Is the function $\sin(\lvert x\rvert)$ differentiable? It is readily shown that the function $\sin(\lvert x\rvert)$  is differentiable when $x\ne 0$.
What I know is that $\lvert x\rvert$ is not differentable at $x=0$. But does $\sin(|x|)$ also follow the same rule?
I believe that it is also not differentiable but why?
 A: We use the fact that
$$
\lim_{x\to 0}\frac{\sin x}{x}=1.
$$
Then, for $h>0$, we have that
$$
\frac{f(h)-f(0)}{h}=\frac{\sin(|h|)-\sin(0)}{h}=\frac{\sin h}{h}\to 1,
$$
while for $h<0$, we have that
$$
\frac{f(h)-f(0)}{h}=\frac{\sin(|h|)-\sin(0)}{h}=\frac{-\sin h}{h}\to -1.
$$
Hence $f$ is not differentiable at $x=0$.
Note. Nevertheless, the function $\dfrac{\sin \lvert x\rvert}{\lvert x\rvert}$ is differentiable everywhere. In fact, it $C^\infty$!
A: It is not at 0.
$x\to 0^+$ 1, $x\to 0^-$ -1
A: Hint
$$ \sin(|x|) = \left\{\begin{array}{l l} \sin(x) & \text{if } x >0 \\ \sin(-x) & \text{if } x <0 \end{array}\right.$$
So clearly this function is differentiable for $x \in (-\infty,0) \cup (0,\infty)$.
A: In terms of intuition (although you should be careful, as this is not a proof), you can note that, for small values of $x$, $\sin(x) \sim x$. So in particular, for small values of $x$ (in which case $|x|$ is also small), we would have that
$$
\sin|x| \sim |x|
$$
and that latter function, as you know, is not differentiable. So (roughly) you should not expect $\sin|x|$ to be either.
Now, in terms of proving this, the other answers give you the right method.
A: Near $x=0$, $sin(x)=x+\bigcirc(x^3)$ so $sin(|x|)=|x|+\bigcirc(x^3)$ giving
$$
\sin(|x|)\prime=|x|\prime+\bigcirc(x^2)
$$
Obviously $|x|\prime$ has a jump from $-1$ to $+1$, whereas the error term is continuous.
It follows $\sin(|x|)\prime$ is a sum of a continuous plus a discontinuous function, and hence discontinuous at $0$.
