It is well known that $\frac{\partial \ \mbox{tr}(\mbox{log}(X))}{\partial X} = X^{-T}$ ($\mbox{log}$ is matrix log, not element-wise), and I'm now trying to compute the following derivative:
$\frac{\partial \ \mbox{tr}(AX \ \mbox{log}(BX))}{\partial X}$.
More generally, I'm interested in the following derivative:
$\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X}$.
Can someone solve these problems? Can these derivatives be written in a closed-form? I will be grateful for any help you can provide.
Let $f:X\rightarrow tr(AX\log(BX))$. For a general $X$ (s.t. $BX$ has no $<0$ eigenvalues), the derivative $Df_X$ has no closed-form
except if $A=B=I$; let $g:X\rightarrow tr(X\log(X))$. Then $Dg_X:H\rightarrow tr(H(\log(X)+I))$, and the gradient is $\nabla(g)(X)=\log(X^T)+I$.
or except if $AB=BA$ and $X=I$ ($B$ has no $<0$ eigenvalues); then $Df_I:H\rightarrow tr(AH\log(B)+ABHB^{-1})=tr(H(A\log(B)+A))$, and the gradient is $\nabla(f)(I)=A^T\log(B^T)+A^T$.