In this explainer, we will learn how to find the derivative of a function using the product rule.

Once we have learned how to differentiate simple functions, we might start to wonder how we can differentiate more complex functions. Generally, more complex functions are created from simpler ones by combining them together in various ways. There are a few basic ways to combine two functions and :

- addition or subtraction: ;
- multiplication or division: or ;
- composition: or .

To be able to differentiate more complex functions, it would be very helpful to have rules that tell us how to differentiate functions combined in these particular ways. At this point in a calculus course, we already know that the derivative of a sum is the sum of the derivatives:

Furthermore, we know that differentiation is actually a linear operation. This means that, in addition to the sum rule, we have the following rule for multiplication by a constant: where is a constant. In this explainer, we will focus on finding the derivatives of products.

Given the rules of differentiation for sums and multiplication by a constant, we might guess that the derivative of a product is the product of the derivatives. This is not actually the correct rule as we will demonstrate in the next example.

### Example 1: Exploring the Derivatives of Products

Consider the functions and .

- Find and .
- Find .
- Given that , find its derivative.

### Answer

**Part 1**

Using the power rule for differentiation, we have and

**Part 2**

Taking the product of and , we have

**Part 3**

Once again, we can apply the power rule to find the derivative of as follows:

The previous demonstrated that, generally,

This poses the question of what the correct formula is. We will begin by considering the geometric interpretation of the product. When we take the product of two values, it is as if we are computing the area of a rectangle. So for two functions and , we can consider the product to represent the area of the rectangle whose side lengths are and as shown in the figure.

We would like to know how this area changes in response to a small change in . If changes by a small amount given by , there will be a corresponding change in both and which we represent as

The area of the resulting rectangle will be . The change in area is therefore given by

The figure below represents this change in area. (The figure shows the effect if the change in area is positive; a similar picture could be drawn to represent negative changes of area.)

Returning to equation (1), we can divide by , which gives us

Taking the limit as , we get the derivative of :

Using the properties of finite limits on continuous functions, we have

Since and , we can rewrite this as

As for the limit , we know that the change in tends to zero as the change in tends to zero. Hence, the value of this limit is zero. Therefore, we have

This result is referred to as the product rule, which we formally state below.

### Product Rule

Given two differentiable functions and , the derivative of their product is given by

This can be written succinctly using prime notation as follows:

### Example 2: Differentiation Using the Product Rule

Find the first derivative of the function .

### Answer

For an example like this, we could either expand the parentheses and then differentiate or apply the product rule to find the derivative and then expand any parentheses if necessary. In this example, we will demonstrate using the product rule which states that where and are both functions of .

We let and .

We begin by finding by applying the power rule as follows:

To find , we first express each term of using indices:

Then, using the power rule, we have

We can now substitute both derivatives into the formula for the product rule, which gives

We can now expand the parentheses as follows:

Finally, we gather together our like terms and rewrite the expression in descending powers of :

### Example 3: Using the Product Rule

Let . If , , , and , find .

### Answer

We begin by applying the product rule to find an expression for . Recall that the product rule states that

Setting and , we have

Substituting these expressions into the product rule, we have

Therefore,

Substituting in the values of , , , and yields

### Example 4: Differentiation Using the Product Rule

Given that , find so that satisfies .

### Answer

Firstly, we need to find an expression for the derivative of . To do this, we will apply the product rule:

Setting and , we find expressions for and . Using the power rule of derivatives, we have

In the question, we have been given the rule to evaluate . Hence,

Substituting these expressions into the product rule, we have

We would like to find the value of which ensures that . To do this, we will substitute into our equation for and set it equal to zero as follows:

We can now solve this equation for . Since for all values of , we can divide through by to get

Hence, guarantees that .

### Example 5: Differentiating Triple Products

The product rule says that . Use this to derive a formula for the derivative .

### Answer

We begin by considering the product as the product of two functions and the function . We can now apply the product rule to the derivative as follows:

We can now apply the product rule to find the derivative of the product as follows:

Hence,

The previous example demonstrates that the product rule can generalize to the product of more than two functions. In fact, we can generalize the product rule to any finite product of functions. In the final example, we will consider a function defined as a triple product of functions.

### Example 6: Differentiating Products

Find the first derivative of at .

### Answer

The function is the product of three functions. Therefore, we might think it is best to apply the product rule for the product of three functions. However, before doing this, it is worth noting that the last two sets of parentheses are the difference of two squares in factored form. Therefore, it might be easier to first expand these parentheses and then apply the product rule to the resulting expression. This is the approach we will take in this example. Hence,

We can now apply the product rule:

Setting and , we have

Substituting these expressions into the product rule, we have

Evaluating at , we have

### Key Points

- To find the derivative of the product of two differential functions and , we can use the product rule which states that This is often written more succinctly using prime notation as follows:
- The product rule can be generalized to the product of an arbitrary number of functions.
- Before applying the product rule, sometimes it is simpler to expand parentheses and simplify the expression.