Derivative of Matrix in Determinant

Question

If we have A, B, and X are all matrices, can we find the derivative of $$ \frac{d}{dX} det(AXB). $$ I notice that $$ \frac{d}{dX}det(X) = det(X)X^{-T}, $$ but I don't know how to pass into the $AXB$. Can anyone help me with this problem? I also notice that $$ \frac{d}{dX} AXB = A \otimes B^T $$ and if we assume $$ Y(X) = AXB, $$ then we can have $$ \frac{d}{dX}Y(X) = det(Y(X))[Y(X)]^{-T} \frac{d}{dX}Y(X). $$ Then I find there is a problem that the size of matrices is not matched with $[Y(X)]^{-T}$ and $\frac{d}{dX}Y(X)$. So I am wondering whether the chain rule in here is not correct?

Answer 1

$ \def\o{{\tt1}}\def\p{\partial} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} $The proposed derivative formula is missing a transpose. It should read $$\eqalign{ \grad{\det(M)}{M} = \det(M)\,M^{-T} \\ }$$ One can turn this into the corresponding formula for the differential of the determinant $$\eqalign{ d\det(M) = \BR{\det(M)\,M^{-T}}:dM \\ }$$ where $(:)$ denotes the Frobenius product, which is a concise notation for the trace $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \big\|A\big\|^2_F \\ }$$ Now substitute $$\eqalign{ M = AXB \qiq \c{dM = A\,dX\,B} \\ }$$ into the differential and recover the gradient $$\eqalign{ d\det(M) &= \BR{\det(M)\,M^{-T}}:\c{\LR{A\,dX\,B}} \\ &= \BR{\det(M)\,A^TM^{-T}B^T}:dX \\ \grad{\det(M)}{X} &= \det(M)\;A^TM^{-T}B^T \\ }$$ or in terms of the original variables $$\eqalign{ \grad{\det(AXB)}{X} &= \det(AXB)\;A^T\LR{AXB}^{-T}B^T \\ }$$ Note that this result holds even when $\{A,B\}$ are rectangular.