Let $$f(x)=e^{Ax+B}, \quad \text{for all }x\in\mathbb R$$ where $A,B$ are $n\times n$ matrice satisfying $AB+BA=0$.
If $B=0$, I know that the derivative is $Ae^{Ax}=e^{Ax}A$. But for the case $B\neq0$, I have no idea. Of course, it is possible to expand $f(x)$ in Taylor series to get a series solution, but I can't simplify it. Is there any other method to get the derivative of $f(x)$? Can the derivative be written in a simple form?
$ \def\o{{\tt1}} \def\K#1{\frac{#1}{k!}} \def\KK#1{\frac{#1}{(2k+1)!}} \def\k{\K{\o}} \def\Sz{\sum_{k=0}^\infty} \def\Sk{\sum_{k=1}^\infty} \def\Sj{\sum_{j=1}^k} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $Define the matrices $$\eqalign{ X &= B+Ax, \qquad F &= \exp(X) \;=\; I+\Sk\K{X^k} \\ }$$ and calculate their derivatives with respect to $x$ (using dot-notation) $$\eqalign{ \dot X &= A, \qquad\qquad \dot F &= \Sk\Sj\K{X^{j-1}A\:X^{k-j}} \\ }$$ That is the standard theoretical approach. However, if you have a reliable algorithm for computing matrix exponentials, then a more practical approach is to simply evaluate a block-triangular matrix $$\eqalign{ \exp\LR{\m{X&A\\0&X}} &= \m{F&\dot F\\0&F} \\ }$$ and extract $\dot F$ from the $(1,2)$ block on the RHS.
NB:$\:$ The commutation relation between $A$ and $B$ is only useful at $x=0$ where $$\eqalign{ X &= B, \qquad \dot F &= \LR{\Sz\KK{B^{2k}}}A \;=\; \fracLR{\sinh(B)}{B}A }$$