On the derivative of the matrix exponential I have the following set up, for column vectors $\alpha,\beta$ and square matrix $A$ I am looking at the derivative $\frac{\partial\alpha^T e^{Ax}\beta}{\partial x}=\alpha^T Ae^{Ax}\beta=\alpha^T e^{Ax}A\beta$, where $e^{Ax}$ represents the matrix exponential.
Can I write this in terms of the original function? So,
$$\frac{\partial\alpha^T e^{Ax}\beta}{\partial x}=\alpha^T Ae^{Ax}\beta=X\alpha^T e^{Ax}\beta$$ 
for some $X$? Under what circumstances is this possible? Thank you!
 A: We have fixed vectors $\alpha$ and $\beta$; let $M = e^{Ax}$. What you're essentially asking is whether the product $\alpha^T(AM)\beta$ can be rewritten as $X(\alpha^TM\beta)$ for some matrix $X$ (which is presumably inependent of $x$).  I claim that this is pretty obviously going to fail in general.
Consider the case in which $\alpha = \beta = (1,0,\dots,0)$.  In this context, note that $\alpha^TM\beta = M_{11}$ (the 1,1 entry of $M$).  You're question amounts to asking whether there exists "some $X$" such that
$$
[AM]_{11} =  X[M]_{11}
$$
First of all, note that $X$ must be a scalar for this multiplication to make sense.  Second, there is no reason we should expect $[AM]_{11}$ to be a multiple of $[M]_{11}$ except, for example, if both $A$ and $M$ are diagonal.  More on that in a minute.  If $X$ is allowed to depend on $x$, we could always define
$$
X(x) =  \frac{[AM(x)]_{11}}{[M(x)]_{11}} = \frac{\alpha^TAM\beta}{\alpha^TM\beta} = \frac{(A\alpha)^TM\beta}{\alpha^TM\beta} = \frac{\alpha^TM(A\beta)}{\alpha^TM\beta}
$$
but that doesn't seem like the kind of thing you're looking for.

Here's a case where this can work: suppose that $\beta$ is an eigenvector of $A$ associated with $\lambda$.  $\beta$ will also be an eigenvector of $e^{Ax}$.  In fact, we have
$$
\alpha^T Ae^{Ax} \beta = \alpha^T e^{Ax} A\beta = \alpha^T e^{Ax}(\lambda \beta) = 
\lambda (\alpha^T e^{Ax} \beta)
$$
A similar trick works if $\alpha^T$ is an eigenvector of $A^T$.
I can think of no other case in which we could expect such behavior.
A: Let $f : \mathbb R \to \mathbb R$ be defined by
$$f (x) := \mathrm u^{\top} \exp (\mathrm A x) \, \mathrm v$$
Taking the derivative,
$$f ' (x) = \lim_{h \to 0} \left( \dfrac{f (x + h) - f (x)}{h} \right) = \cdots = \mathrm u^{\top} \mathrm A \exp (\mathrm A x) \, \mathrm v = \mathrm u^{\top} \exp (\mathrm A x) \, \mathrm A \mathrm v$$
If $\mathrm v$ is a right eigenvector of $\mathrm A$ with eigenvalue $\lambda$, then
$$f ' (x) = \mathrm u^{\top} \exp (\mathrm A x) \, \mathrm A \mathrm v = \mathrm u^{\top} \exp (\mathrm A x) \, (\lambda \mathrm v) = \lambda \, f (x)$$
If $\mathrm u$ is a left eigenvector of $\mathrm A$ with eigenvalue $\mu$, then
$$f ' (x) = \mathrm u^{\top} \mathrm A \exp (\mathrm A x) \, \mathrm v = (\mu \, \mathrm u^{\top}) \exp (\mathrm A x) \, \mathrm v = \mu \, f (x)$$
