I'm having trouble seeing why the $p$-Laplace PDE
$$\text{div}(|Du|^{p-2}Du) = 0$$
is quasi-linear. Any ideas?
For completeness
\begin{align} \text{div}(u) &:= \sum_{i=1}^n \frac{\partial u_i}{\partial x_i} \\ &= \sum_{i=1}^n u_{x_i} \\ |Du| &:= \left( \sum_{i=1}^n u_{x_i}^2 \right)^{1/2} \end{align}
To highlight it we can rewrite the equation with respect to its components, so $$ \operatorname{div}\left(|D u|^{p-2} D u\right)=0 $$ $$ \sum_{i=1}^n \frac{\partial{(|Du|^{p-2}\partial_i u})}{\partial x_i}=$$$$=\sum_{i=1}^n|Du|^{p-2}\partial^2_i u + (p-2)\sum_{i=1}^n (\partial_i u) |Du|^{p-1}\partial_i{(|Du|)}.$$ Let's recall that $$ |Du|=\sqrt{(\partial_1 u)^2+...+(\partial_n u)^2}=\sqrt{\sum_{j=1}^n (\partial_j u)^2}, $$ so $\partial_i(|Du|)=\frac{1}{2|Du|}2\sum_{j=1}^n (\partial_ju)(\partial_{ij}u).$ We can now conclude that the equation takes the form $$ |Du|^{p-2}\Delta u + (p-2)|Du|^{p-2}\sum_{i,j=1}^n(\partial_iu)(\partial_ju)(\partial_{ij}u)=0$$ which is linear in higher order derivatives. This implies the quasilinearity of the PDE.
The computations should be right but maybe I've done some mistake with chain rule because I've done them fast, the important thing is to follow a similar procedure and highlight the linearity in higher order derivatives.