Chain rule for weak derivatives of $f(u)$ where $f'$ is not bounded but $u$ is? Let $f:\mathbb{R} \to \mathbb{R}$ be $C^1$. Suppose $u$ has a weak derivative $u_x$. I want the chain rule
$$\partial_x (f(u)) = f'(u)u_x$$
to hold. We know this holds if $f'$ is bounded. But I don't have that. But I do have $u$ is bounded almost everywhere, so $|f'(u)| \leq C$. Is this then enough for me to use the chain rule?
 A: Definitely, it is!  Let $\Omega\subset\mathbb{R}^n$ be a bounded domain, $n\geqslant 1$, and $u\in W^{1,1}(\Omega)$. Function $\,u\,$ is additionally assumed to be essentially bounded on $\Omega$, so let $\,M=\|u\|_{L^{\infty}(\Omega)}\overset{\rm def}{=}\underset{x\in \Omega}{\rm ess\,sup\,}|u(x)|$.   Choose some cut-off function 
$\,\eta\in C_0^{\infty}(\mathbb{R})\,$ such that
$$
\eta(t)=
\begin{cases}
1,\quad |t|\leqslant M,\\
0,\quad |t|\geqslant M+1,
\end{cases}
$$
and denote $\,g(t)\overset{\rm def}{=}\eta(t)\!\cdot\!f(t)$.  It is clear that function 
$\,g\in C^1(\mathbb{R})\,$ while
$$
\sup_{t\in \mathbb{R}}\bigl(|g(t)|+|g'(t)|\bigr)\leqslant 
\max_{|t|\leqslant M+1}\bigl(|g(t)|+|g'(t)|\bigr),
$$
i.e., function $g$ is bounded and uniformly Lipschitz on $\mathbb{R}$.  
Notice that $\,f\bigl(u(x)\bigr)=g\bigl(u(x)\bigr)\,$ and 
$\,f'\bigl(u(x)\bigr)=g'\bigl(u(x)\bigr)\,$ a.e. in $\Omega$.   Hence, by virtue of 
the theorem 2.1.11 in Ziemer's textbook Weakly Differentiable Functions 
(see p. 48 therein) the chain rule
$$
\partial_x f\bigl(u(x)\bigr)=\partial_x g\bigl(u(x)\bigr)=
g'\bigl(u(x)\bigr)\!\cdot\!\partial_x u(x)=
f'\bigl(u(x)\bigr)\!\cdot\!\partial_x u(x)
$$
holds a.e. in $\Omega$.
