= Solution
At <cosmological horizon exit>, $k$ is proportional to $aH$; the convention-dependent factor $2\pi$ has zero logarithmic derivative. Define the <first Hubble slow-roll parameter> $\epsilon_H=-\dot H/H^2$. Then
$$
\frac{d\log k}{d\log a}=1-\epsilon_H,\qquad\frac d{d\log k}=\frac1{1-\epsilon_H}\frac d{d\log a}\simeq\frac d{d\log a}.
$$
The <chain rule> gives $d/d\log a=(\dot\phi/H)d/d\phi$. In <reduced Planck units>, the <slow-roll equations> are $3H\dot\phi\simeq-V_{,\phi}$ and $3H^2\simeq V$. Therefore the <horizon-exit logarithmic derivative in slow roll> is
$$
\boxed{k\frac d{dk}\simeq a\frac d{da}=\frac{\dot\phi}H\frac d{d\phi}\simeq-\frac{V_{,\phi}}V\frac d{d\phi}.}
$$
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