Lesson Explainer: The Product Rule Mathematics • Higher Education

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 ๐‘“(๐‘ฅ) and ๐‘”(๐‘ฅ):

  1. addition or subtraction: ๐‘“(๐‘ฅ)ยฑ๐‘”(๐‘ฅ);
  2. multiplication or division: ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ) or ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ);
  3. composition: ๐‘“(๐‘”(๐‘ฅ)).

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: (๐‘“ยฑ๐‘”)โ€ฒ=๐‘“โ€ฒยฑ๐‘”โ€ฒ.

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: (๐‘๐‘“)โ€ฒ=๐‘๐‘“โ€ฒ, where ๐‘ 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 first example.

Example 1: Exploring the Derivatives of Products

Consider the functions ๐‘“(๐‘ฅ)=๐‘ฅ and ๐‘”(๐‘ฅ)=๐‘ฅ๏Šจ.

  1. Find ๐‘“โ€ฒ(๐‘ฅ) and ๐‘”โ€ฒ(๐‘ฅ).
  2. Find ๐‘“โ€ฒ(๐‘ฅ)๐‘”โ€ฒ(๐‘ฅ).
  3. Given that ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)=๐‘ฅ๏Šฉ, find its derivative.

Answer

Part 1

Using the power rule for differentiation, dd๐‘ฅ๐‘ฅ=๐‘›๐‘ฅ,๏Š๏Š๏Šฑ๏Šง we have ๐‘“โ€ฒ(๐‘ฅ)=1 and ๐‘”โ€ฒ(๐‘ฅ)=2๐‘ฅ.

Part 2

Taking the product of ๐‘“โ€ฒ and ๐‘”โ€ฒ, we have ๐‘“โ€ฒ(๐‘ฅ)๐‘”โ€ฒ(๐‘ฅ)=2๐‘ฅ.

Part 3

Once again, we can apply the power rule to find the derivative of ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)=๐‘ฅ๏Šฉ as follows: dddd๐‘ฅ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)=๐‘ฅ๐‘ฅ=3๐‘ฅ.๏Šฉ๏Šจ

The previous demonstrated that, generally, dddddd๐‘ฅ(๐‘“(๐‘ฅ)๐‘”(๐‘ฅ))โ‰ ๐‘“๐‘ฅ๐‘”๐‘ฅ.

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 ๐‘ข=๐‘“(๐‘ฅ) and ๐‘ฃ=๐‘”(๐‘ฅ), we can consider the product to represent the area of the rectangle whose side lengths are ๐‘ข and ๐‘ฃ as shown in the figure.

We would like to know how this area changes in response to a small change in ๐‘ฅ. If ๐‘ฅ changes by a small amount given by ฮ”๐‘ฅ, there will be a corresponding change in both ๐‘ข and ๐‘ฃ which we represent as ฮ”๐‘ข=๐‘“(๐‘ฅ+ฮ”๐‘ฅ)โˆ’๐‘“(๐‘ฅ),ฮ”๐‘ฃ=๐‘”(๐‘ฅ+ฮ”๐‘ฅ)โˆ’๐‘”(๐‘ฅ).

The area of the resulting rectangle will be (๐‘ข+ฮ”๐‘ข)(๐‘ฃ+ฮ”๐‘ฃ). The change in area is therefore given by

ฮ”(๐‘ข๐‘ฃ)=(๐‘ข+ฮ”๐‘ข)(๐‘ฃ+ฮ”๐‘ฃ)โˆ’๐‘ข๐‘ฃ=๐‘ขฮ”๐‘ฃ+๐‘ฃฮ”๐‘ข+ฮ”๐‘ขฮ”๐‘ฃ.(1)

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 ฮ”๐‘ฅ, which gives us ฮ”(๐‘ข๐‘ฃ)ฮ”๐‘ฅ=๐‘ขฮ”๐‘ฃฮ”๐‘ฅ+๐‘ฃฮ”๐‘ขฮ”๐‘ฅ+ฮ”๐‘ขฮ”๐‘ฃฮ”๐‘ฅ.

Taking the limit as ฮ”๐‘ฅโ†’0, we get the derivative of ๐‘ข๐‘ฃ: ddlimlim๐‘ฅ(๐‘ข๐‘ฃ)=ฮ”(๐‘ข๐‘ฃ)ฮ”๐‘ฅ=๏€ผ๐‘ขฮ”๐‘ฃฮ”๐‘ฅ+๐‘ฃฮ”๐‘ขฮ”๐‘ฅ+ฮ”๐‘ขฮ”๐‘ฃฮ”๐‘ฅ๏ˆ.๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ

Using the properties of finite limits on continuous functions, we have ddlimlimlimlim๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ขฮ”๐‘ฃฮ”๐‘ฅ+๐‘ฃฮ”๐‘ขฮ”๐‘ฅ+ฮ”๐‘ขฮ”๐‘ฃฮ”๐‘ฅ.๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ

Since limdd๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ฃฮ”๐‘ฅ=๐‘ฃ๐‘ฅ and limdd๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ขฮ”๐‘ฅ=๐‘ข๐‘ฅ, we can rewrite this as ddddddlimdd๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ๐‘ฅ+๐‘ฃ๐‘ข๐‘ฅ+ฮ”๐‘ข๐‘ฃ๐‘ฅ.๏‹ฒ๏—โ†’๏Šฆ

As for the limit lim๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ข, we know that the change in ๐‘ข tends to zero as the change in ๐‘ฅ tends to zero. Hence, the value of this limit is zero. Therefore, we have dddddd๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ๐‘ฅ+๐‘ฃ๐‘ข๐‘ฅ.

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

Rule: Product Rule

Given two differentiable functions ๐‘ข and ๐‘ฃ, the derivative of their product is given by dddddd๐‘ฅ(๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ))=๐‘ข(๐‘ฅ)๐‘ฅ(๐‘ฃ(๐‘ฅ))+๐‘ฃ(๐‘ฅ)๐‘ฅ(๐‘ข(๐‘ฅ)).

This can be written succinctly using prime notation as follows: (๐‘ข๐‘ฃ)โ€ฒ=๐‘ข๐‘ฃโ€ฒ+๐‘ขโ€ฒ๐‘ฃ.

Letโ€™s consider an example where we need to apply the product rule.

Example 2: Differentiation Using the Product Rule

Find the first derivative of the function ๐‘“(๐‘ฅ)=๏€น2๐‘ฅ+๐‘ฅโˆ’5๏…๏€ผ๐‘ฅ+3โˆš๐‘ฅโˆ’3๐‘ฅ๏ˆ๏Šช๏Šจ.

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 dddddd๐‘ฅ(๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ๐‘ฅ+๐‘ฃ๐‘ข๐‘ฅ, where ๐‘ข and ๐‘ฃ are both functions of ๐‘ฅ.

We let ๐‘ข=2๐‘ฅ+๐‘ฅโˆ’5๏Šช and ๐‘ฃ=๐‘ฅ+3โˆš๐‘ฅโˆ’3๐‘ฅ๏Šจ.

We begin by finding dd๐‘ข๐‘ฅ by applying the power rule as follows: dd๐‘ข๐‘ฅ=8๐‘ฅ+1.๏Šฉ

To find dd๐‘ฃ๐‘ฅ, we first express each term of ๐‘ฃ using indices: ๐‘ฃ=๐‘ฅ+3๐‘ฅโˆ’3๐‘ฅ.๏Šจ๏Šฑ๏Šง๏Ž ๏Žก

Then, using the power rule, we have dd๐‘ฃ๐‘ฅ=2๐‘ฅ+3๏€ผ12๏ˆ๐‘ฅโˆ’3(โˆ’1)๐‘ฅ=2๐‘ฅ+32๐‘ฅ+3๐‘ฅ.๏Šฑ๏Šฑ๏Šจ๏Šฑ๏Šฑ๏Šจ๏Ž ๏Žก๏Ž ๏Žก

We can now substitute both derivatives into the formula for the product rule, which gives ๐‘“โ€ฒ(๐‘ฅ)=๏€น2๐‘ฅ+๐‘ฅโˆ’5๏…๏€ฝ2๐‘ฅ+32๐‘ฅ+3๐‘ฅ๏‰+๏€ฝ๐‘ฅ+3๐‘ฅโˆ’3๐‘ฅ๏‰๏€น8๐‘ฅ+1๏….๏Šช๏Šฑ๏Šฑ๏Šจ๏Šจ๏Šฑ๏Šง๏Šฉ๏Ž ๏Žก๏Ž ๏Žก

We can now expand the parentheses as follows: ๐‘“โ€ฒ(๐‘ฅ)=4๐‘ฅ+3๐‘ฅ+6๐‘ฅ+2๐‘ฅ+32๐‘ฅ+3๐‘ฅโˆ’10๐‘ฅโˆ’152๐‘ฅโˆ’15๐‘ฅ+8๐‘ฅ+๐‘ฅ+24๐‘ฅ+3๐‘ฅโˆ’24๐‘ฅโˆ’3๐‘ฅ.๏Šซ๏Šจ๏Šจ๏Šฑ๏Šง๏Šฑ๏Šฑ๏Šจ๏Šซ๏Šจ๏Šจ๏Šฑ๏Šง๏Žฆ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Žฆ๏Žก๏Ž ๏Žก

Finally, we gather together our like terms and rewrite the expression in descending powers of ๐‘ฅ: ๐‘“โ€ฒ(๐‘ฅ)=12๐‘ฅโˆ’15๐‘ฅโˆ’10๐‘ฅ+27๐‘ฅ+92๐‘ฅโˆ’152๐‘ฅโˆ’15๐‘ฅ=12๐‘ฅ+27๐‘ฅโˆš๐‘ฅโˆ’15๐‘ฅ+92โˆš๐‘ฅโˆ’10๐‘ฅโˆ’152โˆš๐‘ฅโˆ’15๐‘ฅ.๏Šซ๏Šจ๏Šฑ๏Šฑ๏Šจ๏Šซ๏Šฉ๏Šจ๏Šจ๏Žฆ๏Žก๏Ž ๏Žก๏Ž ๏Žก

In the next example, we will apply the product rule when the algebriac expression for the functions are not provided.

Example 3: Using the Product Rule

Let ๐‘”(๐‘ฅ)=โˆ’3๐‘“(๐‘ฅ)[โ„Ž(๐‘ฅ)โˆ’1]. If ๐‘“โ€ฒ(โˆ’4)=โˆ’1, โ„Žโ€ฒ(โˆ’4)=โˆ’9, โ„Ž(โˆ’4)=โˆ’6, and ๐‘“(โˆ’4)=โˆ’1, find ๐‘”โ€ฒ(โˆ’4).

Answer

We begin by applying the product rule to find an expression for ๐‘”โ€ฒ. Recall that the product rule states that (๐‘ข๐‘ฃ)โ€ฒ=๐‘ข๐‘ฃโ€ฒ+๐‘ฃ๐‘ขโ€ฒ.

Setting ๐‘ข=โˆ’3๐‘“(๐‘ฅ) and ๐‘ฃ=โ„Ž(๐‘ฅ)โˆ’1, we have ๐‘ขโ€ฒ=โˆ’3๐‘“โ€ฒ(๐‘ฅ),๐‘ฃโ€ฒ=โ„Žโ€ฒ.

Substituting these expressions into the product rule, we have ๐‘”โ€ฒ(๐‘ฅ)=โˆ’3๐‘“(๐‘ฅ)โ„Žโ€ฒ(๐‘ฅ)โˆ’3๐‘“โ€ฒ(๐‘ฅ)[โ„Ž(๐‘ฅ)โˆ’1].

Therefore, ๐‘”โ€ฒ(โˆ’4)=โˆ’3๐‘“(โˆ’4)โ„Žโ€ฒ(โˆ’4)โˆ’3๐‘“โ€ฒ(โˆ’4)[โ„Ž(โˆ’4)โˆ’1].

Substituting in the values of ๐‘“โ€ฒ(โˆ’4)=โˆ’1, โ„Žโ€ฒ(โˆ’4)=โˆ’9, โ„Ž(โˆ’4)=โˆ’6, and ๐‘“(โˆ’4)=โˆ’1 yields ๐‘”โ€ฒ(โˆ’4)=โˆ’3(โˆ’1)(โˆ’9)โˆ’3(โˆ’1)((โˆ’6)โˆ’1)=โˆ’27โˆ’21=โˆ’48.

Product rule can be combined with other rules of differentiation. In the next example, we will apply the product rule together with another provided rule of differentiation.

Example 4: Differentiation Using the Product Rule

Given that dd๐‘ฅ๐‘’=๐‘˜๐‘’๏‡๏—๏‡๏—, find ๐‘˜ so that ๐‘”(๐‘ฅ)=๐‘ฅ๐‘’๏Šจ๏‡๏— satisfies ๐‘”โ€ฒ(1)=0.

Answer

Firstly, we need to find an expression for the derivative of ๐‘”. To do this, we will apply the product rule: (๐‘ข๐‘ฃ)โ€ฒ=๐‘ข๐‘ฃโ€ฒ+๐‘ฃ๐‘ขโ€ฒ.

Setting ๐‘ข=๐‘ฅ๏Šจ and ๐‘ฃ=๐‘’๏‡๏—, we find expressions for ๐‘ขโ€ฒ and ๐‘ฃโ€ฒ. Using the power rule of derivatives, we have ๐‘ขโ€ฒ=2๐‘ฅ.

In the question, we have been given the rule to evaluate ๐‘ฃโ€ฒ. Hence, ๐‘ฃโ€ฒ=๐‘˜๐‘’.๏‡๏—

Substituting these expressions into the product rule, we have ๐‘”โ€ฒ(๐‘ฅ)=๐‘˜๐‘ฅ๐‘’+2๐‘ฅ๐‘’.๏Šจ๏‡๏—๏‡๏—

We would like to find the value of ๐‘˜ which ensures that ๐‘”โ€ฒ(1)=0. To do this, we will substitute ๐‘ฅ=1 into our equation for ๐‘”โ€ฒ and set it equal to zero as follows: 0=๐‘˜๐‘’+2๐‘’.๏‡๏‡

We can now solve this equation for ๐‘˜. Since ๐‘’โ‰ 0๏‡ for all values of ๐‘˜, we can divide through by ๐‘’๏‡ to get 0=๐‘˜+2.

Hence, ๐‘˜=โˆ’2 guarantees that ๐‘”โ€ฒ(1)=0.

In previous examples, we have considered the derivative of the product of two functions. The product rule can be generalized to the product of more than two functions. In the next example, we will derive the product rule for three functions.

Example 5: Differentiating Triple Products

The product rule says that (๐‘“๐‘”)โ€ฒ=๐‘“โ€ฒ๐‘”+๐‘“๐‘”โ€ฒ. Use this to derive a formula for the derivative (๐‘“๐‘”โ„Ž)โ€ฒ.

Answer

We begin by considering the product ๐‘“๐‘”โ„Ž as the product of two functions ๐‘“ and the function ๐‘”โ„Ž. We can now apply the product rule to the derivative as follows: (๐‘“๐‘”โ„Ž)โ€ฒ=(๐‘“(๐‘”โ„Ž))โ€ฒ=๐‘“โ€ฒ๐‘”โ„Ž+๐‘“(๐‘”โ„Ž)โ€ฒ.

We can now apply the product rule to find the derivative of the product ๐‘”โ„Ž as follows: (๐‘”โ„Ž)โ€ฒ=๐‘”โ€ฒโ„Ž+๐‘”โ„Žโ€ฒ.

Hence, (๐‘“๐‘”โ„Ž)โ€ฒ=๐‘“โ€ฒ๐‘”โ„Ž+๐‘“(๐‘”โ€ฒโ„Ž+๐‘”โ„Žโ€ฒ)=๐‘“โ€ฒ๐‘”โ„Ž+๐‘“๐‘”โ€ฒโ„Ž+๐‘“๐‘”โ„Žโ€ฒ.

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 ๐‘“(๐‘ฅ)=๏€น๐‘ฅ+4๏…๏€บ3๐‘ฅโˆš๐‘ฅโˆ’7๏†๏€บ3๐‘ฅโˆš๐‘ฅ+7๏†๏Šฎ at ๐‘ฅ=โˆ’1.

Answer

The function ๐‘“ 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, ๐‘“(๐‘ฅ)=๏€น๐‘ฅ+4๏…๏€บ3๐‘ฅโˆš๐‘ฅโˆ’7๏†๏€บ3๐‘ฅโˆš๐‘ฅ+7๏†=๏€น๐‘ฅ+4๏…๏€น9๐‘ฅโˆ’49๏….๏Šฎ๏Šฎ๏Šฉ

We can now apply the product rule: (๐‘ข๐‘ฃ)โ€ฒ=๐‘ข๐‘ฃโ€ฒ+๐‘ขโ€ฒ๐‘ฃ.

Setting ๐‘ข=๐‘ฅ+4๏Šฎ and ๐‘ฃ=9๐‘ฅโˆ’49๏Šฉ, we have ๐‘ขโ€ฒ=8๐‘ฅ,๐‘ฃโ€ฒ=27๐‘ฅ.๏Šญ๏Šจ

Substituting these expressions into the product rule, we have ๐‘“โ€ฒ(๐‘ฅ)=27๐‘ฅ๏€น๐‘ฅ+4๏…+8๐‘ฅ(9๐‘ฅโˆ’49).๏Šจ๏Šฎ๏Šญ๏Šฉ

Evaluating at ๐‘ฅ=โˆ’1, we have ๐‘“โ€ฒ(โˆ’1)=27(โˆ’1)๏€บ(โˆ’1)+4๏†+8(โˆ’1)๏€บ9(โˆ’1)โˆ’49๏†=(27)(5)+(โˆ’8)(โˆ’58)=599.๏Šจ๏Šฎ๏Šญ๏Šฉ

Key Points

  • To find the derivative of the product of two differential functions ๐‘ข and ๐‘ฃ, we can use the product rule which states that dddddd๐‘ฅ(๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ))=๐‘ข(๐‘ฅ)๐‘ฅ(๐‘ฃ(๐‘ฅ))+๐‘ฃ(๐‘ฅ)๐‘ฅ(๐‘ข(๐‘ฅ)). This is often written more succinctly using prime notation as follows: (๐‘ข๐‘ฃ)โ€ฒ=๐‘ข๐‘ฃโ€ฒ+๐‘ขโ€ฒ๐‘ฃ.
  • The product rule can be generalized to the product of an arbitrary number of functions. For example, the product rule for three functions is written as (๐‘ข๐‘ฃ๐‘ค)โ€ฒ=๐‘ขโ€ฒ๐‘ฃ๐‘ค+๐‘ข๐‘ฃโ€ฒ๐‘ค+๐‘ข๐‘ฃ๐‘คโ€ฒ.
  • The product rule can be combined with other rules of differentiation.

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