= Solution
Now use proper-time dots and $H=\dot a/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\rm Pl}^2)$, then substitute $\dot{\bar\Phi}\simeq-V_{,\Phi}/(3H)$:
$$
2H\dot H\simeq\frac{V_{,\Phi}\dot{\bar\Phi}}{3M_{\rm Pl}^2},
\qquad
-\frac{\dot H}{H^2}\simeq
\frac{V_{,\Phi}^2}{18M_{\rm Pl}^2H^4}
\simeq\frac{M_{\rm Pl}^2}{2}\left(\frac{V_{,\Phi}}V\right)^2
=\epsilon_V.
$$
Thus the Hubble slow-roll parameter $\epsilon_H$ agrees with $\epsilon_V$ to this order. Since $\mathcal P_\psi\propto H_k^2$ and $k=aH$ at horizon exit,
$$
\frac{d\ln\mathcal P_\psi}{dt}=2\frac{\dot H}{H},
\qquad
\frac{d\ln k}{dt}=H+\frac{\dot H}{H}=H(1-\epsilon_H).
$$
Consequently the <tensor spectral index> is
$$
\boxed{n_t=-\frac{2\epsilon_H}{1-\epsilon_H}\simeq-2\epsilon_V}.
$$
The final equality is the requested first-order <slow-roll approximation>; corrections beyond that order also affect the horizon-exit spectrum itself.
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