Find the second distributional derivatives of $\sin|x|$.
$\langle t',\phi\rangle$ = $-\langle t, \phi'\rangle$ [distributional derivative] where $t$ is distribution and $\phi$ is a test function.
If you have any good idea for it, please help. Thank you.
$$(\sin |x|)' = (\sin' |x|) (|x|)' = \cos(|x|)\operatorname{sign}(x)$$ and $$((\cos|x|-1)\operatorname{sign}(x))'=((\cos |x|-1)'\operatorname{sign}(x)+ 0 \delta(x)$$ $$= (-\sin|x|) \operatorname{sign}(x) \operatorname{sign}(x)=-\sin|x|$$
The key argument is that continuous and piecewise $C^1$ implies the derivative is given by the piecewise derivatives