Closed form expression for matrix exponential derivative with respect to scalars

In general, the derivative of the exponential map is given by $$ \frac{d}{dt}e^{Z(t)} = e^{Z}\frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}\frac{dZ(t)}{dt} $$ Thus, for your case of $Z(a) = aX + bY$, we have $$ \begin{align*} \frac{d}{da}e^{Z} &= e^{Z}\frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}(X) \\ & = e^Z \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} \mathrm{ad}_Z^k(X) \end{align*} $$ Where $\mathrm{ad}_Z^k(X)$ denotes $[\overbrace{Z,[Z,\cdots,[Z}^{k \text{ times}},X]\cdots]]$.

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From the other answer based on Hall's text, we also have $$ \left. \frac{d}{da} e^Z \right|_{a = 0} = e^{bY}\left\{ X - \frac 1{2!}b[Y,X] + \frac 1{3!}b^2[Y,[Y,X]] - \cdots \right\} $$

The number $b$ is really irrelevant to your question.

For all $X,Y\in M_n(\mathbb{C})$, we have

$$ \frac{d}{dt}e^{X+tY}\big|_{t=0}=e^X\left\{ Y-\frac{[X,Y]}{2!}+\frac{[X,[X,Y]]}{3!}-\cdots \right\}. $$

More generally, if $X(t)$ is a smooth matrix-valued function, then $$ \frac{d}{dt}e^{X(t)}=e^{X(t)}\left\{ \frac{I-e^{-\operatorname{ad}_{X(t)}}}{\operatorname{ad}_{X(t)}}\bigg(\frac{dX}{dt}\bigg) \right\} $$

See Theorem 5.4 (Derivative of Exponential) and its proof in Brian Hall's Lie Groups, Lie Algebras, and Representations.