= Solution
A comoving <wavenumber> crosses the inflationary Hubble scale at $k=aH$. Consequently $d\ln k=d\ln a+d\ln H\simeq Hdt$ when $H$ changes slowly. Combining with the <slow-roll approximation> gives
$$
\boxed{d\ln k\simeq-\frac{V}{M^2V'}\,d\phi
\simeq\frac{\sigma^4}{12M^2\phi^3}\,d\phi.}
$$
Both $k$ and $\phi$ increase along the rolling branch. If $N$ denotes the remaining <number of e-folds> rather than elapsed expansion, then $d\ln k\simeq-dN$. More precisely $d\ln k=(1-\epsilon_H)d\ln a$; the correction is beyond the constant-Hubble approximation used here.
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