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I'm trying to derive the following equation using a Riemann sum formulation only

$$\frac{d}{d t}e^{A(t)} = \int_0^1 ds \quad e^{sA(t)}(\frac{d A(t)}{dt})e^{(1-s)A(t)}$$

What I've done so far is make sure I understand how to take the derivative of $e^{A(t)}$ for matrix $A$ and continuous parameter $t$:

$$\frac{d}{dt}e^{A(t)} = \frac{d}{dt}(\lim_{N\to \infty} (1 + \frac{A}{N})^N) = \lim_{N\to \infty} \sum_k^{N}(1 + \frac{A}{N})^{N-k} \frac{1}{N}\frac{dA}{dt} (1 + \frac{A}{N})^{k-1}$$

Here's a wiki page that may be relevant.

Now, the formula above makes perfect sense; it's just a matter of accounting for the non-commutativity of $A$ and carrying out the derivative of the limit definition of the exponential correctly. I see how the limit and sum together are implicitly an integral, but I can't seem to figure out the substitution correctly. It's obvious that since we're taking the limit as $N$ goes to infinity that I need to divide $k$ into infinitesimal chunks, so define $s = k/N$. This would make the $1/N$ term in the derivative expression I have above become $\Delta s$ since $k$ is an integer and summing over it would make $\Delta k = 1$ for all $k$. From here, my trouble is in how to get the exponents in the exponential terms correct.

Here's some of my work. I'm not sure if this is correct or not.

Giovanni Brown
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    Why must you pick Riemann sums and this expression of exponentials to be the method? There are many other possible proofs, and something might be easier to understand. – naturallyInconsistent Jan 12 '24 at 04:30
  • Possible duplicate by OP: https://physics.stackexchange.com/q/796986/2451 – Qmechanic Jan 12 '24 at 05:00
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    Hi Giovanni Brown. Welcome to Phys.SE. Please don't repost a closed question in a new entry. Instead, you are supposed to edit the original question within the original entry. – Qmechanic Jan 12 '24 at 05:01
  • @Qmechanic yes I did that by explicitly stating I'd like to derive this in a specific way, specifically a grounds up direct approach without any gamma functions or integral identities nor the adjoint operation – Giovanni Brown Jan 12 '24 at 05:49
  • @naturallyInconsistent the Riemann sum is what is mentally the most understandable to me, then I can move on to other derivations that may be a bit less direct – Giovanni Brown Jan 12 '24 at 05:51
  • Try the argument after Equation (7) in this: https://webhome.phy.duke.edu/~mehen/760/ProblemSets/BCH.pdf – naturallyInconsistent Jan 12 '24 at 07:10
  • @naturallyInconsistent thank you so much man!! – Giovanni Brown Jan 12 '24 at 07:15
  • That is precisely what I was saying in the first comment. Trying to get you to widen your search so that a vastly easier argument is available. – naturallyInconsistent Jan 12 '24 at 07:24

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