$\DeclareMathOperator{\Tr}{Tr}$ I'm trying to find a closed form for $\frac{\partial}{\partial \theta}\Tr(X\log(Y))$ where $X(\theta)$ and $Y(\theta)$ are Hermitian positive definite matrices with trace 1 (i.e. full rank density matrices), parametrized by scalar $\theta$, that in general don't commute.
If it was $\frac{\partial}{\partial \theta}\Tr(X\log(X))$, I could write down the Taylor expansion of the log and using the cyclic property of the trace rearrange all the terms coming from differentiating $X^n$ and pretend like I was doing single variable calculus; which would give me $\Tr(X'\log(X) - X')$, where $X'\equiv\frac{\partial X}{\partial \theta}$.
However, if I were to apply the same approach to $\frac{\partial}{\partial \theta}\Tr(X\log(Y))$, the Taylor expansion gives me a sum of terms like $\Tr(X(Y-1)^n)$ (here $1$ is the identity matrix of the same dimension as $X$ & $Y$):
$$ \Tr(X\log(Y)) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\Tr(X(Y-1)^n) $$
When differentiated, the trace in the $n$-th term produces $$ \Tr(X'(Y-1)^n)+\Tr(XY'(Y-1)^{n-1})+\dots+\Tr(X(Y-1)^k\,Y'\,(Y-1)^{n-k-1})+\dots+\Tr(X(Y-1)^{n-1}Y') $$
The first term can be separated to give $\Tr(X'\log(Y))$. However, the rest of the terms can't be rearranged in a nice way since $X$ breaks the sort of cyclic symmetry we had in the previous case. This is the point I'm stuck at; is there a different approach with which I can manipulate this expression to obtain a closed form derivative? Thanks in advance.
Context: I'm trying to differentiate von Neumann relative entropy in quantum mechanics/information.