# If $F(x) = f(x)g(x)$, what is th nth derivative of F, that is $F^{n}(x)$, if $f$ and $g$ have derivatives of all orders?

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

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 $$f^{(0)}\equiv f$$.