Chain rule in the Sobolev space $W^{1,p}$ 
(Chain rule) Assume $F : \mathbb{R} \to \mathbb{R}$ is $C^1$, with $F'$ bounded. Suppose $U$ is bounded and $u \in W^{1,p}(U)$ for some $1 \le p \le \infty$. Show $$v :=F(u) \in W^{1,p}(U) \quad \text{and} \quad v_{x_i}=F'(u)u_{x_i}.$$

From PDE Evans, 2nd edition: Chapter 5, Exercise 17.
Here is what I understand conceptually so far:
Since $u \in W^{1,p}(U)$, it follows $Du=u'$ exists, with $$\int_U u \phi' dx = -\int_U Du \phi \, dx.$$
I need to show that $D(F(u))=F'(u)Du$ exists, with $$\int_U F(u) \phi' dx = -\int_U D(F(u)) \phi \, dx.$$ Then, I can conclude that $F(u) \in W^{1,p}(U)$.
This is all I know so far; how can I go about making the connection?
 A: Assume first that $1\leq p <\infty$
If $u\in C^\infty(\bar{U})$ then clearly $v=F(u)\in C^1(\bar{U})$ and $\nabla v=F'(u)\nabla u$. 
Now if $u$ is a general $W^{1,p}$ function then take a sequence $u_k \to u$ in $W^{1,p}$ with $u_k\in C^\infty(\bar{U})$ and such that $u_k\to u$ and $\nabla u_k \to \nabla u$ pointwise a.e. in $U$. Then
$$
|F(u)-F(u_k)| \leq M|u-u_k|,
$$
with $M=\| F'\|_\infty$. On the other hand we also have
$$
F'(u_k(x))\nabla u_k(x) \to F'(u(x))\nabla u(x), \qquad \text{ for a.e. } x\in U, 
$$
and moreover $| F'(u_k)\nabla u_k|\leq M|\nabla u_k|$. Since $\nabla u_k \to \nabla u$ in $L^p$, by the dominated convegence theorem, $F'(u_k)\nabla u_k \to F'(u)\nabla u$ in $L^p(U)$. Combining this with the first estimate we get that $F(u)\in W^{1,p}(U)$ and $\nabla F(u)= F'(u)\nabla u$.
If $p=\infty$ then we can simply note that $W^{1,\infty}(U)$ is the space of Lipschitz continuous functions in $U$, so take $u$ a Lipschitz function with Lipschitz constant $N$, and $M$ the Lipschitz constant of $F$ as before, then
$$
|F(u(x))-F(u(y))|\leq M|u(x)-u(y)| \leq NM|x-y|, \qquad \forall x,\ y\in U.
$$
Therefore $F(u)\in W^{1,\infty}(U)$.
Edit: As an extra exercise try to see that the condition $U$ being a $C^1$ domain is not needed when $1\leq p<\infty$ (try to prove that $F(u)\in W^{1,p}(U)$ whenever $u\in C^\infty(U)\cap W^{1,p}(U)$).
A: Here is another proof. Choose a test function $\phi$ whose support is in a compactly contained subset $V$ of $U$. Let $ u^{\epsilon} = \eta_{\epsilon} * u $ be the mollification of $ u $. Then
\begin{align}
 \int_U F(u)\phi_{x_i}dx & = \int_V F(u)\phi_{x_i} \; dx \\
       & = \lim_{\epsilon\to0} \int_V F(u^\epsilon)\phi_{x_i} \; dx \\
       & = -\lim_{\epsilon\to0} \int_V F'(u^\epsilon) u^\epsilon_{x_i} \phi \; dx \\
       & = -\int_V F'(u) u_{x_i} \phi \; dx \\
       & = -\int_U F'(u) u_{x_i} \phi \; dx
\end{align}
Therefore $$ \int_U v\phi_{x_i}\; dx = -\int_U v_{x_i}\phi\; dx $$
where $ v_{x_i} = F'(u) u_{x_i} $. We conclude that $v\in W^{1,p}(U)$.
(Source: Measure Theory and Fine Properties of Functions by Evans, page 130.)
A: As a modification to Jose27 I would argue that you may suppose U (path) connected (otherwise we work on connected components separately). Then you approch $\nabla u$ by $C^\infty$ functions and "take primitives" by some kind of path integral. That way you ensure $u_k \to u$ pointwise and $\nabla u_k \to \nabla u$ in $L_p$ norm right from the construction.
Jose27's last comment is actually the fact that $\nabla$ is a closed operator.
