How Does the n Factor Arise in the QFT Generating Functional Equation?

[binbagsss]

Homework Statement

Hi

I am looking at the attached question part c)

Homework Equations

below

The Attempt at a Solution

so if i take $\frac{\partial^{(n-1)}}{\partial_{(n-1)}}$ of (2) it is clear I can get the $\frac{i}{h} (\lambda_2 +\lambda_4 )$ like-term, but I am unsure about the $nG_{n-1}$ .

There's obviously no other derivatives on the RHS so I will only yield a $G_{n-1}$ and that looks fine, I am a bit confused though, I can yield this from the $Z[J]$ alone on the RHS, whereas the RHS is $Z[J]$ 'multiplied by' (it is already inside the integral) the extra term of $S'[\Phi] + J$ . So I suspect this extra term is the reason we get the $n$ factor but I am unsure how.

Looking at the LHS there is a single $J$ so it looks like this gives a factor of $1$ and then we take across $(n-1)$ from the RHS.

If I take a derivative wrt $J$, on the LHS I can either act on the exponential or the single $J$ (but can only act on this $J$ once,) on the RHS it's the same story, with the difference that on the LHS the $J$ is outside the integral but on the RHS it is inside the integral, I'm trying to use this to deduce where the factor of $n$ comes from but I am struggling..

 

[binbagsss]

bump
many thanks