Assuming that the operator $\hat{A}$ does not necessarily commute with $d \hat A/dt$, what is $\frac{d}{dt}(\exp(\hat A))$?$$\frac{d}{dt}(\exp(\hat A))~?\tag{1}$$ Is it $\exp(\hat A(t)) \frac{d \hat A}{dt}$ or $\frac {d\hat A}{dt} \exp(\hat A(t))$ and why?
EDIT: Actually this washere is my original question: I was asked to show that
$\frac{d(\exp{A})}{dt} \exp{(-A)} = \Sigma \frac{1}{(n+1)!} L^n_A(\frac{dA}{dt})$,$$\frac{d(\exp{A})}{dt} \exp{(-A)} = \sum_{n=0}^{\infty} \frac{1}{(n+1)!} L^n_A(\frac{dA}{dt}),\tag{2}$$ with $L_A(X) = [A,X]$, $L^2_A(X) = [A,[A,X]]$, ...$$L_A(X) = [A,X], \qquad L^2_A(X) = [A,[A,X]],\qquad \ldots \tag{3}$$
I thought the Baker–Campbell–Hausdorff formula said
$\exp{(A)} B \exp{(-A)} = \Sigma \frac{1}{(n+1)!} L^n_A(B)$,$$\exp{(A)} B \exp{(-A)} = \sum_{n=0}^{\infty} \frac{1}{n!} L^n_A(B),\tag{4}$$
and so comparing the two equations above, I would get
$\frac{d\exp(A)}{dt} = \exp{(A)} \frac{dA}{dt}$$$\frac{d\exp(A)}{dt} = \exp{(A)} \frac{dA}{dt}\tag{5}$$ (by plugging in $B = \frac{dA}{dt}$). Now that it seems that this is not true, does it mean that the identity I was asked to prove is also not true?
