Distributional derivative of absolute value function I'm tying to understand distributional derivatives. That's why I'm trying to calculate the distributional derivative of $|x|$, but I got a little confused.
I know that a weak derivative would be $\operatorname{sgn}(x)$, but not only I'm not finding that one in my calculations, I ended up with a wrong solution and in my other attempt I got
stuck pretty quickly.
Could someone tell me where my error in reasoning occurs or what I missed? 
First attempt
$$
(T_{|x|}(\phi))' = T_{|x|'}(\phi)
=\int_{\mathbb{R}_+} \phi(x) dx - \int_{\mathbb{R}_-} \phi(x) dx = const.
$$
Second attempt
$$
(T_{|x|}(\phi))' = -T_{|x|}(\phi') = -\int_\mathbb{R} |x| \phi'(x) dx
=-\int_{\mathbb{R}_+} x \phi'(x) dx + \int_{\mathbb{R}_-} x \phi'(x) dx
$$

Given a test function $\phi$, the goal is to rewrite $-\int |x|\phi'(x)\,dx$ so that it has $\phi$ in it instead of $\phi'$. Split into two integrals over positive and negative half-axes; then integrate by parts. The result: 
$$
\int_{-\infty}^0 x\phi'(x)\,dx - \int_0^{\infty} x\phi'(x)\,dx
 = -\int_{-\infty}^0  \phi(x)\,dx + \int_0^{\infty} \phi(x)\,dx 
= \int_\mathbb{R} \operatorname{sgn}x \,\phi(x)\,dx
$$
which establishes the claim $|x|'=\operatorname{sgn}x$.
It's difficult to evaluate your attempted solutions, because they contain no words, and no steps that I can recognize as integration by parts. 
