Suppose we have $$e^{At}e^{Bt}=F(t),$$ where $$A, B$$ - operators that do not commute. Now I need to take the derivative $$dF(t)/dt.$$ In which order do I write the operators?
$$dF(t)/dt = Ae^{At}e^{Bt} + e^{At}Be^{Bt}$$ or $$dF(t)/dt = e^{At}Ae^{Bt} + e^{At}e^{Bt}B~?$$
They are the same, since any operator commutes with its exponential \begin{align} Ae^{tA} &= A\left(1+tA+\frac{1}{2!}t^2A^2+\dots\right) \\ &= A+tA^2+\frac{1}{2!}t^2A^3+\dots \\ &= \left(1+tA+\frac{1}{2!}t^2A^2+\dots\right)A \\ &= e^{tA}A \end{align} (and, in general, any operator $A$ commutes with every function $f(A)$ of it, for similar reasons).
Expanding on Cosmas Zachos' comment Suppose $G(x)$ is some function which has an expression as a Taylor series.
$G(x) = \sum_i g_i x^i$
It can be proven* by induction that
$$ [X,G(X)]=0 $$
This means $[X,e^{Xt}$]=0$
In particular
$$ Ae^{At} = e^{At}A $$ $$ Be^{Bt} = e^{Bt}B $$
so both of your expressions are equal. What is important is to take care that you don't swap any terms involving $A$'s with any terms involving $B$'s. The answer may have been more complicated.
*At least in a physics usage of the word..
