Sobolev chain rule ($D(\Phi (u))={\Phi }^{\prime }(u)Du$)

If $u\in W_{\mathrm{loc}}^{1,2}$ and $\Phi$ is continuously differentiable with bounded derivative on the range in use, then $D(\Phi (u))={\Phi }^{\prime }(u)Du$ almost everywhere. One proves this first for smooth functions and then uses density of smooth functions in a Sobolev space and boundedness of ${\Phi }^{\prime }$. On compact sets one may truncate $\Phi$ away from the range of a bounded $u$.

If $u\in W_{\mathrm{loc}}^{1,2}$, then $Du=0$ almost everywhere on every level set $\{u=t\}$. Choose smooth truncations ${\Phi }_{\epsilon }(s)$ with $|{\Phi }_{\epsilon }|\le 2\epsilon$, derivative one near zero, bounded derivative, and derivative zero outside $(-2\epsilon ,2\epsilon )$. The Sobolev chain rule gives $D{\Phi }_{\epsilon }(u-t)\to 1_{\{u=t\}}Du$ in local $L^2$ by dominated convergence theorem, while ${\Phi }_{\epsilon }(u-t)\to 0$. The limiting weak derivative is therefore zero.