If $f(x)=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), x \in[-1,1]$. Then,

$f'(x)=\left\{\begin{array}{l}\frac{2}{\sqrt{1-x^2}}, \text { if }|x|<\frac{1}{\sqrt{2}} \\ \frac{-2}{\sqrt{1-x^2}}, \text { if } \frac{1}{\sqrt{2}}<|x|<1\end{array}\right.$

We have,

$f(x) =\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$

$\Rightarrow f(x)= \begin{cases}2 \sin ^{-1} x, & \text { if }-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \pi-2 \sin ^{-1} x, & \text { if } \frac{1}{\sqrt{2}} \leq x \leq 1 \\ -\pi-2 \sin ^{-1} x, & \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}}\end{cases}$

Clearly, f(x) is continuous on [-1, 1] but it is not differentiable at $x= \pm \frac{1}{\sqrt{2}}$ such that

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