I have this math question. That I am stuck on.
If $f$ is a function, let $Df$ be its derivative. For $n\in \mathbb{Z}^+$ let $$ f^{(n)} = \underbrace{D \cdots D }_{n\mathrm{\ times}} f $$ be the $n^\mathrm{th}$ derivative of $f$. In this notation the usual product rule from calculus says that $$ (fg)^{(1)} = fg^{(1)} + f^{(1)} g. $$ Using the product rule, prove the formula for the $n^\mathrm{th}$ derivative of a product $$ (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)}. $$ (Hint: The proof in here is similar to the proof of the Binomial Theorem.)
I can see that for example is $n=2$ the equation becomes $\binom{2}{0}f^{2}+\binom{2}{1}f^{1}g^1+\binom{2}{2}g^2=f^2+2fg+g^2$. I'm not sure how to prove this though. Thanks.
Induction will come down to proving that ...
$$\sum_{k=0}^n \binom{n}{k}\left ( f^{(n-k+1)} g^{(k)} + f^{(n-k)} g^{(k+1)} \right)= \sum_{k=0}^{n+1} \binom{n+1}{k} f^{(n+1-k)} g^{(k)}.$$