 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 ${ϵ}_H=-\dot{H}/H^2$. ThenThe chain rule gives $d/d\log a=(\dot{\varphi }/H)d/d\varphi$. In reduced Planck units, the slow-roll equations are $3H\dot{\varphi }\simeq -V_{,\varphi }$ and $3H^2\simeq V$. Therefore the horizon-exit logarithmic derivative in slow roll is