# 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.

Using the formula I linked above, we come up with the answer $$A'(0) = \int_0^1 e^{\alpha V}We^{(1 - \alpha)V}\,d\alpha$$ I'm not sure how helpful that is, but it's something. Certainly, this is can be approximated in a fairly straightforward fashion.
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 }