Why $x\longmapsto \mathrm{sgn}(x)$ not in $W^{1,p}(\mathbb R)$? Why $x\longmapsto \mathrm{sgn}(x)$ not in $W^{1,p}(\mathbb R)$ for all $p\in [1,+\infty ]?$ Indeed,
$$\int_{-1}^1 sgn(x)\varphi'(x)\mathrm d x=\int (1_{(0,\infty ]}(x)-1_{(-\infty ,0)}(x))\varphi'(x)\mathrm d x=\int(x1_{(0,\infty )}(x)-x1_{(-\infty ,0)}(x))\varphi (x),$$
and thus, for example $-(x1_{(0,\infty )}(x)-x1_{(-\infty ,0)}(x))$ is the weak derivativ of $x\longmapsto \mathrm{sgn}(x)$, no ?
 A: $\text{sgn}(x)$ is not in $L^p({\mathbb{R}})$, hence not in $W^{1, p}({\mathbb R})$ (when $p<+\infty$). Also, $\text{sgn}(x)$ is the derivative of $|x|$ in the distribution sense, but the derivative of $\text{sgn}(x)$ is $2\,\delta_0$, not a $L^p({\mathbb R})$ function (even for $p=+\infty$).
When $p=\infty$, you can take a test function $\psi(x)$ such that $\psi(0)\ne 0$, and the sequence $\phi_n(x) = n\psi(n x)$. A change of variable shows that
$$\|\phi_n\|_{L^1} = \|\psi\|_{L^1}\qquad\text{and}\qquad-\int\text{sgn}(x)\phi_n'(x)d x = 2 n \psi(0)$$ If $\text{sgn}(x)$ was in $W^{1,\infty}$, the integral would be bounded by $C\|\text{sgn}(x)\|_{W^{1,\infty}} \|\phi_n\|_{L^1}$.
A: The 2nd step in your equality chain is not valid. Just use partial integration as follows
$$\begin{align}-\langle \text{sgn'}, \varphi \rangle=\int_{-1}^1 \text{sgn}(x) \varphi'(x)\ dx &=-\int_{-1}^0 \varphi'(x)\ dx+\int_0^1\varphi'(x)\ dx\\
&=-\varphi(0)-\varphi(0) \\ &=-2\varphi(0) \\
&=\langle-2\delta_0,\varphi\rangle \end{align}$$
therefore $\text{sgn}'=2\delta_0$ but $\delta_0$ (the Dirac-Delta-Distribution) is not locally integrable what we demand of weak derivatives; to be exact we defined them to have this property since we want 
$$\int_\Omega u \phi' \ dx = -\int_\Omega u' \phi \ dx, ~\phi \in C_0^\infty(\Omega) $$
to be well-defined i.e. $|\int_\Omega u' \phi \ dx | \leq ||u'||_{L^1(\text{supp}(\phi))} ||\phi||_\infty<\infty$ and since this has to be valid for all test functions we demand 
$u' \in L^1_\text{loc}(\Omega)=\{v:v|_K \in L^1(K) \text{ for all compact subsets } K \subset \Omega\}$.
Also notice that that $L^p(\Omega) \subset L^1_\text{loc}(\Omega)$ for $p\in [1,\infty]$ so from $\text{sgn}'=2\delta_0 \notin L^1_\text{loc}(\Omega)$ we know that $\text{sgn}'=2\delta_0 \notin L^p(\Omega)$ for $p\in [1,\infty]$ but this has to be fulfilled for $\text{sgn}$ to be in $W^{1,p}$,
