= Solution
The <space of test functions> is $\mathcal D(\mathbb R)=C_c^\infty(\mathbb R)$. A sequence $\varphi_j$ converges to $\varphi$ when all functions eventually have <compact support> in one <compact set> $K$ and every derivative converges uniformly:
$$
\|\varphi_j^{(q)}-\varphi^{(q)}\|_\infty\longrightarrow0
\quad\text{for each integer }q\geq0.
$$
A <distribution> is a continuous linear functional on this <space of test functions>. Equivalently, for each <compact set> $K$, there are $C_K$ and a finite integer $m_K$ such that
$$
|\langle u,\varphi\rangle|\leq C_K\max_{0\leq q\leq m_K}\|\varphi^{(q)}\|_\infty
\quad\text{whenever }\operatorname{supp}\varphi\subseteq K.
$$
The integer $m_K$ is permitted to depend on $K$. The usual <weak convergence of distributions> is
$$
\boxed{u_j\longrightarrow u\text{ in }\mathcal D'(\mathbb R)
\quad\Longleftrightarrow\quad
\langle u_j,\varphi\rangle\longrightarrow\langle u,\varphi\rangle
\text{ for every }\varphi\in\mathcal D(\mathbb R).}
$$
Thus the convergence convention on the <distribution> space is its weak dual topology.
For the <principal-value reciprocal distribution>, the symmetric truncations can be written
$$
\left\langle\operatorname{pv}\frac1x,\varphi\right\rangle
=\lim_{\varepsilon\downarrow0}\int_{|x|>\varepsilon}\frac{\varphi(x)}x\,dx
=\int_0^\infty\frac{\varphi(x)-\varphi(-x)}x\,dx.
$$
The numerator is $O(x)$ near zero by the <mean value theorem>, and the integrand vanishes for large $x$ because $\varphi$ has <compact support>. For $\operatorname{supp}\varphi\subset[-R,R]$,
$$
\left|\left\langle\operatorname{pv}\frac1x,\varphi\right\rangle\right|
\leq2R\|\varphi'\|_\infty.
$$
Consequently the <Cauchy principal value> defines a <distribution> of <order of a distribution> at most one.
The function $\log|x|$ has <local integrability>, since $\int_0^1|\log x|\,dx<\infty$, and therefore defines a <distribution>. For its <distributional derivative>, remove $(-\varepsilon,\varepsilon)$ and use <integration by parts> on both remaining intervals:
$$
-\int_{|x|>\varepsilon}\log|x|\,\varphi'(x)\,dx
=\int_{|x|>\varepsilon}\frac{\varphi(x)}x\,dx
+\log\varepsilon\,[\varphi(\varepsilon)-\varphi(-\varepsilon)].
$$
The boundary term is $O(\varepsilon|\log\varepsilon|)$ and tends to zero. The omitted integral of $\log|x|\varphi'$ tends to zero by <local integrability>. Hence the <distributional derivative of the logarithmic modulus> satisfies
$$
\boxed{\frac{d}{dx}\log|x|=\operatorname{pv}\frac1x\quad\text{in }\mathcal D'(\mathbb R).}
$$
Symmetric truncation is essential to this normalization of the <principal-value reciprocal distribution>.
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