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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 $n\in\mathbb{C}$.

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    $\begingroup$ How do you definie $a!$ if $a < 0$? $\endgroup$ – Bib-lost Feb 26 '16 at 20:37
  • $\begingroup$ @Bib-lost Well, its harder to write it out as $n(n-1)(n-2)(n-3)\dots(n-n)\dots$, but they are equivalent. $\endgroup$ – Simply Beautiful Art Feb 26 '16 at 21:08
  • $\begingroup$ So how do you definie $n!$ if $n < 0$? What is (-2)! for example? $\endgroup$ – Bib-lost Feb 26 '16 at 23:19
  • $\begingroup$ @Bib-lost Well, you'd never have to worry about that because $n-n=0$, which occurs if $n$ is a positive integer. Just know that $|(-2)!|=\infty$ so that the denominator is infinity and the whole term becomes $0$. $\endgroup$ – Simply Beautiful Art Feb 27 '16 at 0:40
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    $\begingroup$ Hmm, that does make your question considerably more difficult. Perhaps this is something you should mention in the question itself, as both answers (and I too) supposed $n$ to be a positive integer in your question. $\endgroup$ – Bib-lost Feb 27 '16 at 14:06
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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

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

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