Show differentiability of $e^{\int_0^t A(\tau)\operatorname{d}\tau}$ Consider $A\in C(\mathbb{R},\mathcal{L}(E))$ for some Banach space $E$. For any $t\in\mathbb{R}$ $$B(t):=e^{\int_0^t A(\tau)\operatorname{d}\tau}:= \sum_k \left(\int_0^t A(\tau)\operatorname{d}\tau\right)^k/(k!)$$ is well-defined. How can I show that $B\in C^1(\mathbb{R},\mathcal{L}(E))$?
Is a general result about the differentiability of $e^{f(\cdot)}$ for $f\in C^1(\mathbb{R},\mathcal{L}(E))$ possible?
 A: We begin by considering the general case where $f\in\mathscr{C}^1(\mathbb{R},\mathcal{L}(E))$. It is always true that $e^{f(t)}$ is differentiable, but in general it is not true that $\frac{d}{dt}e^{f(t)}=f'(t)e^{f(t)}.$ 
Since $f(t)$ and $f'(t)$ are both bounded on compact sets in $\mathbb{R}$ that we can pass the limit inside the series to obtain
$$\lim_{h\to 0}\frac{e^{f(t+h)}-e^{f(t)}}{h}=\sum_{k=1}^{\infty}\frac{1}{k!}\frac{d}{dt}(f(t))^k=\sum_{k=1}^{\infty}\frac{1}{k!}\sum_{i=1}^{k}\underbrace{f(t)\cdots\left(\frac{df}{dt}(t)\right)\cdots f(t)}_{\frac{d}{dt}\text{ is in the ith position}},$$
where the last equality follows from the chain rule (and the index $k=0$ drops out since $(f(t))^0=\mathrm{id}$.
So we see that if $\frac{d f}{dt}(t)$ does not commute with $f(t)$ that we cannot simplify this (at least not in any immediate way that I can see right now).
However, if $\frac{df}{dt}(t)$ commutes with $f(t)$, then we deduce that 
$$\underbrace{f(t)\cdots\left(\frac{df}{dt}(t)\right)\cdots f(t)}_{\frac{d}{dt}\text{ is in the ith position}}=(f(t))^{k-1}\frac{df}{dt}(t),$$
and so we recover
$$\frac{d}{dt}e^{f(t)} = \sum_{k=1}^{\infty}\frac{k}{k!}(f(t))^{k-1}\frac{df}{dt}(t)=\frac{df}{dt}(t)\sum_{k=0}^{\infty}\frac{1}{k!}(f(t))^{k}=\left(\frac{df}{dt}(t)\right)e^{f(t)}=e^{f(t)}\left(\frac{df}{dt}(t)\right). $$ In particular, we see that $e^{f}$ is differentiable (in fact $e^{f}\in\mathscr{C}^1(\mathbb{R},\mathcal{L}(E))$.
Even in the case that they do not commute we can find that $e^{f}$ is $\mathscr{C}^1$ with a bit more work. First note that for every $k\geq 1$ we have
$$\left\|\sum_{i=1}^{k}\underbrace{f(t)\cdots\left(\frac{df}{dt}(t)\right)\cdots f(t)}_{(d/dt)\text{ is in the ith position}}\right\|\leq\sum_{i=1}^{k}\|f(t)\|^{k-1}\left\|\frac{df}{dt}(t)\right\|=\|f(t)\|^{k-1}\left\|\frac{df}{dt}(t)\right\|.$$
So the series $\sum_{k=1}^{\infty}\frac{1}{k!}\sum_{i=1}^{k}\underbrace{f(t)\cdots\left(\frac{df}{dt}(t)\right)\cdots f(t)}_{(d/dt)\text{ is in the ith position}}$ is absolutely convergent (the norm is bounded above by $\|\frac{df}{dt}(t)\|e^{\|f(t)\|},$ and so since $E$ is a Banach space we see that the series converges. Hence, the limit of the difference quotient exists and $e^{f}$ is differentiable. It just remains to show that $\frac{d}{dt}e^f$ is continuous, and this should follow easily by considering a sequence $t_n\to t$ and a passage of the limit through the series.
Now let's consider your specific example: fix $A\in\mathscr{C}(\mathbb{R},\mathcal{L}(E))$ and let $f(t)=\int_{0}^{t}A(\tau)~d\tau$. Clearly, $f\in\mathscr{C}^1(\mathbb{R},\mathcal{L}(E))$ with $\frac{df}{dt}(t)=A(t)$. Now we see that $$\left[\frac{df}{dt}(t),f(t)\right]=\left[A(t),\int_{0}^{t}A(\tau)~d\tau\right]=\int_{0}^{t}A(t)A(\tau)~d\tau-\int_{0}^{t}A(\tau)A(t)~d\tau;$$
I'm pretty sure, that this is not necessarily equal to zero (not entirely sure about this though), and so there isn't necessarily a nice formula for the derivative in this case.
