Assuming that the operator $\hat{A}$ does not commute with $d \hat A/dt$, what is $\frac{d}{dt}(\exp(\hat A))$? $\exp(\hat A(t)) \frac{d \hat A}{dt}$ or $\frac {d\hat A}{dt} \exp(\hat A(t))$ and why?

***EDIT***: Actually this was my original question: I was asked to show that

$\frac{d(\exp{A})}{dt} \exp{(-A)} = \Sigma \frac{1}{(n+1)!} L^n_A(\frac{dA}{dt})$, with $L_A(X) = [A,X]$, $L^2_A(X) = [A,[A,X]]$, ...

I thought the Baker–Campbell–Hausdorff formula said

$\exp{(A)} B \exp{(-A)} = \Sigma \frac{1}{(n+1)!} L^n_A(B)$,

and so comparing the two equations above, I would get

$\frac{d\exp(A)}{dt} = \exp{(A)} \frac{dA}{dt}$ (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?