Calculating derivative of a special matrix exponential I am trying to calculate the following derivative:
Let $V,W$ be fixed $n \times n$ real matrices. Define $A(t)=e^{V+tW}$.
What is $A'(0)$?
I do not assume $V,W$ commute. (If they commute, it becomes trivial since then $A(t)=e^Ve^{tW}$).
Here is the problem as I see it:
Define $A(t)=e^{\alpha(t)}$ where $\alpha(t)$ is a (smooth) path inside $Mat(n,\mathbb{R})$.
$e^{\alpha(t)}=\sum_{k=0}^\infty \frac{\alpha(t)^k}{k!} \Rightarrow (e^{\alpha(t)})'=\sum_{k=0}^\infty \frac{(\alpha(t)^k)'}{k!}$
But the formula $(\alpha^k)'=k(\alpha^{k-1})\alpha'$ is not necessarily valid if $\alpha,\alpha'$ do not commute.
Here is what happens when we try to calculate $(\alpha^2)'$:
By the Leibniz rule: $(\alpha^2)'=(\alpha \cdot \alpha)'=\alpha' \cdot \alpha + \alpha \cdot \alpha' \neq 2\alpha \cdot \alpha'$ if $\alpha,\alpha'$ do not commute.
The general formula is* $(\alpha(t)^n)'=\sum_{k=1}^n\alpha(t)^{k-1}\alpha'(t)\alpha(t)^{n-k}$, so plugging this into the series formula of matrix exponential does not give you something managable, at least not immediately.
Is there a way to claculate (or at least give an estimate of) the derivative $A'(0)$? Can we deduce if it's zero?

Another approach is maybe to use the Baker–Campbell–Hausdorff formula
but so far it didn't work for me.

*Note that while in general it seems that there is no simpler closed expression for $(\alpha^n)'$, maybe there is till hope that the specific "exponential series" with this term plugged in is tractable in some sense.

You can use a block triangular matrix to calculate the Frechet derivative using the method of Kenney & Laub
$${\rm F}\Bigg(\begin{bmatrix}X&E\\0&X\end{bmatrix}\Bigg) =  \begin{bmatrix}F&L\\0&F\end{bmatrix} $$
where $F = {\rm F}(X)$ and $$L = \lim_{h\rightarrow\,0}\, \frac{{\rm F}(X+hE)-{\rm F}(X)}{h}$$
For your particular case
$$\eqalign{
 {\rm F}(X) &= \exp(X) \cr
 X &= V \cr
 E &= W \cr
 L &= A'(0) \cr
}$$
