Talk:Leibniz's Rule

We assume that $f$ and $g$ are defined on an open interval $I$. If we assume that $f$ and $g$ are $n$-times differentiable on $I$ (instead of assuming that they are differentiable only at $x\in I$, then the formula is true for all $x\in I$. This turns $x$ into a dummy variable that could be removed from the proof. Perhaps this would be cleaner? Then we are looking at:

\[(fg)^{(n)}=\sum_{k=0}^n{{n\choose k}f^{(k)}g^{(n-k)}}\]


 * Maybe it makes it look cleaner, but it would also make it weaker. Defining Leibniz's Rule for individual points states it in its strongest form possible, as it is then always possible to apply the rule to all the points in the interval if the conditions are so met. --prime mover 17:10, 20 November 2011 (CST)