Prove Leibniz 's formula for the nth derivitive of a product by induction https://www.dropbox.com/s/qerdl1fspwvpo7a/diff%20an%20induction.png
ok so I am stuck when I have to differentiate the $n=z$ (or some other random letter) equation which is the $n=z+1$ step.
My problem is that I don't know how to manipulate the equation to get the result I need. I start by differentiating inside the sum and using the product rule in the process. I can make two sums here because of the $2$ terms the product rule gives but that is as far as I can go. I think that I need to use the sum properties used in the binomial theorem proof by induction however I don't see how. I have spent a lot of time on this so please just give me at least a few steps with the same notation as in the question.
 A: Well, to begin with, by your description, I suppose that you've gotten to this point: $$\begin{align}\frac{d^{k+1}}{dx^{k+1}}(uv) &= \sum_{k=0}^n\binom{n}{k}\left[\frac{d^ku}{dx^k}\frac{d^{n+1-k}v}{dx^{n+1-k}}+\frac{d^{k+1}u}{dx^{k+1}}\frac{d^{n-k}v}{dx^{n-k}}\right]\\ &= \sum_{k=0}^n\binom{n}{k}\frac{d^ku}{dx^k}\frac{d^{n+1-k}v}{dx^{n+1-k}}+\sum_{k=0}^n\binom{n}{k}\frac{d^{k+1}u}{dx^{k+1}}\frac{d^{n-k}v}{dx^{n-k}}.\end{align}$$ Am I correct? Reindexing the right sum by $k\mapsto k-1,$ we get $$\begin{align}\frac{d^{k+1}}{dx^{k+1}}(uv) &= \sum_{k=0}^n\binom{n}{k}\frac{d^ku}{dx^k}\frac{d^{n+1-k}v}{dx^{n+1-k}}+\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{d^ku}{dx^k}\frac{d^{n-(k-1)}v}{dx^{n-(k-1)}}\\ &= \sum_{k=0}^n\binom{n}{k}\frac{d^ku}{dx^k}\frac{d^{n+1-k}v}{dx^{n+1-k}}+\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{d^ku}{dx^k}\frac{d^{n+1-k}v}{dx^{n+1-k}}.\end{align}$$ Now, recall the following: $$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}k\\\binom n0=1=\binom{n+1}0\\\binom nn=1=\binom{n+1}{n+1}$$ See if you can take it the rest of the way from there.
A: The formula can be more easily written as
$$D^n(uv)=\sum_{k=0}^n \binom{n}{k} D^ku D^{n-k}v$$
where $D^0u=u$.
For $n=0$ there's nothing to prove. Assume it holds for $n$; then
\begin{align}
D^{n+1}(uv)&=D\biggl(\sum_{k=0}^n\binom{n}{k} D^ku D^{n-k}v\biggr)\\
&=\sum_{k=0}^n \binom{n}{k} D(D^ku D^{n-k}v)\\
&=\sum_{k=0}^n \binom{n}{k}(D^{k}u D^{n-k+1}v+D^{k+1}u D^{n-k}v)\\
&=\sum_{k=0}^n \binom{n}{k}D^{k}u D^{n-k+1}v
 +\sum_{k=0}^n \binom{n}{k}D^{k+1}u D^{n-k}v
\end{align}
Do a change of indices and recall the fundamental property of binomial coefficients.
It's really the same as the proof of the binomial theorem.
