Derivatives of exponential operator I'm reading the paper (eq.(14) and eq.(10)) and got curious how the paper uses this equation:
$\frac{\partial}{\partial c}\exp(-i\Delta t (X+cY)) = \exp(-i\Delta t (X+cY))(-iY\Delta t + \frac{\Delta t^2}{2}[X+cY, Y] + \frac{i\Delta t^3}{6}[X+cY, [X+cY,Y]]+ \cdots )$
Can anybody help me deriving the equation?
 A: The standard identity for the derivative of the exponential map is
$$
\partial_c e^{M(c)}= e^{M} \left (1-\frac{1}{2}[M,\bullet]+ \frac{1}{6}[M,[M,\bullet]]+... \right ) \partial_c M,
$$
where $\bullet$ pipes the argument on the right in case you were not familiar with the adjoint map.
So, just plug in, $M= -i\Delta t (X+cY)$,
$$
\partial_c e^{-i\Delta t (X+cY)} \\ = e^{-i\Delta t (X+cY)} \Delta t \Bigl (-i Y +[\Delta t (X+cY),  Y ]/2 + i[\Delta t (X+cY) , [\Delta t (X+cY),Y]]/6 +...\Bigr  )  ,
$$
amounting to your result.
A: OP's sought-for identity is
$$\begin{align}e^{-\hat{A}}\frac{d}{d\lambda}e^{\hat{A}} ~=~& \int_0^1\!ds~e^{-s\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}}\cr
~\stackrel{(3)}{=}~& \int_0^1\!ds~e^{-s~{\rm ad}\hat{A}}\frac{d\hat{A}}{d\lambda}  \cr
~=~& \int_0^1\!ds\sum_{n=0}^{\infty}\frac{(-s~{\rm ad}\hat{A})^n}{n!}\frac{d\hat{A}}{d\lambda}\cr
~\stackrel{(4)}{=}~& \sum_{n=0}^{\infty}\frac{(-{\rm ad}\hat{A})^n}{(n+1)!}\frac{d\hat{A}}{d\lambda}  ,\end{align}\tag{1}$$
where we have defined the adjoint map
$${\rm ad}\hat{A}~\equiv ~[\hat{A},~\cdot~],\tag{2}$$
used the identity
$$  e^\hat{X} \hat{Y} e^{-\hat{X}}~=~e^{{\rm ad}\hat{X}}\hat{Y},\tag{3}$$
and used the integral
$$ \int_0^1\!ds~s^n~=~\frac{1}{n+1}.\tag{4}$$
The first equality in eq. (1) is proven in my Phys.SE answer here.
