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.)
Here's my work for it so far:
$$(fg)^{(1)} = \sum_{k=0}^{1}\binom{1}{k}f^{(1-k)}g^k=f^{(1)}g^{(0)}+f^{(0)}g^{(1)}$$
We assume $P(m)$ is true (induction assumption):
$$(fg)^{(m)}=\sum_{k=0}^{m}\binom{m}{k}f^{(m-k)}g^k$$
We want to show that $P(m+1)$ is also true:
$$(fg)^{(m+1)}=\sum_{k=0}^{m+1}\binom{m+1}{k}f^{(m+1-k)}g^k$$
I'm not sure how to connect the induction assumption with $P(m+1)$ thanks.
Hint:
Observe that
$$D\left[{m\choose k}f^{(m-k)}g^{(k)}\right]={m \choose k}f^{(m-k)}g^{(k+1)}+{m \choose k}f^{(m-k+1)}g^{(k)}$$ And use the identity $${n \choose r}+{n \choose r+1}={n+1 \choose r+1}.$$