Derivative of matrix involving trace and log I'm stuck on this problem.
Let $X\in\mathbb{R}^{n\times n}$, compute the following matrix derivatives
$$\frac{\partial}{\partial X}\mathrm{tr}(\log(XA)\log(XA)^\top),$$
$$\frac{\partial}{\partial X}\mathrm{tr}(B\log(XA)), $$
where $\log(\cdot)$ is the matrix logarithm (not element-wise) and $A,B\in\mathbb{R}^{n\times n}$ are constant matrices.
Thanks in advance for your suggestions!
 A: I assume you are able to compute the derivative without trace. And the rest part, actually, is not hard. Try to compute it componentwisely, then
$$\begin{align}
\frac{\partial}{\partial X^i_j}\text{tr}f(X)=&
\frac{\partial}{\partial X^i_j}f^k_l(X)\delta^l_k\\
=&\frac d{dx}f^k_m(x)|_{x=X}\frac{\partial}{\partial X^i_j}X^m_l\delta^l_k\\
=&\frac d{dx}f^k_m(x)|_{x=X}\delta^m_i\delta^j_l\delta^l_k\\
=&\frac d{dx}f^j_i(x)|_{x=X}
\end{align}$$
The conclusion is

$$\frac{\partial}{\partial X}\text{tr}f(X)=\left(\frac d{dx}f(x)|_{x=X}\right)^T$$

A: Clearly the calculation is quasi-unfeasible. In particular, the answer of GMB is absolutely false (cf. my comment above). Who has given  a good point to his answer ?
Yet, one can give a compact answer if $X=A^{-1}$ because $D\log_I=id$.
Let $f(X)=tr(B\log(XA))$. Then $Df_{A^{-1}}:H\rightarrow tr(BHA)$. Thus $\nabla (f)_{A^{-1}}=(AB)^T$.
EDIT: Let $h(X)=tr(\log(I+X))$ where $\log$ denotes the principal logarithm and $||X||<1$. Then $h(X)=tr(\sum_{k=1}^{\infty}(-1)^{k+1}/k.X^k)
 =\sum_{k=1}^{\infty}(-1)^{k+1}/k.tr(X^k)$ and $Dh_X:H\rightarrow \sum_{k=1}^{\infty}(-1)^{k+1}/k.tr(kX^{k-1}H)$ (because $tr(UV)=tr(VU)$)$=tr(\sum_{k=1}^{\infty}(-1)^{k-1}X^{k-1}H)=tr((I+X)^{-1}H)$. In other words $\nabla(h)_X={(I+X)^{-1}}^T$. More generally, if $X$ has no eigenvalues in $\mathbb{R}^-$, then let $g(X)=tr(\log(X))$. One has $Dg_X:H\rightarrow tr(X^{-1}H)$ and $\nabla(g)_X={X^{-1}}^T$.
Here $f(X)=tr(B\log(XA))=tr(B\log(Y))$; then we must derive $u(Y)=tr(BY^k)$. Unfortunately $Du_Y:K\rightarrow tr(B(kY^{k-1})K)$ is false in general ; yet, it is true if, for instance, $BY=YB$.
 Finally, when $BXA=XAB$, $Df_X:H\rightarrow tr(B(XA)^{-1}HA)$ (because $K=HA$)$=tr((XA)^{-1}BHA)=tr(X^{-1}BH)$ and $\nabla(f)_X=(X^{-1}B)^T$ does not depend on $A$. 
A: Here's an incomplete answer:
$\log A = - \sum_{i = 1}^{\infty} \, \frac{1}{i}(I - A)^i$
So
$d \log A = \sum_{i = 0}^{\infty} \, (I - A)^i \, dA$
If you know the chain and product rules, deriving over the $\log A$ term should be the only tricky part.
A: $
\def\l{\lambda}\def\p{\partial}
\def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)}
\def\trace#1{\operatorname{tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\notimplies{\;{\large\not}\!\!\!\!\!\implies}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}
$Denote the function $\Phi(\l)$ of a scalar $\l\,$ and its derivative
$$\phi(\l) = \frac{d\Phi}{d\l}$$
and write Shuchang's matrix result as a differential expression
$$\eqalign{
d\trace{{\large\Phi}(X)} &= \phi\!\LR{X^T} :dX \\
}$$
where the colon denotes the Frobenius product
$$\trace{A^TB} \;=\; A:B \;=\; A^T:B^T $$
Let's also define the matrix variables
$$\eqalign{
Y &= XA
 &\implies\quad Y^{-1} = A^{-1}X^{-1} \\
L &= \log(Y) &\;\notimplies\quad L = \log(X) + \log(A) \\
}$$
Consider a function to which Shuchang's result can be applied, e.g.
$$\eqalign{
\phi &= \trace{L} \\
 &= \trace{\log(Y)} \\
d\phi
 &= \LR{Y^T}^{-1}:dY \\
 &= \LR{Y^{-1}}^T:\LR{dX\,A} \\
 &= \LR{Y^{-1}}^TA^T:dX \\
 &= \LR{AY^{-1}}^T:dX \\
 &= \LR{X^{-1}}^T:dX \\
\grad{\phi}{X} &= \LR{X^T}^{-1} \\
}$$
But differentiation of the functions posed in this question get stuck after the very first step !
$$\eqalign{
\phi &= \trace{B^TL} = B:L \\
d\phi &= B:dL \\
\\
\psi &= \trace{LL^T} = L:L \\
d\psi &= 2L:dL \\
}$$
The problem is that no simple expression for $dL\,($in terms of $dX)$ exists. Note that Shuchang's result applies to the trace of  $dL$ and not to $dL$ itself, so it cannot be used here.
The standard approach is to employ a power series, but one much more complicated than that proposed by GMB $\,($who naively assumed that $A$ commutes with $dA).\,$
While the power series approach yields a solution, it's complicated, has a tiny radius of convergence, and when it does converge it is so slow that it's practically useless.
Since high quality algorithms exist for evaluating matrix logarithms, a better approach is to calculate $dL$ via the logarithm of a block-triangular matrix
$$\eqalign{
&\m{L&dL\\0&L} =
 \log\L(\m{Y&dY\\0&Y}\R) =
 \log\L(\m{X&dX\\0&X}\cdot\m{A&0\\0&A}\R) \\
\\
d\phi = &\m{0&B\\0&0}:\log\L(\m{X&dX\\0&X}\cdot\m{A&0\\0&A}\R) \\
\\
d\psi = &\m{0&2L\\0&0}:\log\L(\m{X&dX\\0&X}\cdot\m{A&0\\0&A}\R) \\
}$$
These expressions are straightforward and accurate. All of the problematic non-commutative behavior is handled by the block structure of the function argument.
Despite the fact that $dX$ appears within a function argument,
these formulas represent a practical way to calculate
directional derivatives. Simply set $dX$ parallel to the direction of interest and scale it to have unit norm.
In summary, there are no closed-form solutions for
$\,\large\grad{\phi}{X}\,$ or $\,\large\grad{\psi}{X}$
