Deriving Equation for Multi-Variable Case in Zee's Quantum Field Theory

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I've decided to slowly work my way through Zee's quantum field theory in a nutshell over the vacation. I'm confused by the second question of the book which deals with matrix differentiation.

Homework Statement

Derive the equation \langle x_i x_j \cdots x_k x_l \rangle = \sum_{\textrm{wick}} (A^{-1})_{ab}\cdot (A^{-1})_{cd} where {a,b,...,c,d} is a permutation of {i,j,...,k,l}

Homework Equations

\langle x_i x_j \cdots x_k x_l \rangle = \frac{\int d\mathbf{x} e^{1/2 \mathbf{x}^TA \mathbf{x}}x_i x_j \cdots x_k x_l}{\int d\mathbf{x} e^{1/2 \mathbf{x}^TA \mathbf{x}}}

The Attempt at a Solution

I have no trouble deriving this equation for the 1-variable case. For the multi-variable case Zee suggests repeatedly differentiating the equation

\int d\mathbf{x} e^{-1/2 \mathbf{x}^TA\mathbf{x} + J\mathbf{x}} = [(2\pi)^N /\det A]e^{(1/2) J A^{-1}J}

with respect to J. How does one differentiate wrt a matrix anyway?

 

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Doh! It just occurred to me that J is a vector and therefore we can differentiate componentwise. I still don't quite get how to prove this, however. I can show that the claim is true for small n. Is some kind of inductive argument required here??

 

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I figured it out: For future knowledge, the trick is to use that the matrix A must be symmetric, and thus the derivative of the \frac{1}{2}\mathbf{x}^t A^{-1} \mathbf{x} with respect to x_i, say can be written

\sum_n (A^{-1})_{in} x_n.