Calculating higher-derivatives of quantum effective potential (Weinberg QFT chapter 16 problem 2) Couple a set of classical currents, denoted by $J$, to a set of fields $\phi$. Let $iW[J]$ be the sum of all connecte vacuum-vacuum amplitudes.
The Weinberg Quantum Theory of Fields chapter 16 says that the quantum effectivve potential is defined as the Legendre transform of $W[J]$, i.e $$\Gamma[\phi]=-\int d^4x \phi^r(x) J_r(x)+W[J].$$
Also it says $$\phi^r(x)=\frac{\delta W[J]}{\delta J_r(x)}, \qquad -J_r(x)=\frac{\delta \Gamma[\phi]}{\delta \phi^r(x)}.$$
Now I would like to calculate the general formula for $\delta^3W[J]/\delta J_n(x)\delta J_m(y) \delta J_l(z)$ and $\delta^4W[J]/\delta J_n(x)\delta J_m(y) \delta J_l(z) \delta J_k(w)$ in terms of variational derivatives of $\Gamma[\phi]$ w.r.t $\phi$.
I tried to use the fact that the Legendre transform of a function has the derivative inverse to that of the original function and the chain rule. However, I cannot figure out how to calculate terms like $\delta (\frac{\delta \Gamma[\phi]}{\delta \phi^r(x)})^{-1}/\delta J_m(y)$.
Could anyone please help me? This is actually the problem 2 of Weinberg chapter 16. 
 A: Let's start with the first derivative:
$$
\frac{\delta W[J]}{\delta J(x)}=\phi(x)
$$
Then the second derivative:
$$
\begin{align}
\frac{\delta^2 W[J]}{\delta J(x)\delta J(y)}&= \frac{\delta \phi(x)}{\delta J(y)} \\
&= \Big[\frac{\delta J(y)}{\delta \phi(x)}\Big]^{-1} \\
&= \Big[-\frac{\delta^2 \Gamma[\phi]}{\delta \phi(x)\delta \phi(y)}\Big]^{-1} \\
&= -\Gamma_2(x,y)^{-1}
\end{align}
$$
where I have used $\Gamma_2(x,y)$ to denote the expression inside the bracket. Here note that the inverse is to be understood in the sense of matrix inverse. More specifically:
$$
\int d^4z\; \Gamma_2(x,z)^{-1}\Gamma_2(z,y) =\delta^4(x-y)
$$
We have the following identity:
$$
\delta(\Gamma_2^{-1}\Gamma_2)=0 \implies\delta \Gamma_2^{-1}=-\Gamma_2^{-1}\delta\Gamma_2 \Gamma_2^{-1}
$$
Again, the product is to be understood in the matrix sense. I've omitted the integrations for brevity.
Now we are ready for the third derivative. Using the expression above for the second derivative:
$$
\begin{align}
\frac{\delta^3 W[J]}{\delta J(x)\delta J(y)\delta J(z)}&= -\frac{\delta \Gamma_2(x,y)^{-1}}{\delta J(z)} \\
&=\int d^4w_1\;d^4w_2\;\Gamma_2(x,w_1)^{-1} \frac{\delta \Gamma_2(w_1,w_2)}{\delta J(z)} \Gamma_2(w_2,y)^{-1} \\
&=\int d^4w_1d^4w_2d^4w_3\Gamma_2(x,w_1)^{-1} \Gamma_2(w_2,y)^{-1}\frac{\delta \Gamma_2(w_1,w_2)}{\delta \phi(w_3)} \frac{\delta \phi(w_3)}{\delta J(z)}\\
&=-\int d^4w_1d^4w_2d^4w_3\Gamma_2(x,w_1)^{-1} \Gamma_2(w_2,y)^{-1}\Gamma_3(w_1,w_2,w_3) \Gamma_2(w_3,z)^{-1}\\
\end{align}
$$
That is the final result. In terms of Feynman diagrams, $G_2(x,y)=-\Gamma_2(x,y)^{-1}$ is the sum of all connected 2-point graphs, $\Gamma_3(w_1,w_2,w_3)$ is the sum of 3-point 1PI graphs. The result for the third derivative:
$$
\begin{align}
G_3(x,y,z)&=\int d^4w_1\;d^4w_2\;d^4w_3\;G_2(x,w_1)\Gamma_3(w_1,w_2,w_3)G_2(w_2,y) G_2(w_3,z)\\
\end{align}
$$
just means that the sum of all 3-point connected graphs can be computed from the 3-point 1PI graph and 2-point connected graphs. This result can be represented by the following diagram (taken from Lectures of Sidney Coleman).

Similarly, you can derive the expression for the fourth derivative (or guess the result by drawing Feynman diagrams).
