compute the following weak derivative $D(|u|^{p/2-1} u)=p/2|u|^{p/2-1} Du$ in Sobolev space

Question

Let $u(x)\in W_0 ^{1, p} (\Omega) $ with $p >2$ ,prove $D(|u|^{p/2-1} u)=p/2|u|^{p/2-1} Du$ Moreover $|u|^{p/2-1} u \in W^{1,2}_0(\Omega)$.

I can show the result that if $D(|u|^{p/2-1} u)=p/2|u|^{p/2-1} Du$ then we have $|u|^{p/2-1} u \in W^{1,2}(\Omega)$.Rest of part seems not easy to deduce:It may go like this,since $u(x)\in W_0 ^{1, p} (\Omega)$ exist some $u_m\to u$ in $W^{1,p}(\Omega)$. such that $u_m \in C^\infty_c(\Omega)$

1. If we can prove $D(|u_m|^{p/2-1} u_m)=p/2|u_m|^{p/2-1} Du_m$ (Which I have no idea how to show it)

2. Since $\text{supp}(|u_m|^{p/2-1} u_m )\subset \text{supp}(u_m)$ which is compact,hence $|u_m|^{p/2-1} u_m $ lies in $W^{1,2}_0(\Omega)$

3. Finally if $|u_m|^{p/2-1} u_m \to |u|^{p/2-1} u$ in $W^{1,2}(\Omega)$,then we are done.Since subspace $W^{1,2}_0(\Omega)$ is closed the limit must lies in it.

The main question may be the first one (1).Since the chain rule is not so easy to use here.

Ok I have worked the first one out,which is very easy to deduce due to the result from here and product rule

All the problem lies on the third one (3)