We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation:
$$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$
$$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$
Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be considered as a visible "response" of the system on the source and it usually denoted as a new variable, called the "classical field". We would like to find it when there are no external sources: $J=0$.
For that, one then does the Legendre transform trick, arriving at the effective action:
$$\Gamma[\phi_{cl}] = - E - J\phi_{cl}\quad\quad\frac{\delta\,\Gamma}{\delta \phi_{cl}} = - J$$
Remembering our goal to find $\phi_{cl}$ at $J=0$, we arrive at the equation.
$$\frac{\delta\,\Gamma}{\delta \phi_{cl}} = 0$$
Adding an extra assumption that $\phi_{cl}$ is space and time independent: $\phi_{cl}(x) = v$, the effective action functional $\Gamma[\phi_{cl}]$ is then reduced to effective potential $V_{eff}(v)$ and the equation becomes.
$$\frac{dV_{eff}}{dv} = 0$$
Now, as David Vercauteren correctly pointed out, $V_{eff}(v)$ is not the same function as $V(\phi)$. But usually it is a good first approximation, because we usually consider systems where the "real" quantum field fluctuates weakly around its vacuum: $\phi(x)=v+\eta(x)$ with $\eta$ being small.
