Explainer: The Quotient Rule

In this explainer, we will learn how to find the derivative of a function using the quotient 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 which 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 knowthat 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. Additionally, we have the product rule stated 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.

In this explainer, we will focus on the rule for differentiating quotients. One way to think about a quotient 𝑎𝑏 is to think of it as a product of 𝑎 and 1𝑏. We will consider this approach first.

Example 1: Finding the Derivative of a Quotient Using the Product Rule

Find the first derivative of 𝑦=3𝑥2𝑥+17𝑥 with respect to 𝑥.

Answer

Let us consider 𝑦 as the product of two functions: 𝑢=3𝑥2𝑥+17,𝑣=𝑥.

In doing this, we can apply the product rule, (𝑢𝑣)=𝑢𝑣+𝑢𝑣, to find the derivative. Firstly, we need to find 𝑢 and 𝑣 using the power rule of differentiation: dd𝑥𝑥=𝑛𝑥; we have 𝑢=6𝑥2,𝑣=12𝑥=12𝑥.

Substituting the expressions for 𝑢, 𝑢, 𝑣, and 𝑣 into the product rule, we have dd𝑦𝑥=3𝑥2𝑥+172𝑥+6𝑥2𝑥=3𝑥+2𝑥172𝑥+6𝑥2𝑥.

By multiplying the numerator and denominator of the second fraction by 2𝑥, we can rewrite this as a single fraction: dd𝑦𝑥=3𝑥+2𝑥172𝑥+12𝑥4𝑥2𝑥=9𝑥2𝑥172𝑥.

Certainly, there is a simpler way to differentiate this function. We can begin by simplifying the expression for the function to 𝑦=3𝑥2𝑥+17𝑥.

From this point, we can simply use the power rule to differentiate each term.

We can now generalize the method of expressing a quotient as a product to derive a general formula for the derivative of a quotient. We start by considering 𝑢𝑣 as the product of two functions 𝑢 and 1𝑣. We can now apply the product rule as follows:

dddddd𝑥𝑢𝑣=𝑢𝑥1𝑣+𝑢𝑥1𝑣.(1)

Therefore, we only need to know how to evaluate dd𝑥1𝑣 and then we will have a general formula for derivatives of quotients. Let us consider the effect of a small change in 𝑥 on the value of 1𝑣. If 𝑥 changes by a small amount given by Δ𝑥, there will be a corresponding change in 𝑣 which we denote Δ𝑣. Therefore, the change in 1𝑣 can be expressed as Δ1𝑣=1𝑣+Δ𝑣1𝑣.

Expressing this as a single fraction, we get Δ1𝑣=𝑣(𝑣+Δ𝑣)𝑣(𝑣+Δ𝑣)=Δ𝑣𝑣(𝑣+Δ𝑣).

Dividing through by Δ𝑥 yields ΔΔ𝑥=𝑣(𝑣+Δ𝑣).

Taking the limit as Δ𝑥0, we get the derivative of 1𝑣: ddlimlim𝑥1𝑣=ΔΔ𝑥=𝑣(𝑣+Δ𝑣).

Using the properties of finite limits on continuous functions, we can rewrite this as ddlimlimlimlim𝑥1𝑣=Δ𝑣Δ𝑥𝑣(𝑣+Δ𝑣)=Δ𝑣Δ𝑥𝑣+𝑣Δ𝑣.

Since limddΔ𝑣Δ𝑥=𝑣𝑥 and limΔ𝑣=0, we have dddd𝑥1𝑣=1𝑣𝑣𝑥.

We can now substitute this back into the product rule in equation (1), which gives us

dddddd𝑥𝑢𝑣=𝑢𝑥1𝑣𝑢𝑣𝑣𝑥.(2)

Geometrically, we can visualize this 𝑢𝑣 as representing the area of a rectangle whose sides are of lengths 𝑢 and 1𝑣 as shown in the figure.

By changing 𝑥 by a small value Δ𝑥, the area of the rectangle will change. Assuming that both 𝑢 and 𝑣 are increasing, the change in 𝑢 will increase the area of the rectangle by 1𝑣Δ𝑢, whereas the change in 1𝑣 will decrease the area of the rectangle by 𝑢Δ1𝑣. We will also need to subtract Δ𝑢Δ1𝑣 from the increase due to 𝑢 in order to find the area of the new rectangle.

Often, the quotient rule (equation (2)) is expressed as a single fraction as follows: dd𝑥𝑢𝑣=𝑣𝑢𝑣.dddd

Unfortunately, this form can somewhat disguise the geometric interpretation. In spite of this, it is the form that is most often used in practice. Below is a statement of the quotient rule in full.

Quotient Rule

Given two differentiable functions 𝑢 and 𝑣, the derivative of their quotient is given by dd𝑥=𝑢(𝑥)𝑣(𝑥)=𝑣(𝑥)(𝑢(𝑥))𝑢(𝑥)(𝑣(𝑥))(𝑣(𝑥)).dddd

This can be written succinctly using prime notation as follows: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.

There is an alternative way to derive the quotient rule without appealing to the product rule. We can consider the changes in 𝑢 and 𝑣 are Δ𝑢 and Δ𝑣, as a result of a small change in Δ𝑥. Then, the corresponding change in 𝑢𝑣 is given by Δ𝑢𝑣=𝑢+Δ𝑢𝑣+Δ𝑣𝑢𝑣.

Expressing this as a single fraction, we have Δ𝑢𝑣=𝑣(𝑢+Δ𝑢)𝑢(𝑣+Δ𝑣)𝑣(𝑣+Δ𝑣)=𝑣Δ𝑢𝑢Δ𝑣𝑣(𝑣+Δ𝑣).

Dividing by Δ𝑥, we have ΔΔ𝑥=𝑣𝑢𝑣(𝑣+Δ𝑣).

Taking the limit as Δ𝑥0, we get the derivative of 𝑢𝑣 as follows: ddlimlim𝑥𝑢𝑣=ΔΔ𝑥=𝑣𝑢𝑣(𝑣+Δ𝑣).

Using the rules of finite limits for continuous functions, we have ddlimlimlim𝑥𝑢𝑣=𝑣Δ𝑢Δ𝑥𝑢Δ𝑣Δ𝑥𝑣+𝑣Δ𝑣.

Since limddΔ𝑢Δ𝑥=𝑢𝑥, limddΔ𝑣Δ𝑥=𝑣𝑥, and limΔ𝑣=0, we have dd𝑥𝑢𝑣=𝑣𝑢𝑣.dddd

We will now consider a number of examples where we apply the quotient rule to find derivatives.

Example 2: Using the Quotient Rule to Find Derivatives

Find the first derivative of 𝑦=8𝑥+53𝑥+22.

Answer

To find the derivative of 𝑦, we will apply the quotient rule: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.

We set 𝑢=8𝑥+5 and 𝑣=3𝑥+22. Before we can apply the quotient rule, we need to calculate the derivatives of 𝑢 and 𝑣. Using the power rule of differentiation, we have 𝑢=8,𝑣=3.

Substituting these expressions into the quotient rule, we have dd𝑦𝑥=8(3𝑥+22)3(8𝑥+5)(3𝑥+22).

Expanding the parentheses in the numerator, we have dd𝑦𝑥=24𝑥+17624𝑥15(3𝑥+22)=161(3𝑥+22).

Example 3: Differentiating Quotient Functions

Differentiate 𝑓(𝑥)=4𝑥5𝑥+83𝑥4.

Answer

We will apply the quotient rule, 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣, to find the derivative of 𝑓. We begin by setting 𝑢=4𝑥5𝑥+8 and 𝑣=3𝑥4. Now we need to find the derivatives of 𝑢 and 𝑣. We can do this using the power rule as follows: 𝑢=8𝑥5,𝑣=3.

Substituting these expressions into the quotient rule, we have 𝑓(𝑥)=(8𝑥5)(3𝑥4)34𝑥5𝑥+8(3𝑥4).

We now expand the parentheses in the numerator and simplify as follows: 𝑓(𝑥)=24𝑥47𝑥+2012𝑥+15𝑥24(3𝑥4)=12𝑥32𝑥4(3𝑥4).

Example 4: Using the Quotient Rule

Suppose 𝑓(𝑥)=𝑥+𝑎𝑥𝑎 and 𝑓(2)=2. Determine 𝑎.

Answer

Since we have been given the value of the derivative of 𝑓 at a particular point, we first need an expression for the derivative of 𝑓. We can find such an expression using the quotient rule: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.

Setting 𝑢=𝑥+𝑎 and 𝑣=𝑥𝑎, we have 𝑢=1,𝑣=1.

Substituting these expressions into the quotient rule, we have 𝑓(𝑥)=(𝑥𝑎)(𝑥+𝑎)(𝑥𝑎)=2𝑎(𝑥𝑎).

To find the value of 𝑎, we will use the fact that 𝑓(2)=2. Substituting 𝑥=2 into the expression for our derivative, we have 𝑓(2)=2𝑎(2𝑎).

Hence, 2=2𝑎(2𝑎).

Multiplying both sides of the equation by (2𝑎) and dividing by 2 yield (2𝑎)=𝑎.

We can now expand the parentheses as follows: 44𝑎+𝑎=𝑎.

Hence, by subtracting 𝑎 from both sides of the equation, we get 𝑎5𝑎+4=0.

This expression can be factored by inspection, which gives (𝑎4)(𝑎1)=0; alternatively, we could have used the quadratic formula or another method. Whatever approach taken, we get either solution of 𝑎=4 and 𝑎=1.

Example 5: Evaluating the Derivative at a Point Using the Quotient Rule

Let 𝑔(𝑥)=𝑓(𝑥)4(𝑥)5. Given that 𝑓(2)=1, 𝑓(2)=8, (2)=2, and (2)=5, find 𝑔(2).

Answer

We begin by using the quotient rule, 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣, to find an expression for the derivative of 𝑔. Let 𝑢=𝑓(𝑥) and 𝑣=4(𝑥)5. We begin by finding the derivatives of 𝑢 and 𝑣 as follows: 𝑢=𝑓(𝑥),𝑣=4(𝑥).

Substituting these into the quotient rule, we have 𝑔(𝑥)=(4(𝑥)5)𝑓(𝑥)(4(𝑥))𝑓(𝑥)(4(𝑥)5)=4𝑓(𝑥)(𝑥)(4(𝑥)+5)𝑓(𝑥)(4(𝑥)+5).

Setting 𝑥=2, we have 𝑔(2)=4𝑓(2)(2)(4(2)+5)𝑓(2)(4(2)+5).

Since 𝑓(2)=1, 𝑓(2)=8, (2)=2, and (2)=5, we have 𝑔(2)=4(1)(5)(4(2)+5)(8)(4(2)+5)=449.

Example 6: Differentiating a Combination of Rational Functions Using the Quotient Rule

If 𝑦=𝑥+5𝑥5𝑥5𝑥+5, find dd𝑥𝑦.

Answer

When asked to differentiate a function like this, we could differentiate each term using the quotient rule. However, it is often simpler to express a sum of fractions as a single fraction and then apply the quotient rule once. This is the approach we will demonstrate here. We begin by rewriting the expression for 𝑦 as a single fraction: 𝑦=(𝑥+5)(𝑥5)(𝑥5)(𝑥+5).

We can now expand the parentheses in the numerator and denominator and simplify: 𝑦=𝑥+10𝑥+25𝑥10𝑥+25𝑥25=20𝑥𝑥25. We can now differentiate 𝑦 using the quotient rule, 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣, by setting 𝑢=20𝑥 and 𝑣=𝑥25. We begin by finding the derivatives of 𝑢 and 𝑣 as follows: 𝑢=20,𝑣=2𝑥.

We now substitute these expressions into the quotient rule to get dd𝑥𝑦=20𝑥252𝑥(20𝑥)(𝑥25)=20𝑥500(𝑥25).

Key Points

  1. To find the derivative of the quotient of two differential functions 𝑢(𝑥) and 𝑣(𝑥), we can use the quotient rule which states that dd𝑥𝑢(𝑥)𝑣(𝑥)=𝑣(𝑥)(𝑢(𝑥))𝑢(𝑥)(𝑣(𝑥))(𝑣(𝑥)).dddd This is often written more succinctly using prime notation as follows: 𝑢𝑣=𝑣𝑢𝑣𝑢𝑣.
  2. Before applying the quotient rule, it is worth checking whether it is possible to simplify the expression for the function. This is particularly relevant when the function is expressed as the sum or difference of two rational expressions.

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