Derivative of exponential map and the adjoint action

Question

According to wikipedia the derivative of the exponential map (here I will talk about the matrix exponential) is:

$$
\frac{d}{dt}\exp[A(t)] = \exp[A(t)]\frac{1-e^{-\text{ad}_{A}}}{\text{ad}_{A}} \frac{d A}{d t}
$$

where ad$_X Y = [X,Y]$. I believe that I understand the proof in the article but I just want to confirm a few things based off of this expression:

1. The expression in the middle: $$ \frac{1-e^{-\text{ad}_{A}}}{\text{ad}_{A}} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k-1)!} (ad_A)^k$$
   is not an operator, correct? The operator is
   $$ \frac{1-e^{-\text{ad}_{A}}}{\text{ad}_{A}} \frac{d A}{d t}$$ right?

2. If I wanted to generalize this to a function of multiple variables I would get: $$ \frac{\partial}{\partial \varepsilon}\exp[A(\varepsilon,t)] = \exp[A(\varepsilon,t)]\frac{1-e^{-\text{ad}_{A}}}{\text{ad}_{A}} \frac{\partial A}{\partial \varepsilon} $$
   correct?

3. If A is a Hermitian operator (i.e. $A^\dagger = A$), and $U(\varepsilon,t)=\exp[-iA(\varepsilon,t)]$
   is a unitary operator then
   $$\frac{\partial}{\partial \varepsilon}U^\dagger(\varepsilon,t) = \frac{\partial}{\partial \varepsilon}\exp[iA(t)]= \exp[iA(t)]\frac{1-e^{-\text{ad}_{iA}}}{\text{ad}_{iA}} \frac{d (iA)}{d t}$$
   and similarly
   $$\frac{\partial}{\partial \varepsilon}U(\varepsilon,t) = \frac{\partial}{\partial \varepsilon}\exp[-iA(t)] = -\frac{1-e^{-\text{ad}_{iA}}}{\text{ad}_{iA}} \frac{d (iA)}{d t} \exp[-iA(t)].$$
   Do these expressions make sense, or am I missing something? I am working in quantum physics and am not an expert in Lie group theory so I want to double-check that I am at least on the right track.

Thanks!

Answer 1

1. You are correct.
2. Yes, of course.
3. You took a bad turn in your straightforward plug-in, and then made it worse. Ignore t, and stick to your active variable, $$ \frac{\partial}{\partial \varepsilon}\exp[iA(\varepsilon)] = i\exp[iA(\varepsilon)]\frac{1-e^{-\text{ad}_{iA}}}{\text{ad}_{iA}} \frac{\partial A}{\partial \varepsilon}\\ =i\exp[iA(\varepsilon)]\left ( \frac{\partial A}{\partial \varepsilon}-{i\over 2} [A, \frac{\partial A}{\partial \varepsilon}]+...\right ) ~~ .$$ More crucially, $$ \frac{\partial}{\partial \varepsilon}\exp[-iA(\varepsilon)] = -i\exp[-iA(\varepsilon)]\frac{e^{\text{ad}_{iA}}-1}{\text{ad}_{iA}} \frac{\partial A}{\partial \varepsilon} ~~.$$
4. A more practical/popular equivalent version of the leading identity, the one you often use in physics, expanding to arbitrary order in the exponents, is, instead, $$ \frac{\partial}{\partial \varepsilon}\exp[A(\varepsilon)] = \int _0^s \!\!ds ~\exp[(1-s)A(\varepsilon)] ~\frac{\partial A}{\partial \varepsilon} ~\exp[ sA(\varepsilon)] ~ ,$$ also in the same article.