Let $X$ be a Banach space and $A:\mathbb{R} \to L(X,X)$, i.e $A(t)$ is an linear and bounded operator. Furthermore there holds: $A(s)A(t)=A(t)A(s), \; \forall s,t \in \mathbb{R}$.
Now define $u(t):=exp(-\int_{s}^t A(r) \,dr)\; u_0$, with $u_0 \in X$ arbitrary and $exp(B):=\sum_{k=0}^{\infty} \frac{B^k}{k!}$ for $B \in X$.
I want to show that $u'(t)=-A(t)u(t)$:
$$u'(t)=exp(-\int_{s}^t A(r) \,dr)\, (-A(t))\,u_0=-A(t)\,exp(-\int_{s}^t A(r) \,dr)\, u_0$$
but why does the last equality holds? Why commutes everything?