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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?

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2 Answers 2

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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..

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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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