= Solution
The <distributional derivative> is defined by $\langle F',\phi\rangle=-\langle F,\phi'\rangle$ for every <test function>. For the <Heaviside function>, compact support gives
$$
\langle\mathcal H',\phi\rangle=-\int_0^\infty\phi'(x)\,dx=\phi(0)=\langle\delta,\phi\rangle.
$$
Thus $\mathcal H'=\delta$. Splitting the integral for $|x|$ at zero and using <integration by parts> gives its first derivative as the <sign function>. The jump of that function at zero is two, so
$$
\boxed{D|x|=\operatorname{sgn}(x),\qquad D^2|x|=2\delta}
$$
as <distributions>. The value assigned to the <sign function> at the single point zero does not affect its associated <distribution>.
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