MIT has a cute little problem in their calculus problem set, which you can see here. I like this problem because it takes something complicated and shows that it actually reduces down to something very simple.

To start this problem off let’s suppose we have a function that we can write as the product of two other functions, . There are many such functions. For example, .

To find the derivative we would need to use the power rule. To get warmed up let’s take two derivatives of .

(that’s just the product rule definition).

Applying the rule again we get four terms, two from the first product and two from the second:

We can combine terms to get:

This process can continue and indeed the problem set gives us Libniz’s formula for the n-th derivative, which is:

This formula may look daunting at first, but it’s really pretty simple. Note that we are not raising our functions to powers, means the -st derivative. Let’s apply Libniz’s formula to the second derivative of above.

This is exactly what we got above, the only difference is the notation.

The problem now asks us to calculate if .

Before we begin let’s make a few observations that will help us along the way. First, in this example our is and is .

Next, let me remind you about two well known facts of . As an example, I’ll take to be . Then we just have the following, where I will adopt the same notation used in Libniz’s formula.

So what have we learned? Well, first we know that all derivatives higher than order are 0. In our example, was 3, and by the fourth derivative we got 0. Of course the derivative of 0 is 0, so from here on out (for all derivatives of order greater than 3), we’re going to get 0.

The second thing to notice is that the p-th derivative is actually . Again, in this case was 3 and for our answer was , otherwise known as .

These two rules also hold for . True, we have to use the chain rule, but of course the derivative of is just 1.

Given what we’ve learned let’s move forward writing out the first few terms of Leibniz’s formula with our given .

So far all of these terms are 0 since the derivatives of are of higher order than . We can ask when they will stop being 0. This occurs on the -st term when the derivative is , which is just . So we know that’s the first non-zero term. Let’s start with that term and write out the next few terms.

Notice what happens here. For derivatives of order higher than the term is 0. So we are left with only a single term! That term is:

But this can be simplified using our rules from above. We can replace with since this is the p-th derivative of and with since it’s the q-th derivative of .

And now we can expand the binomial term. Recall that .

See what I mean! The derivatives of the product of two fairly general functions just turns into a few multiplications.