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(π‘₯) and 𝑔(π‘₯):

  1. addition or subtraction: 𝑓(π‘₯)±𝑔(π‘₯);
  2. multiplication or division: 𝑓(π‘₯)𝑔(π‘₯) or 𝑓(π‘₯)𝑔(π‘₯);
  3. composition: 𝑓(𝑔(π‘₯)) or 𝑔(𝑓(π‘₯)).

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 𝑒 and 𝑣, the derivative of their product is given by ddddddπ‘₯(𝑒(π‘₯)𝑣(π‘₯))=𝑒(π‘₯)π‘₯(𝑣(π‘₯))+𝑣(π‘₯)π‘₯(𝑒(π‘₯)).

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

In this explainer, we will focus on the rule for differentiating quotients. One way to think about a quotient π‘Žπ‘ is to think of it as a product of π‘Ž 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=βˆ’3π‘₯βˆ’2π‘₯+17√π‘₯ with respect to π‘₯.

Answer

Let us consider 𝑦 as the product of two functions: 𝑒=βˆ’3π‘₯βˆ’2π‘₯+17,𝑣=π‘₯.

In doing this, we can apply the product rule, (𝑒𝑣)=𝑒𝑣+𝑒𝑣, to find the derivative. Firstly, we need to find π‘’οŽ˜ and π‘£οŽ˜ using the power rule of differentiation: ddπ‘₯π‘₯=𝑛π‘₯; we have 𝑒=βˆ’6π‘₯βˆ’2,𝑣=βˆ’12π‘₯=βˆ’12√π‘₯.

Substituting the expressions for 𝑒, π‘’οŽ˜, 𝑣, and π‘£οŽ˜ into the product rule, we have dd𝑦π‘₯=βˆ’βˆ’3π‘₯βˆ’2π‘₯+172√π‘₯+βˆ’6π‘₯βˆ’2√π‘₯=3π‘₯+2π‘₯βˆ’172√π‘₯+βˆ’6π‘₯βˆ’2√π‘₯.

By multiplying the numerator and denominator of the second fraction by 2π‘₯, we can rewrite this as a single fraction: dd𝑦π‘₯=3π‘₯+2π‘₯βˆ’172√π‘₯+βˆ’12π‘₯βˆ’4π‘₯2√π‘₯=βˆ’βˆ’9π‘₯βˆ’2π‘₯βˆ’172√π‘₯.

Certainly, there is a simpler way to differentiate this function. We can begin by simplifying the expression for the function to 𝑦=βˆ’3π‘₯βˆ’2π‘₯+17π‘₯.

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 𝑒𝑣 as the product of two functions 𝑒 and 1𝑣. We can now apply the product rule as follows:

ddddddπ‘₯𝑒𝑣=𝑒π‘₯1𝑣+𝑒π‘₯ο€Ό1π‘£οˆ.(1)

Therefore, we only need to know how to evaluate ddπ‘₯ο€Ό1π‘£οˆ and then we will have a general formula for derivatives of quotients. Let us consider the effect of a small change in π‘₯ on the value of 1𝑣. If π‘₯ changes by a small amount given by Ξ”π‘₯, there will be a corresponding change in 𝑣 which we denote Δ𝑣. Therefore, the change in 1𝑣 can be expressed as Ξ”ο€Ό1π‘£οˆ=1𝑣+Ξ”π‘£βˆ’1𝑣.

Expressing this as a single fraction, we get Ξ”ο€Ό1π‘£οˆ=π‘£βˆ’(𝑣+Δ𝑣)𝑣(𝑣+Δ𝑣)=βˆ’Ξ”π‘£π‘£(𝑣+Δ𝑣).

Dividing through by Ξ”π‘₯ yields ΔΔπ‘₯=βˆ’π‘£(𝑣+Δ𝑣).οŠ§ο“ο‹²ο“ο‹²ο—

Taking the limit as Ξ”π‘₯β†’0, we get the derivative of 1𝑣: ddlimlimπ‘₯ο€Ό1π‘£οˆ=ΔΔπ‘₯=οƒβˆ’π‘£(𝑣+Δ𝑣).ο‹²ο—β†’οŠ¦οŠ§ο“ο‹²ο—β†’οŠ¦ο‹²ο“ο‹²ο—

Using the properties of finite limits on continuous functions, we can rewrite this as ddlimlimlimlimπ‘₯ο€Ό1π‘£οˆ=βˆ’Ξ”π‘£Ξ”π‘₯𝑣(𝑣+Δ𝑣)=βˆ’Ξ”π‘£Ξ”π‘₯𝑣+𝑣Δ𝑣.ο‹²ο—β†’οŠ¦ο‹²ο—β†’οŠ¦ο‹²ο—β†’οŠ¦οŠ¨ο‹²ο—β†’οŠ¦

Since limddο‹²ο—β†’οŠ¦Ξ”π‘£Ξ”π‘₯=𝑣π‘₯ and limο‹²ο—β†’οŠ¦Ξ”π‘£=0, we have ddddπ‘₯ο€Ό1π‘£οˆ=βˆ’1𝑣𝑣π‘₯.

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

ddddddπ‘₯𝑒𝑣=𝑒π‘₯1π‘£βˆ’π‘’π‘£π‘£π‘₯.(2)

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

By changing π‘₯ by a small value Ξ”π‘₯, the area of the rectangle will change. Assuming that both 𝑒 and 𝑣 are increasing, the change in 𝑒 will increase the area of the rectangle by 1𝑣Δ𝑒, whereas the change in 1𝑣 will decrease the area of the rectangle by 𝑒Δ1π‘£οˆ. We will also need to subtract Δ𝑒Δ1π‘£οˆ from the increase due to 𝑒 in order to find the area of the new rectangle.

Often, the quotient rule (equation (2)) is expressed as a single fraction as follows: ddπ‘₯𝑒𝑣=π‘£βˆ’π‘’π‘£.ddddο‘ο—ο“ο—οŠ¨

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 𝑒 and 𝑣, the derivative of their quotient is given by ddπ‘₯=𝑒(π‘₯)𝑣(π‘₯)=𝑣(π‘₯)(𝑒(π‘₯))βˆ’π‘’(π‘₯)(𝑣(π‘₯))(𝑣(π‘₯)).ddddο—ο—οŠ¨

This can be written succinctly using prime notation as follows: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£.

There is an alternative way to derive the quotient rule without appealing to the product rule. We can consider the changes in 𝑒 and 𝑣 are Δ𝑒 and Δ𝑣, as a result of a small change in Ξ”π‘₯. Then, the corresponding change in 𝑒𝑣 is given by Δ𝑒𝑣=𝑒+Δ𝑒𝑣+Ξ”π‘£βˆ’π‘’π‘£.

Expressing this as a single fraction, we have Δ𝑒𝑣=𝑣(𝑒+Δ𝑒)βˆ’π‘’(𝑣+Δ𝑣)𝑣(𝑣+Δ𝑣)=π‘£Ξ”π‘’βˆ’π‘’Ξ”π‘£π‘£(𝑣+Δ𝑣).

Dividing by Ξ”π‘₯, we have ΔΔπ‘₯=π‘£βˆ’π‘’π‘£(𝑣+Δ𝑣).

Taking the limit as Ξ”π‘₯β†’0, we get the derivative of 𝑒𝑣 as follows: ddlimlimπ‘₯𝑒𝑣=ΔΔπ‘₯=οƒπ‘£βˆ’π‘’π‘£(𝑣+Δ𝑣).ο‹²ο—β†’οŠ¦ο‘ο“ο‹²ο—β†’οŠ¦ο‹²ο‘ο‹²ο—ο‹²ο“ο‹²ο—

Using the rules of finite limits for continuous functions, we have ddlimlimlimπ‘₯𝑒𝑣=βŽ›βŽœβŽœβŽπ‘£Ξ”π‘’Ξ”π‘₯βˆ’π‘’Ξ”π‘£Ξ”π‘₯𝑣+π‘£Ξ”π‘£βŽžβŽŸβŽŸβŽ .ο‹²ο—β†’οŠ¦ο‹²ο—β†’οŠ¦οŠ¨ο‹²ο—β†’οŠ¦

Since limddο‹²ο—β†’οŠ¦Ξ”π‘’Ξ”π‘₯=𝑒π‘₯, limddο‹²ο—β†’οŠ¦Ξ”π‘£Ξ”π‘₯=𝑣π‘₯, and limο‹²ο—β†’οŠ¦Ξ”π‘£=0, we have ddπ‘₯𝑒𝑣=π‘£βˆ’π‘’π‘£.ddddο‘ο—ο“ο—οŠ¨

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=8π‘₯+53π‘₯+22.

Answer

To find the derivative of 𝑦, we will apply the quotient rule: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£.

We set 𝑒=8π‘₯+5 and 𝑣=3π‘₯+22. Before we can apply the quotient rule, we need to calculate the derivatives of 𝑒 and 𝑣. Using the power rule of differentiation, we have 𝑒=8,𝑣=3.

Substituting these expressions into the quotient rule, we have dd𝑦π‘₯=8(3π‘₯+22)βˆ’3(8π‘₯+5)(3π‘₯+22).

Expanding the parentheses in the numerator, we have dd𝑦π‘₯=24π‘₯+176βˆ’24π‘₯βˆ’15(3π‘₯+22)=161(3π‘₯+22).

Example 3: Differentiating Quotient Functions

Differentiate 𝑓(π‘₯)=4π‘₯βˆ’5π‘₯+83π‘₯βˆ’4.

Answer

We will apply the quotient rule, 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£, to find the derivative of 𝑓. We begin by setting 𝑒=4π‘₯βˆ’5π‘₯+8 and 𝑣=3π‘₯βˆ’4. Now we need to find the derivatives of 𝑒 and 𝑣. We can do this using the power rule as follows: 𝑒=8π‘₯βˆ’5,𝑣=3.

Substituting these expressions into the quotient rule, we have 𝑓(π‘₯)=(8π‘₯βˆ’5)(3π‘₯βˆ’4)βˆ’3ο€Ή4π‘₯βˆ’5π‘₯+8(3π‘₯βˆ’4).

We now expand the parentheses in the numerator and simplify as follows: 𝑓(π‘₯)=24π‘₯βˆ’47π‘₯+20βˆ’12π‘₯+15π‘₯βˆ’24(3π‘₯βˆ’4)=12π‘₯βˆ’32π‘₯βˆ’4(3π‘₯βˆ’4).

Example 4: Using the Quotient Rule

Suppose 𝑓(π‘₯)=π‘₯+π‘Žπ‘₯βˆ’π‘Ž and 𝑓(2)=βˆ’2. Determine π‘Ž.

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: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£.

Setting 𝑒=π‘₯+π‘Ž and 𝑣=π‘₯βˆ’π‘Ž, we have 𝑒=1,𝑣=1.

Substituting these expressions into the quotient rule, we have 𝑓(π‘₯)=(π‘₯βˆ’π‘Ž)βˆ’(π‘₯+π‘Ž)(π‘₯βˆ’π‘Ž)=βˆ’2π‘Ž(π‘₯βˆ’π‘Ž).

To find the value of π‘Ž, we will use the fact that 𝑓(2)=βˆ’2. Substituting π‘₯=2 into the expression for our derivative, we have 𝑓(2)=βˆ’2π‘Ž(2βˆ’π‘Ž).

Hence, βˆ’2=βˆ’2π‘Ž(2βˆ’π‘Ž).

Multiplying both sides of the equation by (2βˆ’π‘Ž) and dividing by βˆ’2 yield (2βˆ’π‘Ž)=π‘Ž.

We can now expand the parentheses as follows: 4βˆ’4π‘Ž+π‘Ž=π‘Ž.

Hence, by subtracting π‘Ž from both sides of the equation, we get π‘Žβˆ’5π‘Ž+4=0.

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

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

Let 𝑔(π‘₯)=𝑓(π‘₯)βˆ’4β„Ž(π‘₯)βˆ’5. Given that 𝑓(βˆ’2)=βˆ’1, 𝑓(βˆ’2)=βˆ’8, β„Ž(βˆ’2)=βˆ’2, and β„Ž(βˆ’2)=5, find 𝑔(βˆ’2).

Answer

We begin by using the quotient rule, 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£, to find an expression for the derivative of 𝑔. Let 𝑒=𝑓(π‘₯) and 𝑣=βˆ’4β„Ž(π‘₯)βˆ’5. We begin by finding the derivatives of 𝑒 and 𝑣 as follows: 𝑒=𝑓(π‘₯),𝑣=βˆ’4β„Ž(π‘₯).

Substituting these into the quotient rule, we have 𝑔(π‘₯)=(βˆ’4β„Ž(π‘₯)βˆ’5)𝑓(π‘₯)βˆ’(βˆ’4β„Ž(π‘₯))𝑓(π‘₯)(βˆ’4β„Ž(π‘₯)βˆ’5)=4𝑓(π‘₯)β„Ž(π‘₯)βˆ’(4β„Ž(π‘₯)+5)𝑓(π‘₯)(4β„Ž(π‘₯)+5).

Setting π‘₯=βˆ’2, we have 𝑔(βˆ’2)=4𝑓(βˆ’2)β„Ž(βˆ’2)βˆ’(4β„Ž(βˆ’2)+5)𝑓(βˆ’2)(4β„Ž(βˆ’2)+5).

Since 𝑓(βˆ’2)=βˆ’1, 𝑓(βˆ’2)=βˆ’8, β„Ž(βˆ’2)=βˆ’2, and β„Ž(βˆ’2)=5, we have 𝑔(βˆ’2)=4(βˆ’1)(5)βˆ’(4(βˆ’2)+5)(βˆ’8)(4(βˆ’2)+5)=βˆ’449.

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

If 𝑦=π‘₯+5π‘₯βˆ’5βˆ’π‘₯βˆ’5π‘₯+5, find ddπ‘₯𝑦.

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=(π‘₯+5)βˆ’(π‘₯βˆ’5)(π‘₯βˆ’5)(π‘₯+5).

We can now expand the parentheses in the numerator and denominator and simplify: 𝑦=π‘₯+10π‘₯+25βˆ’ο€Ήπ‘₯βˆ’10π‘₯+25π‘₯βˆ’25=20π‘₯π‘₯βˆ’25. We can now differentiate 𝑦 using the quotient rule, 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£, by setting 𝑒=20π‘₯ and 𝑣=π‘₯βˆ’25. We begin by finding the derivatives of 𝑒 and 𝑣 as follows: 𝑒=20,𝑣=2π‘₯.

We now substitute these expressions into the quotient rule to get ddπ‘₯𝑦=20ο€Ήπ‘₯βˆ’25ο…βˆ’2π‘₯(20π‘₯)(π‘₯βˆ’25)=βˆ’20π‘₯βˆ’500(π‘₯βˆ’25).

Key Points

  1. To find the derivative of the quotient of two differential functions 𝑒(π‘₯) and 𝑣(π‘₯), we can use the quotient rule which states that ddπ‘₯𝑒(π‘₯)𝑣(π‘₯)=𝑣(π‘₯)(𝑒(π‘₯))βˆ’π‘’(π‘₯)(𝑣(π‘₯))(𝑣(π‘₯)).ddddο—ο—οŠ¨ This is often written more succinctly using prime notation as follows: 𝑒𝑣=π‘£π‘’βˆ’π‘£π‘’π‘£.
  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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