Show that if $A, B \in \mathbb{C}^{p \times p}$, then
$$
\lim _{(s, t) \rightarrow(0,0)} \frac{e^{t A} e^{s B} e^{-t A} e^{-s B}-I_p}{s t}=A B-B A \text {. }
$$
Let
$$
F(t)=e^{t A}=I_p+t A+t^2 \frac{A^2}{2!}+\cdots .
$$
Then
$$
F(0)=I_p
$$
and
$$
\begin{aligned}
\frac{F(t+h)-F(t)}{h} & =\frac{e^{(t+h) A}-e^{t A}}{h} \\
& =e^{t A}\left(\frac{e^{h A}-I_p}{h}\right),
\end{aligned}
$$
which tends to
$$
e^{t A} A=A e^{t A}
$$
as $h$ tends to zero, thanks to the bound (13.6). Thus, the derivative
$$
F^{\prime}(t)=\lim _{h \rightarrow 0} \frac{F(t+h)-F(t)}{h}=A F(t) .
$$
The same definition is used for the derivative of any suitably smooth matrix valued function $F(t)=\left[f_{i j}(t)\right]$ with entries $f_{i j}(t)$ and implies that
$$
F^{\prime}(t)=\left[f_{i j}^{\prime}(t)\right], \quad \text { and correspondingly } \quad \int_a^b F(s) d s=\left[\int_a^b f_{i j}(s) d s\right] ;
$$
i.e., differentiation and integration of a matrix valued function is carried out on each entry in the matrix separately.