Showing that for $u\in W^{1,p}(U)$ that $u^+$ is also in $W^{1,p}(U)$ Evans - Edition 1 - Chapter 5 - Problem 17 - p292

I want to show that if $u\in W^{1,p}(U)$, then $u^+,u^-$ are in $W^{1,p}(U)$, and that:
  $$Du^+=\begin{cases}Du&a.e\,u>0\\0&a.e. \,u\leq 0 \end{cases}$$

They give the hint $u^+=\lim_{\epsilon\to 0} F_\epsilon(u)$ for:
$$F_\epsilon(z)=\begin{cases}(z^2+\epsilon^2)^{1/2}-\epsilon&z\geq 0\\0&z<0\end{cases}$$
So to show this, I need to show that each weak first derivative exists right?
So I let $\phi\in C^\infty_c(U)$ be a test function, and take: If $u>0$
$$\int_U u^+\phi_{x_i}dx=\lim_{\epsilon \to 0} \int_U F_\epsilon(u)\phi_{x_i}dx=\lim_{\epsilon\to0}\int_U ((u^2+\epsilon^2)^{1/2}-\epsilon)\phi_{x_i} dx$$$$ = \left( \int_U u\phi_{x_i}\right) dx=\int u_{x_i}\phi dx$$
So for $u>0$ we have $Du^+ =(D^{(1,0,\dots,0)}u^+,\dots,D^{(0,\cdots,0,1)}u^+)=(u_{x_1},\dots,u_{x_n})=Du$
and for
$u<0,$
we have"
$$\int_U u^+\phi_{x_i}dx=\lim_{\epsilon \to 0} \int_U 0\phi_{x_i}dx=0?$$

I think I have done this wrong, and I am not sure how to proceed for $u<0$, I guess the weak derivative is just $0$ as desired?

 A: $\newcommand{\C}{\mathscr{C}}\DeclareMathOperator{\supp}{supp}$It's a bit strange to see you casing on whether $u > 0$ or $u < 0$ since you are integrating against a test function $\phi\in\C^\infty_c(U)$ and it is certainly possible to have $u$ take on both positive and negative values on $U$.
Here is an approach that you can take instead: Note that $F_\epsilon\in\C^1(\mathbb{R})$, $F_\epsilon'$ is bounded, and $F(0)=0$ (you need this to show that the composition is still in $L^p$ if your domain has infinite measure), and so you can prove the chain rule (using mollifications of $u$ or absolute continuity on lines), i.e. that $F_\epsilon\circ u\in W^{1,p}(U)$ and $$D(F_\epsilon\circ u)=(F_\epsilon'\circ u)~Du.$$
Now consider a test function $\phi\in\C^\infty_c(U)$ and we can compute using the chain rule
$$\int_{U}(F_\epsilon\circ u)\frac{\partial\phi}{\partial x_i}~dx = -\int_{U}(F_\epsilon'\circ u)\frac{\partial u}{\partial x_i}\phi~dx. $$
Now compute $$F_\epsilon'(z)=\begin{cases}\frac{z}{\sqrt{z^2 + \epsilon^2}} & z\geq 0,\\ 0 & z < 0,\end{cases}$$
and so $F_\epsilon'\to\chi_{\mathbb{R}^+}$ as $\epsilon\to 0$.
Finally, taking $\epsilon\to 0$ yields 
$$\int_{U}u^+\frac{\partial\phi}{\partial x_i}~dx = -\int_{U}\chi_{u > 0}\frac{\partial u}{\partial x_i}\phi~dx,$$
which is exactly the desired result $\frac{\partial u^+}{\partial x_i}=\chi_{u > 0}\frac{\partial u}{\partial x_i}.$
