I have a doubt in the proof of Gagliardo-Nirenberg-Sobolev inequality that is found in Evans' book "Partial Differential Equations" (Theorem 1, pages 277-279).
In this part of the proof, $ \gamma>1 $ is a constant and $ u\in C^1(\mathbb{R}^n) $ is a function with compact support (i.e., $ u\in C^1_c(\mathbb{R}^n) $). By definition, $ D|u|^\gamma $ is the gradient of $ |u|^\gamma $ (I'm not sure if it is the ordinary derivative or the weak derivative, but I think that's the ordinary derivative).
Why is $ D|u|^\gamma $ equal to $ \gamma\cdot|u|^{\gamma-1}\cdot|Du| $? It is similar to the chain rule, but I can't understand the partial derivatives of $ |u|^\gamma $, because the function $ |x| $ isn't usually differentiable.
(PS.: I'm not sure about my English level, but I think that my question can be understood)
For Evans, $Du$ always refers to the weak gradient. However, in this case $\vert u \vert^\gamma$ is differentiable in the classical sense, so $Du$ also denotes the classical gradient (since when a function is classically differentiable then it is weakly differentiable and the two notions of derivatives agree).
Indeed, let $g : \mathbb R \to \mathbb R$ defined by $z \mapsto \vert z \vert^\gamma$. Then one can check from first principles that $g':\mathbb R \to \mathbb R$ and is given by $$g'(z) = \begin{cases} \gamma z \vert z \vert^{\gamma - 2}, &\text{if } z\neq 0, \\ 0, &\text{if } z=0. \end{cases} $$ Note that $g'$ is continuous at $z=0$ since $\gamma>1$, so $g\in C^1(\mathbb R)$. Since this is the case, we can just say $g'(z)=\gamma z \vert z \vert^{\gamma-2}$ with the understanding that $g'(0)=0$. Hence, the chain rule applies to $g(u(x))$ and $$D \vert u \vert^\gamma = g'(u(x))D u=\gamma u(x) \vert u(x) \vert^{\gamma -2} D u .\tag{$\ast$} $$
Note carefully that what you have written ($D \vert u \vert^\gamma =\gamma \vert u(x) \vert^{\gamma -1}\vert D u\vert$) is incorrect (indeed, the LHS is a vector and the RHS is a scalar) and is a misquote from Evans. What he actually claims is that $$ \big \vert D \vert u \vert^\gamma \big \vert =\gamma \vert u(x) \vert^{\gamma -1}\vert D u\vert.$$ This follows immediately from ($\ast$).