Let $f(x)=|x|+|\sin x|, x \in(-\pi / 2, \pi / 2)$. Then,

differentiable everywhere except at x = 0

We have,

$f(x) =|x|+|\sin x|$  for all  $x \in(-\pi / 2, \pi / 2)$

$\Rightarrow f(x) = \begin{cases}-x-\sin x & \text { for }-\pi / 2<x<0 \\ x+\sin x & \text { for } 0 \leq x<\pi / 2\end{cases}$

Clearly, $\lim\limits_{x \rightarrow 0^{-}} f(x)=f(0)=\lim\limits_{x \rightarrow 0^{+}} f(x)$

So, f(x) is continuous at x = 0

Also, f(x) is continuous on $(-\pi / 2,0) \cup(0, \pi / 2)$.

Thus, f(x) is continuous on $(-\pi / 2, \pi / 2)$

Now,

(LHD at x = 0)

$=\left\{\frac{d}{d x}(-x-\sin x)\right\}_{\text {at } x=0}=\{-1-\cos x\}_{\text {at } x=0}=-1-1=-2$

and,

(RHD at x = 0)

$=\left\{\frac{d}{d x}(x+\sin x)\right\}_{\text {at } x=0}=\{1+\cos x\}_{\text {at } x=0}=1+1=2$

Clearly, (LHD at x = 0) ≠ (RHD at x = 0)

Clearly, f(x) is differentiable on $(-\pi / 2,0) \cup(0, \pi / 2)$.

Hence, f(x) is everywhere differentiable except at x = 0