This problem seems really problematic, checking the pattern from the first $5$ derivatives of F, I proceeded to form a general formula, but failed. I did this -

$f^{n}(x)g(x) + nf^{n-1}(x)g'(x) + \frac{n(n-1)}{2}f^{n-2}(x)g''(x) + \frac{2 + 3(n-2)(n-3)}{2}f^{n-3}(x)g'''(x) + ...... + g^n(x)$

First $5$ derivatives of F are given here

Can someone please help me find the correct answer to this question?


2 Answers 2


This is called Leibnitz' rule. The pattern of the coefficients is the same as in the binomial theorem, essentially for the same reason. Just google Leibnitz rule..


General Leibniz Rule generalizes the product rule as follows:

It states that if $f$ and $g$ are $n$-times differentiable functions, then the product $fg$ is also $n$-times differentiable and its $n$th derivative is given by

$$F^{(n)}=(fg)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(n-k)} g^{(k)} \text{ where } {n \choose k}={n!\over k! (n-k)!}$$ is the binomial coefficient and $f^{(0)}\equiv f$.


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