Leibnitz theorem

The proof of Leibnitz theorem is based on the concept of induction and the product rule for differentiation. This article explains how to prove the generic formula for the n-th derivative of the product of two functions.

Statement of Leibnitz Theorem:

For two differentiable functions f(x) and g(x), the n-th derivative of their product is given by:

We will prove this formula by induction.

Base case (n = 1):

For n = 1, the formula becomes the first derivative of a product:

This is exactly the standard product rule and it matches the formula with the binomial coefficient

  . Thus the base case holds true.

Inductive step: Assume the formula holds for some 𝑛 = 𝑘 .We assume that:

Now, we want to prove that the formula holds for n=k+1. To begin, we differentiate both sides of the inductive hypothesis (the k-th derivative of f(x) and g(x)) with respect to x.

We now apply the product rule to each term in the sum. For a given term:

This yields:

Thus the sum over all i becomes:

Now we adjust the indices of the two sums:

The first section includes the term: d^(i+1)/dx^(i+1). f(x) has a summation over i from 0 to k, but we change the index of summation from 1 to k+1. Therefore, this term becomes:

In the second part, the terms involving d^(i)/dx^(i) f (x) and d^(k-i+1)/dx^(k-i+1) g (x) simply shift to:

Finally, combining these sums yields the necessary formula: n=k+1

This completes the induction.

We have demonstrated that the formula is true for all  n≥ 1 using the mathematical induction principle. Thus, it is established that the Leibnitz rule applies to the n-th derivative of a product of two functions.

Leibnitz theorem is commonly employed in calculus, particularly when differentiating products of functions numerous times. It is useful for calculating higher-order derivatives of polynomial products, trigonometric functions, and other complex functions.