x \leq 1 \\ \pi-2 \tan ^{-1} x, & \text { if } x>1 \\ -\pi-2 \tan ^{-1} x, & \text { if } x<-1\end{cases}$

$\Rightarrow f'(x) = \begin{cases}\frac{2}{1+x^2}, & \text { if }-1<x<1 \\ \frac{-2}{1+x^2}, & \text { if }|x|>1\end{cases}$

Thus, f(x) is not differentiable at $x= \pm 1$ and f'(x) is as given above.