Explainer: The Product Rule

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 𝑓(𝑥) and 𝑔(𝑥):

  1. addition or subtraction: 𝑓(𝑥)±𝑔(𝑥);
  2. multiplication or division: 𝑓(𝑥)𝑔(𝑥) or 𝑓(𝑥)𝑔(𝑥);
  3. composition: 𝑓(𝑔(𝑥)) or 𝑔(𝑓(𝑥)).

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: (𝑓±𝑔)=𝑓±𝑔.

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: (𝑐𝑓)=𝑐𝑓, where 𝑐 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 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥.

  1. Find 𝑓(𝑥) and 𝑔(𝑥).
  2. Find 𝑓(𝑥)𝑔(𝑥).
  3. Given that 𝑓(𝑥)𝑔(𝑥)=𝑥, find its derivative.


Part 1

Using the power rule for differentiation, dd𝑥𝑥=𝑛𝑥, we have 𝑓(𝑥)=1 and 𝑔(𝑥)=2𝑥.

Part 2

Taking the product of 𝑓 and 𝑔, we have 𝑓(𝑥)𝑔(𝑥)=2𝑥.

Part 3

Once again, we can apply the power rule to find the derivative of 𝑓(𝑥)𝑔(𝑥)=𝑥 as follows: dddd𝑥𝑓(𝑥)𝑔(𝑥)=𝑥𝑥=3𝑥.

The previous demonstrated that, generally, dddddd𝑥(𝑓(𝑥)𝑔(𝑥))𝑓𝑥𝑔𝑥.

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 𝑢=𝑓(𝑥) and 𝑣=𝑔(𝑥), we can consider the product to represent the area of the rectangle whose side lengths are 𝑢 and 𝑣 as shown in the figure.

We would like to know how this area changes in response to a small change in 𝑥. If 𝑥 changes by a small amount given by Δ𝑥, there will be a corresponding change in both 𝑢 and 𝑣 which we represent as Δ𝑢=𝑓(𝑥+Δ𝑥)𝑓(𝑥),Δ𝑣=𝑔(𝑥+Δ𝑥)𝑔(𝑥).

The area of the resulting rectangle will be (𝑢+Δ𝑢)(𝑣+Δ𝑣). 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 Δ𝑥, which gives us Δ(𝑢𝑣)Δ𝑥=𝑢Δ𝑣Δ𝑥+𝑣Δ𝑢Δ𝑥+Δ𝑢Δ𝑣Δ𝑥.

Taking the limit as Δ𝑥0, we get the derivative of 𝑢𝑣: ddlimlim𝑥(𝑢𝑣)=Δ(𝑢𝑣)Δ𝑥=𝑢Δ𝑣Δ𝑥+𝑣Δ𝑢Δ𝑥+Δ𝑢Δ𝑣Δ𝑥.

Using the properties of finite limits on continuous functions, we have ddlimlimlimlim𝑥(𝑢𝑣)=𝑢Δ𝑣Δ𝑥+𝑣Δ𝑢Δ𝑥+Δ𝑢Δ𝑣Δ𝑥.

Since limddΔ𝑣Δ𝑥=𝑣𝑥 and limddΔ𝑢Δ𝑥=𝑢𝑥, we can rewrite this as ddddddlimdd𝑥(𝑢𝑣)=𝑢𝑣𝑥+𝑣𝑢𝑥+Δ𝑢𝑣𝑥.

As for the limit limΔ𝑢, we know that the change in 𝑢 tends to zero as the change in 𝑥 tends to zero. Hence, the value of this limit is zero. Therefore, we have dddddd𝑥(𝑢𝑣)=𝑢𝑣𝑥+𝑣𝑢𝑥.

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

Product Rule

Given two differentiable functions 𝑢 and 𝑣, the derivative of their product is given by dddddd𝑥(𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑥(𝑣(𝑥))+𝑣(𝑥)𝑥(𝑢(𝑥)).

This can be written succinctly using prime notation as follows: (𝑢𝑣)=𝑢𝑣+𝑢𝑣.

Example 2: Differentiation Using the Product Rule

Find the first derivative of the function 𝑓(𝑥)=2𝑥+𝑥5𝑥+3𝑥3𝑥.


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 dddddd𝑥(𝑢𝑣)=𝑢𝑣𝑥+𝑣𝑢𝑥, where 𝑢 and 𝑣 are both functions of 𝑥.

We let 𝑢=2𝑥+𝑥5 and 𝑣=𝑥+3𝑥3𝑥.

We begin by finding dd𝑢𝑥 by applying the power rule as follows: dd𝑢𝑥=8𝑥+1.

To find dd𝑣𝑥, we first express each term of 𝑣 using indices: 𝑣=𝑥+3𝑥3𝑥.

Then, using the power rule, we have dd𝑣𝑥=2𝑥+312𝑥3(1)𝑥=2𝑥+32𝑥+3𝑥.

We can now substitute both derivatives into the formula for the product rule, which gives 𝑓(𝑥)=2𝑥+𝑥52𝑥+32𝑥+3𝑥+𝑥+3𝑥3𝑥8𝑥+1.

We can now expand the parentheses as follows: 𝑓(𝑥)=4𝑥+3𝑥+6𝑥+2𝑥+32𝑥+3𝑥10𝑥152𝑥15𝑥+8𝑥+𝑥+24𝑥+3𝑥24𝑥3𝑥.

Finally, we gather together our like terms and rewrite the expression in descending powers of 𝑥: 𝑓(𝑥)=12𝑥15𝑥10𝑥+27𝑥+92𝑥152𝑥15𝑥=12𝑥+27𝑥𝑥15𝑥+92𝑥10𝑥152𝑥15𝑥.

Example 3: Using the Product Rule

Let 𝑔(𝑥)=3𝑓(𝑥)[(𝑥)1]. If 𝑓(4)=1, (4)=9, (4)=6, and 𝑓(4)=1, find 𝑔(4).


We begin by applying the product rule to find an expression for 𝑔. Recall that the product rule states that (𝑢𝑣)=𝑢𝑣+𝑣𝑢.

Setting 𝑢=3𝑓(𝑥) and 𝑣=(𝑥)1, we have 𝑢=3𝑓(𝑥),𝑣=.

Substituting these expressions into the product rule, we have 𝑔(𝑥)=3𝑓(𝑥)(𝑥)3𝑓(𝑥)[(𝑥)1].

Therefore, 𝑔(4)=3𝑓(4)(4)3𝑓(4)[(4)1].

Substituting in the values of 𝑓(4)=1, (4)=9, (4)=6, and 𝑓(4)=1 yields 𝑔(4)=3(1)(9)3(1)((6)1)=2721=48.

Example 4: Differentiation Using the Product Rule

Given that dd𝑥𝑒=𝑘𝑒, find 𝑘 so that 𝑔(𝑥)=𝑥𝑒 satisfies 𝑔(1)=0.


Firstly, we need to find an expression for the derivative of 𝑔. To do this, we will apply the product rule: (𝑢𝑣)=𝑢𝑣+𝑣𝑢.

Setting 𝑢=𝑥 and 𝑣=𝑒, we find expressions for 𝑢 and 𝑣. Using the power rule of derivatives, we have 𝑢=2𝑥.

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

Substituting these expressions into the product rule, we have 𝑔(𝑥)=𝑘𝑥𝑒+2𝑥𝑒.

We would like to find the value of 𝑘 which ensures that 𝑔(1)=0. To do this, we will substitute 𝑥=1 into our equation for 𝑔 and set it equal to zero as follows: 0=𝑘𝑒+2𝑒.

We can now solve this equation for 𝑘. Since 𝑒0 for all values of 𝑘, we can divide through by 𝑒 to get 0=𝑘+2.

Hence, 𝑘=2 guarantees that 𝑔(1)=0.

Example 5: Differentiating Triple Products

The product rule says that (𝑓𝑔)=𝑓𝑔+𝑓𝑔. Use this to derive a formula for the derivative (𝑓𝑔).


We begin by considering the product 𝑓𝑔 as the product of two functions 𝑓 and the function 𝑔. We can now apply the product rule to the derivative as follows: (𝑓𝑔)=(𝑓(𝑔))=𝑓𝑔+𝑓(𝑔).

We can now apply the product rule to find the derivative of the product 𝑔 as follows: (𝑔)=𝑔+𝑔.

Hence, (𝑓𝑔)=𝑓𝑔+𝑓(𝑔+𝑔)=𝑓𝑔+𝑓𝑔+𝑓𝑔.

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 𝑓(𝑥)=𝑥+43𝑥𝑥73𝑥𝑥+7 at 𝑥=1.


The function 𝑓 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, 𝑓(𝑥)=𝑥+43𝑥𝑥73𝑥𝑥+7=𝑥+49𝑥49.

We can now apply the product rule: (𝑢𝑣)=𝑢𝑣+𝑢𝑣.

Setting 𝑢=𝑥+4 and 𝑣=9𝑥49, we have 𝑢=8𝑥,𝑣=27𝑥.

Substituting these expressions into the product rule, we have 𝑓(𝑥)=27𝑥𝑥+4+8𝑥(9𝑥49).

Evaluating at 𝑥=1, we have 𝑓(1)=27(1)(1)+4+8(1)9(1)49=(27)(5)+(8)(58)=599.

Key Points

  1. To find the derivative of the product of two differential functions 𝑢 and 𝑣, we can use the product rule which states that dddddd𝑥(𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑥(𝑣(𝑥))+𝑣(𝑥)𝑥(𝑢(𝑥)). This is often written more succinctly using prime notation as follows: (𝑢𝑣)=𝑢𝑣+𝑢𝑣.
  2. The product rule can be generalized to the product of an arbitrary number of functions.
  3. Before applying the product rule, sometimes it is simpler to expand parentheses and simplify the expression.

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