I am trying to write an expression to $\partial_t \|\nabla u\|_{L^p(\Omega)}^p.$ Here $\Omega$ is a smooth domain, the function $u$ has no regularity problems (you can take it smooth) and the normal derivative of $u$ vanishes at the boundary.
Here is how I tried: \begin{align}\partial_t \|\nabla u\|_{L^p(\Omega)}^p &= \int_\Omega \partial_t|\nabla u|^pdx = p \int_\Omega |\nabla u|^{p-2}\nabla u \cdot \nabla u_t dx \\ \end{align} then by using the divergence theorem we have \begin{align}\partial_t \|\nabla u\|_{L^p(\Omega)}^p &= -p\int_\Omega u_t\mbox{div}(|\nabla u|^{p-2}\nabla u)dx \end{align} and then using the product rule for the divergence \begin{align}\partial_t \|\nabla u\|_{L^p(\Omega)}^p &= -p\int_\Omega u_t\left[|\nabla u|^{p-2}\Delta u + \nabla(|\nabla u|^{p-2})\cdot \nabla u\right]dx\end{align}.
My question is: is there any identity that would allow me to simplify the term $$-p\int_\Omega u_t\nabla(|\nabla u|^{p-2})\cdot \nabla udx.$$ Applying the chain rule for the gradient gives me a next step, but I still have a $\nabla(|\nabla u|)$ which I still do not know how to deal with.
I have the impression that either there is an identity that could help me or maybe the path I am following so far is not the ideal one.
