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At leading order the scalar curvature power and total tensor power areApplying the preceding $d/d\ln k$ relation gives the scalar spectral index $n-1=-6{ϵ}_V+2{\eta }_V$ and the tensor spectral index $n_T=-2{ϵ}_V$. For this hilltop,Thus, fifty e-folds before the end,The tiny $-6{ϵ}_V$ contribution and finite-endpoint corrections have been omitted in the quoted leading scalar tilt. The tensor tilt is nearly zero but slightly negative; treating $H$ as exactly constant in its spectrum would incorrectly discard its leading nonzero value.

Now use proper-time dots and $H=\dot{a}/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$, then substitute $\dot{\bar{\Phi }}\simeq -V_{,\Phi }/(3H)$:Thus the Hubble slow-roll parameter ${ϵ}_H$ agrees with ${ϵ}_V$ to this order. Since ${\mathcal{P}}_{\psi }\propto H_k^2$ and $k=aH$ at horizon exit,Consequently the tensor spectral index isThe final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.