Derivative of the product of operators and Derivative of exponential

$$A(\lambda+\epsilon)B(\lambda+\epsilon) = (A(\lambda) + \epsilon \dot{A} )(B(\lambda) +\epsilon \dot B ) = A(\lambda)B(\lambda) + \epsilon(\dot AB+A\dot B) + o(\epsilon^2)$$

Here we will only consider the added last subquestion (v4):

$$ \frac{d}{d\lambda}e^{\hat{A}} ~=~ \int_0^1\!ds~e^{(1-s)\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .\tag{1}$$

The identity (1) follows by setting $t=1$ in the following identity

$$ e^{-t\hat{A}} \frac{d}{d\lambda}e^{t\hat{A}} ~=~ \int_0^t\!ds~e^{-s\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .\tag{2}$$

To prove equation (2), first note that (2) is trivially true for $t=0$. Secondly, note that a differentiation wrt. $t$ on both sides of (2) produces the same expression

$$e^{-t\hat{A}}[\frac{d}{d\lambda},\hat{A}]e^{t\hat{A}}~=~e^{-t\hat{A}}\frac{d\hat{A}}{d\lambda}e^{t\hat{A}},\tag{3}$$

where we use the fact that

$$\frac{d}{dt}e^{t\hat{A}}~=~\hat{A}e^{t\hat{A}}~=~e^{t\hat{A}}\hat{A}.\tag{4}$$

So the two sides of eq.(2) must be equal. $\Box$