**Required math: calculus**

**Required physics: none**

We’ve seen that the product rule for derivatives is, for two functions and :

This rule can be generalized to give Leibniz’s formula for the derivative of a product:

where the notation and 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 , say . Second, we *assume* that the formula is true for a value . From this assumption we then must prove that the formula is also true for . It doesn’t matter what is as long as it is taken to be equal to or greater than the value of used in step 1. The idea is that if we can show that the truth of the formula for a particular value of always implies its truth for the next value of , then as long as we can demonstrate the truth of the formula for *some* value of (as we do in step 1), the inductive reasoning implies it is true for all 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 gives us the identity which is clearly true. So now we prove the inductive step. Assuming the formula is true for gives us

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

In the first term, we can shift the summation index by replacing by to get

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

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

Putting over a common denominator:

Since we can combine the two outlying terms in 6 to get the final result

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

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Amin HosseiniThank the person who has done this proof .