Why is the functional derivative of the Lagrangian wrt. field evaluated at the classical field equal to negative of source current at lowest order?

Question

I am having difficulty showing this equation in Peskin & Schroeder's Introduction to Quantum Field Theory (Section 11.4 p.340):

We wish to compute $\Gamma$ as a function of $\phi_{\text{cl}}$. But the functional $Z[J]$ depends on $\phi_{\text{cl}}$ through its dependence on $J$. Thus, we must find, at least implicitly, a relation between $J(x)$ and $\phi_{\text{cl}}(x)$. At the lowest order in perturbation theory, that relation is just the classical field equation: $$\left.\frac{\delta \mathcal{L}}{\delta \phi}\right\vert_{\phi = \phi_{\text{cl}}} + J(x) = 0 \quad \text{(to lowest order).}$$

I tried to differentiate the $\lambda \phi ^4$ Lagrangian and plugging in the classical field $\phi _{\mbox{cl}}$, but I am not seeing how to expand that to lowest order and show that it is $-J(x)$. The classical field is also the vacuum expectation value:

$$\phi _{\mbox{cl}}=\dfrac{\int [d\phi ]\phi \exp{(i/\hbar) \int d^4x (\mathcal L(\phi)+J \phi)}}{\int [d\phi ] \exp{(i/\hbar) \int d^4x (\mathcal L(\phi)+J \phi)}}.$$

Answer 1

1. The first equation follows from the fact that $$\Gamma[\phi_{\rm cl}] ~=~ S[\phi_{\rm cl}] +{\cal O}(\hbar) ,$$ so that $$ -J~=~\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}}~=~\frac{\delta S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}}+{\cal O}(\hbar).$$

2. The second equation follows from $$\langle \phi \rangle_J ~=~ \frac{1}{Z[J]} \frac{\hbar}{i}\frac{\delta Z[J]}{\delta J}~=~ \frac{\delta W_c[J]}{\delta J} ~=~ \phi_{\rm cl}.$$

For more details, see e.g. this Phys.SE post.