N-th derivative of a product in binomial expansion?

Question

I believe that the following is true:

$$\frac{d^n}{dx^n}f(x)g(x)=\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}f^{(n-m)}(x)g^{(m)}(x)$$

The rational part of the summation is binomial expansion constants and $f^{(n)}(x)=\frac{d^n}{dx^n}f(x)$

I have tested it for some values of $n$ where $f$ and $g$ are either polynomials or exponential functions and it appears to hold true.

The question is whether or not the above is true with a proof.

For those who concern, $n$ may or may not be a positive integer or even an integer at all because I wish to use this in Fractional Calculus allowing $n\in\mathbb{C}$.

Answer 1

Yes, this is a famous formula about the Nth-derivative of a product,
it's from Leibniz (so says my textbook at least).
It can be proved by induction without any obstacles.

It's just that the sum is finite.

See also: General Leibniz Rule

Answer 2

This is the General Leibniz rule. https://en.m.wikipedia.org/wiki/General_Leibniz_rule