Derivative of scalar-valued function with kronekcer product: $\log(\operatorname{det}(AA^T \otimes BB^T))$ Let A and B be $m\times p$ and $n\times q$ matrices, respectively. Define the $mn \times mn$ matrix C as $$C=AA' \otimes BB'$$ with $A'$ denoting the transpose of $A$. Consider the scalar-valued function $f(C)= \log|C|$ with $|\cdot|$ denoting the matrix determinant. I am looking for the derivatives $\partial f(C)/\partial A$ and $\partial f(C)/\partial B$.
Any idea?
 A: $
\def\l{f}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\rank#1{\operatorname{rank}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$First, introduce the Frobenius product, which can be thought of as a product notation for the trace
$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A\cdot B^T} \\
A:A &= \|A\|^2_F \\
}$$
Then consider various combinations of Kronecker, Frobenius and Matrix products
$$\eqalign{
&\LR{A\otimes B}\,\cdot\LR{E\otimes F} = \LR{A\cdot E}\otimes\LR{B\,\cdot F}  \\
&\LR{A\otimes B}:\LR{E\otimes F} = \LR{A:E}\otimes\LR{B:F}
   = \LR{A:E}\LR{B:F}  \\
&\LR{A\otimes B}^+ = \LR{A^+\otimes B^+} \\
&\LR{A\otimes B}^T \,=\, \LR{A^T\otimes B^T} \\
&A:B = B:A = A^T:B^T \\
&C:\LR{A\cdot B} = \LR{C\cdot B^T}:A = \LR{A^T\cdot C}:B \\
}$$
In the expressions above, the Matrix product has been shown explicitly, but it's usually omitted $$AB=A\cdot B$$
One final formula is for the rank of a matrix in terms of its pseudoinverse
$$\eqalign{
\rank B &= \trace{B^+B} = B^+:B^T \\
 &= B^T\LR{BB^T}^{-1}:{B^T} \\
 &= \LR{BB^T}^{-1}:\LR{BB^T} \\
\\
}$$

Use Jacobi's formula to calculate the gradient of the function with respect to $A$
$$\eqalign{
C &= AA^T\otimes BB^T \;=\; C^T \\
dC &= \LR{A\;dA^T+dA\;A^T}\otimes BB^T \\
\\
\l &= \log(\det(C)) \\
d\l &= d\trace{\log(C)} \\
 &= C^{-1}:dC \qquad\qquad \big\{{\rm Jacobi}\big\} \\
 &= \LR{\LR{AA^T}^{-1}\otimes\LR{BB^T}^{-1}}
  :\BR{\LR{A\;dA^T+dA\;A^T}\otimes BB^T} \\
 &= \rank{B}\;\LR{AA^T}^{-1}:\LR{A\;dA^T+dA\;A^T} \\
 &= 2\rank{B}\;\BR{\LR{AA^T}^{-1}A}:dA \\
 &= 2n\,\LR{A^+}^T:dA \\
\\
\grad{\l}{A}
 &= 2n\,\LR{A^+}^T \\
}$$
The calculation for the gradient wrt $B$ is exactly the same
$$\eqalign{
\grad{\l}{B} &= 2m\,\LR{B^+}^T
 \qquad\qquad\qquad\qquad\qquad
 \qquad\qquad\qquad\qquad \quad
\\
}$$
Update
On second thought, it might be simpler to start from the relationship
$$\det(C) = {\det\!\LR{AA^T}^n}\;{\det\!\LR{BB^T}^m}$$
