# Generalized product rule: Leibniz’s formula

Required math: calculus

Required physics: none

We’ve seen that the product rule for derivatives is, for two functions ${f(x)}$ and ${g(x)}$:

$\displaystyle (fg)'=f'g+fg' \ \ \ \ \ (1)$

This rule can be generalized to give Leibniz’s formula for the ${n^{th}}$ derivative of a product:

$\displaystyle (fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)f^{(k)}g^{(n-k)} \ \ \ \ \ (2)$

where the notation ${f^{(n)}\equiv d^{n}f/dx^{n}}$ and ${\left(\begin{array}{c} n\\ k \end{array}\right)=\frac{n!}{k!(n-k)!}}$ is the binomial coefficient.

The proof of this formula is a nice little exercise in proof by mathematical induction. An inductive proof requires two steps. First, we must show that the formula is true for one particular value of ${n}$, say ${n=0}$. Second, we assume that the formula is true for a value ${n=m}$. From this assumption we then must prove that the formula is also true for ${n=m+1}$. It doesn’t matter what ${m}$ is as long as it is taken to be equal to or greater than the value of ${n}$ used in step 1. The idea is that if we can show that the truth of the formula for a particular value of ${n}$ always implies its truth for the next value of ${n}$, then as long as we can demonstrate the truth of the formula for some value of ${n}$ (as we do in step 1), the inductive reasoning implies it is true for all ${n}$ greater than or equal to the specific value.

The first step in the proof is called the anchor step and the second step the inductive step.

In our case, taking ${n=0}$ gives us the identity ${(fg)^{(0)}=fg}$ which is clearly true. So now we prove the inductive step. Assuming the formula is true for ${n=m}$ gives us

$\displaystyle (fg)^{(m)}=\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k \end{array}\right)f^{(k)}g^{(m-k)} \ \ \ \ \ (3)$

Taking the derivative of this, and applying the regular product rule, gives

$\displaystyle (fg)^{(m+1)}=\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k \end{array}\right)(f^{(k+1)}g^{(m-k)}+f^{(k)}g^{(m-k+1)}) \ \ \ \ \ (4)$

In the first term, we can shift the summation index by replacing ${k}$ by ${k-1}$ to get

$\displaystyle (fg)^{(m+1)}=\sum_{k=1}^{m+1}\left(\begin{array}{c} m\\ k-1 \end{array}\right)f^{(k)}g^{(m-k+1)}+\sum_{k=0}^{m}\left(\begin{array}{c} m\\ k \end{array}\right)f^{(k)}g^{(m-k+1)} \ \ \ \ \ (5)$

Note the limits on the two sums. We can now combine the sums in the regions where their indexes overlap to get

$\displaystyle (fg)^{(m+1)}=fg^{(m+1)}+\sum_{k=1}^{m}\left[\left(\begin{array}{c} m\\ k-1 \end{array}\right)+\left(\begin{array}{c} m\\ k \end{array}\right)\right]f^{(k)}g^{(m-k+1)}+f^{(m+1)}g \ \ \ \ \ (6)$

We can work out the sum of the two binomial coefficients by expanding them in terms of factorials:

$\displaystyle \left(\begin{array}{c} m\\ k-1 \end{array}\right)+\left(\begin{array}{c} m\\ k \end{array}\right)=\frac{m!}{(k-1)!(m-k+1)!}+\frac{m!}{k!(m-k)!} \ \ \ \ \ (7)$

Putting over a common denominator:

 $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{(k)m!}{k!(m-k+1)!}+\frac{(m-k+1)m!}{k!(m-k+1)!}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{(m+1)!}{k!(m-k+1)!} \ \ \ \ \ (9)$

$\displaystyle =\left(\begin{array}{c} m+1\\ k \end{array}\right) \ \ \ \ \ (10)$

Since ${\left(\begin{array}{c} m+1\\ 0 \end{array}\right)=\left(\begin{array}{c} m+1\\ m+1 \end{array}\right)=1}$ we can combine the two outlying terms in 6 to get the final result

$\displaystyle (fg)^{(m+1)}=\sum_{k=0}^{m+1}\left(\begin{array}{c} m+1\\ k \end{array}\right)f^{(k)}g^{(m-k+1)} \ \ \ \ \ (11)$

This completes the inductive proof and establishes Leibniz’s formula.