Calculus (IX): How To Really Understand The Product Rule

Welcome to the ninth calculus series entry, where we take a look at the first principles behind the product rule of differentiation. In my previous essay, I covered the sum rule and the difference rule of differentiation.

Suppose that we have the following functions:

As you can see, u(x) and v(x) are separately functions of the independent variable ‘x’. We get the function y(x) by multiplying u(x) and v(x). Now, how shall we go about differentiating y(x)? Let us find out.

This essay is supported by Generatebg

The Intuitive Method

We cannot just differentiate u(x) and v(x) separately with respect to ‘x’ and multiply the results. If we do that we are sure to miss the cross multiplied terms. So, why don’t we start by multiplying the components of y(x) first:

Now, we just have a single expression in terms of ‘x’. If we differentiate this expression with respect to ‘x’ using the power rule, we get the derivative dy/dx as follows:

This is a good start. Now we know what the derivative of the function is. But even though we have managed to differentiate a function that is a product of two separate functions, we have still not covered the underlying principles in the abstract sense.

To do that, we shall use first principles that we have been applying from the beginning of this series (if you haven’t checked out my previous essays in this series, I strongly recommend that you do).

---

The Product Rule Emerges from First Principles

If we increase ‘x’ by an infinitesimal amount of ‘dx’, ‘y’ increases by ‘dy’, ‘u’ increases by ‘du’, and ‘v’ increases by ‘dv’ (where ‘dy’, ‘du’, and ‘dv’ are also infinitesimal amounts). We get the corresponding expression as follows:

When we carry out the multiplication operation on the right-hand side of the equation, we get the following result:

The last term (du*dv) is an infinitesimal term of the second order of smallness. So, we may neglect it in the limit of ‘x’ tending to zero. We may then subtract [y = (u*v)] from both sides of the resulting expression as follows:

If we divide the above expression by ‘dx’ throughout, we get the derivative of ‘y’ with respect to ‘x’:

This is the general expression for the product rule of calculus. Let us see how exactly it works.

---

Interpreting the Product Rule of Calculus

The product rule essentially conveys the following steps in order to differentiate a function that is a product of two functions:

1. Multiply each function with the derivative of the other separately. To this end, treat any other function that is currently not being differentiated as a constant. For example, when differentiating ‘u’, treat ‘v’ as a constant and when differentiating ‘v’, treat ‘u’ as a constant.

2. Sum all the resulting products to obtain the derivative of the original encompassing function.

We may test out how this works using our original example:

There we go; we have obtained the same result as before, but this time we successfully applied the product rule based on first principles. In my next essay, I will be covering the quotient rule of differentiation.

---

Reference and credit: Silvanus Thompson.

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you:

• Is ‘0.99999…’ Really Equal To ‘1’?
• How To Benefit From Computer Science In Real Life (I)
• The Hindsight Bias: Cause And Effect

If you would like to support me as an author, consider contributing on Patreon.