Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 20 / a / Solution

For functions with the required derivatives, the Leibniz rule isAt $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}{g}^{(n+1-k)}$ after regrouping is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$ by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\left(\genfrac{}{}{0ex}{}{n}{p+q}\right)\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)=n!/(p!q!r!)$. Consequentlywhere $p,q,r$ range over nonnegative integers. This is the three-factor generalized Leibniz rule.