Closed form expression for matrix exponential derivative with respect to scalars I'd like to evaluate generic expressions of the following form:
$$\frac{d}{da}\exp\left[aX + bY\right]$$
where $a,b$ are scalars and $X,Y$ are arbitrary complex matrices. Replacing the exponential with its defining series gives us an infinite series of terms which are each symmetrized by the derivative:
$$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d}{da}\left(aX+bY\right)^{n}$$
For say $n=3$ this gives a term of the form
$$X(aX+bY)^{2}+(aX+bY)X(aX+bY)+(aX+bY)^{2}X$$
I don't see a clean way to resum such a series into a nice analytic form, but I thought maybe there was a clean result for this up to commutators. Any insights or literature suggestions are very welcome.
 A: 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 $[\overbrace{Z,[Z,\cdots,[Z}^{k \text{ times}},X]\cdots]]$.

From the other answer based on Hall's text, we also have
$$
\left. \frac{d}{da} e^Z \right|_{a = 0} = e^{bY}\left\{
X - \frac 1{2!}b[Y,X] + \frac 1{3!}b^2[Y,[Y,X]] - \cdots
\right\}
$$
A: 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.
A: If $X$ and $Y$ commute, then
$$
\frac{d}{da}e^{aX+bY}=\frac{d}{da}e^{aX}e^{bY}=Xe^{aX+bY}.
$$
A: Let me add a different way of writing the result.
If $X(t)$ is an operator-valued function, then
$$ \frac{\mathrm d}{\mathrm dt} \mathrm e^{X(t)} = \int_0^1 \mathrm e^{(1-\lambda) X(t)} \frac{\mathrm dX}{\mathrm dt} \mathrm e^{\lambda X(t)}\; \mathrm d\lambda . \tag{*} $$
Proof.
We begin with the identity
$$ \mathrm e^{-\Lambda X} \frac{\mathrm d}{\mathrm dt} \mathrm e^{\Lambda X} = \int_0^\Lambda \mathrm e^{-\lambda X} \frac{\mathrm dX}{\mathrm dt} \mathrm e^{\lambda X}\; \mathrm d\lambda . $$
These two expressions are equal because they satisfy the same initial value problem as functions of $\Lambda$.
Setting $\Lambda = 1$ gives the desired result (*).

Applied to your example:
$$ \frac{\mathrm d}{\mathrm da} \mathrm e^{aX + bY} = \int_0^1 \mathrm e^{(1-\lambda)(aX+bY)}\, X\, \mathrm e^{\lambda(aX+bY)}\; \mathrm d\lambda . $$
