If $y=\sin ^{-1}(\sin x)$, then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$, is

non-existent

We have,

$y=\sin ^{-1}(\sin x)= \begin{cases}x, & -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \pi-x, & \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \\ x-2 \pi, & \frac{3 \pi}{2} \leq x \leq 2 \pi\end{cases}$ and so on

We observe that

(LHD at x = $\pi / 2$) = 1 and (RHD at x = $\pi / 2$) = -1

Hence, $\left(\frac{d y}{d x}\right)_{x=\pi / 2}$ does not exist.