Is there any meaning to an "infinite derivative"? I've been thinking about this: say you have an infinitely differentiable function. Then you can form a sequence $f(x), f'(x), f''(x), \cdots, f^{(n)}(x), \cdots$ and attempt to take its limit. For some functions this is easy: for $e^x$ the limit will be itself, $\sin x$ and $\cos x$ won't have a limit, every polynomial will eventually be $0$, etc.
I was wondering, does this have any use? Is there any interesting interpretation of the infinite derivative, and does it existing or not tell us anything about the function?
 A: If we consider just analytic functions....
Suppose $f(x)$ is analytic at $x=0$. Then we can write it as a Taylor series
$$ f(x) = \sum_{n=0}^{\infty} f^{(n)}(0) \frac{x^n}{n!} $$
Furthermore, we can express its derivatives
$$ f^{(k)}(x) = \sum_{n=0}^{\infty} f^{(n+k)}(0) \frac{x^n}{n!} $$
If we insist that the convergence of the sequence $f^{(k)}$ be sufficiently nice (I'm quite rusty, but I think we want the limit to converge uniformly), then the limit of the sequence of Taylor series (if it exists) can be computed by taking the limit of the coefficients, when :
$$ f^{(\infty)}(x) = \sum_{n=0}^{\infty} \left( \lim_{k\to \infty} f^{(n+k)}(0) \right) \frac{x^n}{n!} $$
Of particular note is that the limit is completely independent of $n$; if we define
$$a = \lim_{k \to \infty} f^{(k)}(0) $$
then, when it exists, we have
$$ f^{(\infty)}(x) = a e^x$$
Hopefully you'll find this partial analysis useful.
This answer has a reference that asserts the same conclusion with just the hypothesis that $f$ is infinitely differentiable and the sequence of derivatives converges uniformly.
