Hello all; above is my question! :)
I've gone through all the way up to the final "and hence deduce that". Up to this point, the question has been fairly straightforward, but I have no idea how to deduce the final equalities from the derivative.
I know that it is possible to show that $f(z) = e^z e^w e^{-(z+w)}$, for fixed $w$, has $f'(z) = 0$ for all $z$, and as such is constant. Choosing $z = -w$ gives $f(-w) = 1$, and as such $f(z) = 1$ for all $z$; therefore, $e^z e^w = e^{z+w}$.
Unfortunately, I don't know how to adapt this for operator-valued functions (since the derivative isn't $0$); I don't even know if this is the right approach for operator-valued functions.
As always, assistance, not just the answer, is most appreciated!
(Reposted from comments at questioner's suggestion)
Hint: You're looking for an operator which, if you differentiate it with respect to $\lambda$, gives you the original operator times $\lambda[A,B]$. You've got one operator which satisfies that already; can you find a second one?