Cool Calculus: Extending the Product Rule to the n-th Derivative

The n-th derivative of a product looks remarkably like a binomial expansion

Keith McNulty

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We all know from high school how to get the derivative of a product of two functions u(x) and v(x):

In this article, I’m going to show an expression for the n-th derivative of a product — this expression will look remarkably like a binomial expansion. Recall the binomial expansion of an n-th power looks like this:

Another way of writing this is:

where

which is a well-known terminology for binomial coefficients.

The expression for the n-th derivative of a product which I am going to prove is as follows:

where the index (k) means k-th derivative, and where the zero-th derivative is the function itself.

Proof by induction

I’m going to use an induction argument to prove this result. Those of you who frequently read my articles will know that I like to describe induction as a neverending chain of dominos lined up to knock each other over one by one. To know that all dominos will fall, you need to know that the first one will fall (the induction start) and that whenever a domino falls it will cause the next one to fall (the induction step).

To use induction to prove our result above, we first need to prove that it is true for n=1 (the induction start). Then we assume it is true for n=k and we use this to prove that it is true for n=k+1 (the induction step).

For the case n=1:

so the induction start is proved because it is the precise statement of the product rule for the first derivative.

Now for the induction step, first we assume for n=k that:

Now we use the product rule to take the first derivative of this:

Now we can rewrite the first part of the sum above to obtain:

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Keith McNulty

Pure and Applied Mathematician. LinkedIn Top Voice in Tech. Expert and Author in Data Science and Statistics. Find me on LinkedIn, Twitter or keithmcnulty.org