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At horizon exit $k=aH$, $d\log k/dt=H(1-{ϵ}_H)$, where ${ϵ}_H=-\dot{H}/H^2$ is the Hubble slow-roll parameter. To leading slow-roll order,The slow-roll curvature power spectrum is proportional to $V^3/V^{\prime 2}$. Therefore the scalar spectral index in potential slow-roll parameters isUsing the potential slow-roll parameter and the second potential slow-roll parameter givesThe neglected horizon-exit derivative correction contributes at higher slow-roll order.

The slow-roll approximation neglects $\ddot{\varphi }$ relative to $3H\dot{\varphi }$ and kinetic energy relative to the potential. It is self-consistent on the attractor when the potential slow-roll parameter $ϵ\ll 1$ and the second potential slow-roll parameter $|\eta |\ll 1$. ThusHere, as in the inflaton equations, $M_{\mathrm{pl}}$ denotes the reduced Planck mass, not the unreduced mass used in the thermal calculation. The slow-roll curvature power spectrum becomesso

To differentiate with respect to horizon-exit scale, write $\mathcal{N}=\ln a$, increasing with physical time. Along the slow-roll trajectory,The difference between differentiating with respect to $\ln k$ and $\mathcal{N}$ contributes only at second slow-roll order to the tilt. Since $\ln {\Delta }_{\mathcal{R}}^2=3\ln V-2\ln |V^{\prime }|+\mathrm{constant}$,Therefore the scalar spectral index in potential slow-roll parameters isAll background quantities in these expressions are evaluated when the particular mode exits the Hubble radius.