# The derivative of a product of functions

### Erin Horst

This task provides a geometric representation of the derivative of the product of two functions. After constructing the model, the learner can derive the formula for the Product Rule.

Supplies you will need:

• a "wall-to-floor" (WAF) coordinate system
• scissors
• clear tape

The Product Rule: If and are both differentiable, then

Before we begin, let us first discuss a WAF coordinate system. A WAF coordinate system contains an x-axis and two y-axes. The two y-axes are at right angles to each other, as seen in the illustration below.

We will use the WAF coordinate system to geometrically explore the derivate of the product of two functions.

To begin, let's sketch the curves

on the "wall" and

on the "floor" of the WAF coordinate system.

Next, cut out a rectangle with dimensions and . Now, tape your rectangle to the WAF coordinate system where it coincides with the function, as seen in the illustration below where the yellow rectangle represents your cut out rectangle.

Now, cut out five more rectangles for andfor various values of x. It is recommended that you create at least five more rectangles, for x having the values 1, 3, 4, 5, and 6. By cutting out more and more rectangles, you can begin to visualize the solid created if there were infinately many rectangles, as seen in the illustration below.

Now, consider the function that defines, for any given x, the area of the cross section. This cross section is also one of the rectangles you previously cut out. Let's call this function . Then, .

Next, construct two more rectangles for some value of x and x+h as illustrated below. You need not tape your rectangles on your model, but consider where they would be placed.

Now, how can we see geometrically what occurs to the function when we take different values x and x+h. Consider our rectangles just constructed.

Now, compute the lengths of the sides of each rectangle, A(x) and A(x+h).

Now, can you determine what is?

Notice that . Now, find in terms of f and g. Do you get one of the following?

Now, assuming that the required limits exist and behave as we would expect, we can obtain the product rule from the last equation, as follows:

then follows . We have now derived the Product Rule!

It may useful to check that we can use A(x) and A'(x) to compute values of f(x)g(x) and the derivative of f(x)g(x). As an example, let x =1/2, then the steps are as follows:

and and .

Then,and .

Then,and .

Finally, .

This task was adapted from ---- (1990) Derivative of a product: Exploration 2 of concepts of calculus for middle school teachers. Middle School Math Project. Salem, OR: The Math Learning Center.

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