= Solution
On each interval $[k,k+1]$, $k\in\mathbb Z$, the <function> is $d(x)=x-k$ on $[k,k+\tfrac12]$ and $d(x)=k+1-x$ on $[k+\tfrac12,k+1]$. Its ordinary <derivative> exists off $\tfrac12\mathbb Z$ and is the <locally integrable function>
$$
s(x)=\begin{cases}
1,&k<x<k+\tfrac12,\\
-1,&k+\tfrac12<x<k+1,
\end{cases}\qquad k\in\mathbb Z.
$$
Assign arbitrary values at the endpoints. For a <test function> $\phi$, split $-\int d\phi'$ at these corners and apply <integration by parts> on each piece. The adjacent <boundary terms> cancel because $d$ is continuous, and only finitely many pieces meet the <compact support> of $\phi$. Hence
$$
\boxed{\langle d',\phi\rangle=\int_{\mathbb R}s(x)\phi(x)\,dx,\qquad d'=s.}
$$
The <distributional derivative> is exactly the <regular distribution> of the ordinary <derivative> defined <almost everywhere>. There are no <Dirac delta distributions> in this first <distributional derivative>: corners change slope, while a jump of the <function> itself would create a point mass.
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