= Solution
Let $\delta=4-D$ and $F(g)=g-Cg^2/\delta+O(g^3)$. Since changing the arbitrary <renormalization scale> must leave the bare theory fixed,
$$
0=\mu\frac{d}{d\mu}\big[\mu^\delta F(g)\big]
=\mu^\delta\big[\delta F(g)+\beta(g)F'(g)\big].
$$
Write $\beta(g)=-\delta g+bg^2+O(g^3)$. Substitution gives, through order $g^2$,
$$
0=\delta g-Cg^2+(-\delta g+bg^2)
\left(1-\frac{2Cg}{\delta}\right)
=(b+C)g^2+O(g^3).
$$
Therefore $b=-C$, and the specified residue gives the <renormalization-group beta function>
$$
\boxed{\beta(g)=-(4-D)g+\frac3{16\pi^2}g^2+O(g^3).}
$$
The derivative is at fixed bare parameters. It does not say the physical theory changes when $\mu$ changes; the <running coupling> compensates the scale used to describe it. The coefficient also depends on defining the derivative with respect to $\log\mu$, rather than $\log\mu^2$; using the latter would divide both terms by two.
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