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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

The slow-roll equations turn the slow-roll curvature power spectrum intoThe spectral tilt is the logarithmic derivative of this dimensionless spectrum. Apply the horizon-exit logarithmic derivative in slow roll:With the potential slow-roll parameter $ϵ$ above and the second potential slow-roll parameter $\eta =V_{,\varphi \varphi }/V$,This is the first-order departure of the scalar spectral index from unity, in reduced Planck units.