# Is there proof of $\frac{dTr(\log(A))}{dA}=A^{-1}$ when A is symmetric using index notation.

So I am a physicist and I encountered the following derivative in my study of the SYK model:

$$\frac{dTr(\log(A))}{dA}$$ where $$A$$ is a symmetric matrix.

I know that Tr$$(\log(X))=\log(\det(X))$$ and found a proof that $$\frac{d\log(\det(X))}{dX}=X^{-T}$$ for a positive definite matrix X.

My question is if there is a known way to proof that $$\frac{dTr(\log(A))}{dA}=A^{-1}$$ using index notation? For example I think the following proofs that $$\frac{dTr(XY)}{dX}=Y^T$$ by looking at the indices (using Einstein summation convention): $$\frac{d}{dX_{ij}}X_{kl}Y_{lk}=\delta_{ik}\delta_{jl}Y_{lk}=Y_{ji}$$. But I am not able to (find a) proof that $$\frac{dTr(\log(A))}{dA}=A^{-1}$$ (where A is symmetric) in a similar way and have no idea if its even in principle possible.

\eqalign{ f &= f(x) = \sum_{k=0}^\infty \alpha_k x^k \quad\implies f' &= \frac{df}{dx} \cr } applied to a matrix argument $$A$$ yields a matrix value $$F = f(A)$$.
The differential of the trace of such a function is given by \eqalign{ d{\,\rm Tr}(F) &= f'(A^T):dA \quad\implies \frac{\partial{\,\rm Tr}(F)}{\partial A} = f'(A^T) \cr }
Consider a simple function like $$f(x)=x^3$$ \eqalign{ {\rm Tr}(F) &= I:f(X) \cr &= I:XXX \cr d\,{\rm Tr}(X^3) &= I:dX\,X\,X + I:X\,dX\,X + I:X\,X\,dX \cr &= 3(X^T)^2:dX \cr &= f'(X^T):dX \cr } Applying this procedure to each term in a Taylor expansion, you'll find that this result holds for any analytic function, not just the logarithm.
Further, thanks to Cayley-Hamilton, such a Taylor series is finite, containing only $$n$$ terms for $$A\in{\mathbb C}^{n\times n}$$
NB: In some steps above, colons are used as a convenient product notation for the trace, i.e. $${\rm Tr}(A^TB) = A:B = A_{ij} B_{ij}$$ Note that the RHS is the index notation equivalent of the colon product. Using that equivalence to replace the colons in this post will generate a proof via index notation.