Let $t\mapsto A(t)\in C(\mathbb{R},M_n(\mathbb{R}))$, it seems "obvious" that we should have:
\begin{equation} \lim_{h\rightarrow 0} \frac{\exp\left\{\int_t ^{t+h}A(s)\,\mathrm{d}s\right\}-I_n}{h}=A(t)\end{equation}
However, I'm having a hard time finding rigorous proof of this result. I don't even know where to start?
Use a taylor expansion of $\exp(t)$, since $A$ is continuous in $\mathbb{R}$, we observe $|\int_{t}^{t+h}A(s)ds| \leq \max_{s \in [t,t+h]}\|A(s)\| h < \infty$. Then \begin{align} \exp \left \{ {\int_{t}^{t+h}A(s)ds} \right\} = I_n+ {\int_{t}^{t+h}A(s)ds} + O(h^2) \end{align} Consequently \begin{align} \frac{\exp \left \{ {\int_{t}^{t+h}A(s)ds} \right\} - I_n}{h} = \frac{1}{h}{\int_{t}^{t+h}A(s)ds} + O(h) \end{align} By Fundamental theorem of calculus \begin{align} \lim_{h \to 0 }\frac{\exp \left \{ {\int_{t}^{t+h}A(s)ds} \right\} - I_n}{h} = \frac{d}{dh}\int_{t}^{t+h}A(s)ds \left |_{h=0} = A(t) \right.. \end{align}