= Solution
The <first Hubble slow-roll parameter> measures the fractional change of $H$ per <number of e-folds>, and the <second Hubble slow-roll parameter> measures the fractional change of the first parameter. Since $d\log a/dt=H$,
$$
\boxed{\epsilon=-\frac{\dot H}{H^2},\qquad\eta=\frac{\dot\epsilon}{H\epsilon}.}
$$
For a canonical <inflaton>, $\epsilon=\dot\phi^2/(2M_{\rm Pl}^2H^2)\geq0$. Expanding <cosmic inflation> means $\ddot a>0$, equivalently $\boxed{\epsilon<1}$ because $\ddot a/a=H^2(1-\epsilon)$. Small $\eta$ is not required for accelerated expansion itself, but helps maintain the <slow-roll approximation> over many e-folds. Its logarithmic definition assumes $\epsilon>0$; in exact de Sitter expansion it is undefined unless interpreted as a limit.
For conventional <slow-roll inflation>, assume $\dot\phi^2\ll V$, $|\ddot\phi|\ll3H|\dot\phi|$, a smooth positive <scalar potential>, and slowly varying small parameters. The <Friedmann equation> and <inflaton equation of motion> then reduce to
$$
\boxed{H^2\simeq\frac{V}{3M_{\rm Pl}^2},\qquad3H\dot\phi\simeq-V_{,\phi}.}
$$
Substitute these in the exact expression for $\epsilon$:
$$
\epsilon\simeq\frac{V_{,\phi}^2}{18M_{\rm Pl}^2H^4}=\frac{M_{\rm Pl}^2}{2}\left(\frac{V_{,\phi}}V\right)^2\equiv\epsilon_V.
$$
To obtain the <potential and Hubble slow-roll parameter relation>, differentiate $\log\epsilon_V$ along the slow-roll trajectory $\dot\phi/H\simeq-M_{\rm Pl}^2V_{,\phi}/V$:
$$
\eta\simeq2\left(\frac{V_{,\phi\phi}}{V_{,\phi}}-\frac{V_{,\phi}}V\right)\frac{\dot\phi}{H}
=4\epsilon_V-2M_{\rm Pl}^2\frac{V_{,\phi\phi}}V.
$$
Hence
$$
\boxed{\epsilon\simeq\epsilon_V,\qquad2\epsilon-\frac\eta2\simeq\eta_V=M_{\rm Pl}^2\frac{V_{,\phi\phi}}V.}
$$
The second quantity on the right is the <second potential slow-roll parameter>, not the logarithmic Hubble-flow $\eta$ itself. Consistency usually requires $\epsilon_V\ll1$ and $|\eta_V|\ll1$; identities involving division by $V_{,\phi}$ are used on rolling intervals and interpreted by limiting values at stationary points.
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