On the derivative of the exponential of adjoint action on a Lie algebra I am studying the Drinfel'd-Sokolov hierarchies. And here is what confuses me. Let $\mathfrak{g}$ be a (simple) finite-dimensional Lie algebra, let $u(x), v(x) \in C^\infty(\mathbb{R},\mathfrak{g})$. We can define the exponential of $F\in\operatorname{End}(\mathfrak{g})$ as
$\exp(F) := \mathrm{id} + F + \frac{1}{2!}(F\circ F)+\cdots \in \operatorname{GL}(\mathfrak{g})$
Question: how to calculate $\frac{d}{dx} \exp(\operatorname{ad} u(x))(v(x))$?
I believe there is an explicit formula but after long time searching, I cannot find any. And I hope I can get some inspiration here.
 A: Every Lie group is locally isomorphic to a matrix Lie group. So in calculations like these it is in practice sufficient to do them in matrix groups. Moreover, you defined the exponential using power series which kind of implies that you are actually taking the exponential to live not in $G$ but in $\mathrm{Aut}(\mathfrak{g})$.
The result is
$$
 e^{\mathrm{ad}_{u(x)}} ([u'(x), v(x)] + v'(x)).
$$
A: It looks as though this is a standard lemma: https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula.
A: From wikipedia we know that
$$\frac{d}{dx} \exp(u(x)) = \exp(u(x))  \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} (\operatorname{ad} u(x))^k \frac{d u(x)}{dx}.$$
Here in my question we have a discussion on differentiating $\exp(\operatorname{ad} X)$. My calculation shows that we actually have similar formula:
$$\frac{d}{dx} \exp(\operatorname{ad} u(x)) = \exp(\operatorname{ad} u(x)) \operatorname{ad} \left( \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} (\operatorname{ad} u(x))^k \frac{d u(x)}{dx}  \right).$$
