The function f(x)=|sin x|,

We know that, $|x|$ is not differentiable at $x=0$.

Therefore, $|\sin x|$ is not differentiable when $\sin x=0$.

$\sin x=0$

$\Rightarrow x=n \pi, n \in Z$

$\Rightarrow x=\ldots,-2 \pi,-\pi, 0, \pi, 2 \pi, \ldots$

Now, the only value of $x$ lying in given interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ at which the function $f(x)=|\sin x|$ is not differentiable is 0 .

Thus, the function $f(x)=|\sin x|, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is not differentiable at $x=0$.

The function $f(x)=|\sin x|,\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is not differentiable at $x=$ ___0___.