From the power series definition of the exponential function in the real or complex numbers you can quickly prove the sum rule by considering the function $$ f(x) = \exp(a + b - x) \exp(x), $$ proving that f'(x) = 0, then using that to show that f(0) = f(b), which gives you $\exp(a + b) = \exp(a) \exp(b)$.
I have been interested in extending this argument to arbitrary Banach algebras. I was talking with someone else about it, and our idea was to use the function $$ f(t) = \exp((1 - t)(A + B) + tA) \exp(tB) $$ where $t$ is a real number, and, assuming that $AB = BA$, the previous argument seems like it works. However, in looking at it further I realized that applying the chain rule would require figuring out what the derivative in an arbitrary Banach algebra would be, and I can't think of any definition of the derivative that I know of that would work. So, my question basically is, is there a way to salvage this proof? Is there some kind of derivative that will work here? If not, is there some other way to make it work? I know that the traditional way of proving this involves fiddling with series, but I was wanting to see if I could make this argument work.