Let $X,Y\in\mathbb{R}^{n\times n}$ and define $$ F(a,t) = \exp[(aX+Y)t]. $$ Then, compute $$ \frac{\partial F}{\partial a} = ? $$
My attempt: Let $M_a = aX + Y$, then $$ \frac{\partial F}{\partial a} = \frac{\partial}{\partial a} ( I_n + M_a t + M_a^2 t^2 / 2! + M_a^3 t^3 / 3! + \dots ). $$ Since $\partial M_a^k /\partial a = \sum_{j=0}^{k-1} M_a^j X M_a^{k-1-j}$, we obtain $$ \frac{\partial F}{\partial a} = \sum_{k=1}^\infty \sum_{j=0}^{k-1} M_a^j X M_a^{k-1-j} \frac{t^k}{k!}. $$ The question, therefore, is that whether it's possible to write this power series as a function?
$\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}\def\p#1#2{\frac{d #1}{d #2}}$Given a scalar function, like $f(x)=\exp(x)$, and the matrices $$\eqalign{ M = M(a)&= aXt + Yt,\qquad &\dot M = \dot M(a) = \p{M}{a} = Xt \\ F = F(a) &= f(M), &\dot F = \dot F(a) = \p{F}{a} \\ }$$ Evaluating the function with a block-triangular argument yields $$\eqalign{ B \;=\; f\left(\m{M&\dot M\\{0}&M}\right) \;=\; \m{F&\dot F\\0&F} \\ }$$ Block-matrix analogs of the standard basis vectors $$\eqalign{ e_1 &= \m{1\\0} \quad &e_2 = \m{0\\1} \\ E_1 &= e_1\otimes I_n \qquad &E_2 = e_2\otimes I_n \\ }$$ can be used to extract the upper-right block, which one might call a closed-form solution $$\eqalign{ \dot F = E_1^T B\, E_2 \\ }$$