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.
 A: A scalar function
$$\eqalign{
f &= f(x) = \sum_{k=0}^\infty \alpha_k x^k \quad\implies
f' &= \frac{df}{dx} \cr
}$$
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.
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.
