There is a curious formula relating the derivation of an exponential in some, possibly noncommutative (associative) algebra, namely $$ D(e^K) = \int_0^1 e^{(1-s)K}D(K)e^{sK} ds $$ where $D$ is a derivation (it verifies the usual Leibniz rule) and $K$ is a nilpotent element (there exists some $n\in \mathbb{N}$ such that $K^n = 0$).
The first idea would be to write
$$
D(e^K) = D\left( \displaystyle\sum_{l\geq 0} \frac{K^l}{l!} \right)
=
\displaystyle\sum_{l\geq0} \frac{1}{l!} D(K^l)
$$
Now we apply the Leibniz rule:
$$
\begin{array}{rcl}
D(K^l) &=& D(K)K^{l-1} + K D(K^{l-1})
\\
&=& D(K)K^{l-1} + K D(K) K^{l-2} + K^2D(K^{l-2})
\\
&=& \dots
\\
&=& \displaystyle\sum_{j=0}^l K^{j-1} D(K) K^{l-j}
\end{array}
$$
Then we arrived to
$$
D(e^K) = \sum_{l\geq 0} \sum_{j=0}^l \dfrac{1}{l!} K^{j-1} D(K)K^{l-j}
$$
The latter sum is over all integer points inside, say, the first octant of the plane. The horizontal direction being the $l$-direction and the vertical one is the $j$-direction.
Using Fubini's theorem, we transform the sum of the vertical slices of the first octant into the sum of its horizontal slices:
$$
\sum_{l\geq 0} \sum_{j=0}^l \dfrac{1}{l!} K^{j-1} D(K)K^{l-j} = \sum_{j \geq 0} \sum_{l \geq j} \dfrac{1}{l!} K^{j-1} D(K)K^{l-j}
$$
Then set $m= l-j$ to get
$$
\sum_{j \geq 0} \sum_{m \geq 0} \dfrac{1}{(m+j)!} K^{j-1} D(K)K^{m}
$$
I guess the coefficient $(m+j)!$ should be related with some integration over the $1$-simplex $\{(1-s,s): s\in [0,1]\}$ but I couldn't find yet.
Any hint to finish the proof of the formula? Or maybe a clever approach ?
The formula is usually called Duhamel's formula in the math literature and it is also valid for general $K$ (without the requirement of being nilpotent). Here is an elementary derivation.
Consider
\begin{align} \frac{d}{dt}e^{(A+B)t}e^{-At} &=e^{(A+B)t}(A+B)e^{-At}-e^{(A+B)t}Ae^{-At} \\ &=e^{(A+B)t}Be^{-At}, \end{align}
now integrate both sides from zero to one:
$$ e^{A+B}e^{-A}- 1 =\int_{0}^{1}dt \, e^{(A+B)t}Be^{-At} $$ from which we obtain
$$ e^{A+B}-e^{A}=\int_{0}^{1}dt \, e^{(A+B)t}Be^{(1-t)A}. $$
Now set $B=\epsilon D(A)$ and, assuming that $A$ is differentiable, we obtain
\begin{align} De^{A} & =\lim_{\epsilon\to0}\frac{e^{(A+\epsilon DA)}-e^{A }}{\epsilon} \\ &=\int_{0}^{1}dt\, e^{At}D(A)e^{(1-t)A} . \end{align}