From Eq. 51 of the matrix cookbook we know that
$\frac{\partial \log\det (AXB)}{\partial X} = (X^{-1})^\top$,
where $\det(X)$ is the determinant of $X$.
I was wondering what is the derivative of $\frac{\partial \log\det (AXB + C)}{\partial X}$. Is it still $ (X^{-1})^\top$?
Thanks!
Define the matrix variable $$\eqalign{ Y &= AXB+C \cr }$$ Write the function in terms of this new variable, then find its differential and gradient $$\eqalign{ \phi &= \log\det Y \cr d\phi &= Y^{-T}:dY = Y^{-T}:A\,dX\,B = A^TY^{-T}B^T:dX \cr \frac{\partial\phi}{\partial X} &= A^TY^{-T}B^T = \Big(B(AXB+C)^{-1}A\Big)^T \cr }$$ When $(C=0)$ and $(A,B)$ are invertible, this reduces to the result that you found in the cookbook.
In some of the steps above, a colon was used to denote the trace/Frobenius product, i.e. $$A:B = {\rm tr}(A^TB)$$