Solution
= Solution
For a monotonic inflaton trajectory, the <potential slow-roll parameter> satisfies
$$
\left|\frac{d\phi}{dN}\right|
=M_{\rm Pl}\sqrt{2\epsilon_V(\phi)}.
$$
The remaining <slow-roll e-fold count> is the integral from the present field value to its end value along the direction of motion:
$$
\boxed{\widetilde N(\phi)
=\int_\phi^{\phi_e}
\frac{d\widetilde\phi}
{M_{\rm Pl}\sqrt{2\epsilon_V(\widetilde\phi)}}}.
$$