Lesson Explainer: The Quotient Rule | Nagwa Lesson Explainer: The Quotient Rule | Nagwa

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Lesson Explainer: The Quotient Rule Mathematics • Second Year of Secondary School

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 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 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: (๐‘“ยฑ๐‘”)=๐‘“ยฑ๐‘”.๏Ž˜๏Ž˜๏Ž˜

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: (๐‘๐‘“)=๐‘๐‘“,๏Ž˜ where ๐‘ is a constant. Additionally, we have the product rule stated 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)=๐‘ข๐‘ฃ+๐‘ข๐‘ฃ.๏Ž˜๏Ž˜๏Ž˜

In this explainer, we will focus on the rule for differentiating quotients. One way to think about a quotient ๐‘Ž๐‘ is to think of it as a product of ๐‘Ž and 1๐‘. We will consider this approach first.

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

Find the first derivative of ๐‘ฆ=โˆ’3๐‘ฅโˆ’2๐‘ฅ+17โˆš๐‘ฅ๏Šจ with respect to ๐‘ฅ.

Answer

Let us consider ๐‘ฆ as the product of two functions: ๐‘ข=โˆ’3๐‘ฅโˆ’2๐‘ฅ+17,๐‘ฃ=๐‘ฅ.๏Šจ๏Šฑ๏Ž ๏Žก

In doing this, we can apply the product rule, (๐‘ข๐‘ฃ)=๐‘ข๐‘ฃ+๐‘ข๐‘ฃ,๏Ž˜๏Ž˜๏Ž˜ to find the derivative. Firstly, we need to find ๐‘ข๏Ž˜ and ๐‘ฃ๏Ž˜ using the power rule of differentiation: dd๐‘ฅ๐‘ฅ=๐‘›๐‘ฅ;๏Š๏Š๏Šฑ๏Šง we have ๐‘ข=โˆ’6๐‘ฅโˆ’2,๐‘ฃ=โˆ’12๐‘ฅ=โˆ’12โˆš๐‘ฅ.๏Ž˜๏Ž˜๏Šฑ๏Šฉ๏Žข๏Žก

Substituting the expressions for ๐‘ข, ๐‘ข๏Ž˜, ๐‘ฃ, and ๐‘ฃ๏Ž˜ into the product rule, we have dd๐‘ฆ๐‘ฅ=โˆ’โˆ’3๐‘ฅโˆ’2๐‘ฅ+172โˆš๐‘ฅ+โˆ’6๐‘ฅโˆ’2โˆš๐‘ฅ=3๐‘ฅ+2๐‘ฅโˆ’172โˆš๐‘ฅ+โˆ’6๐‘ฅโˆ’2โˆš๐‘ฅ.๏Šจ๏Šฉ๏Šจ๏Šฉ

By multiplying the numerator and denominator of the second fraction by 2๐‘ฅ, we can rewrite this as a single fraction: dd๐‘ฆ๐‘ฅ=3๐‘ฅ+2๐‘ฅโˆ’172โˆš๐‘ฅ+โˆ’12๐‘ฅโˆ’4๐‘ฅ2โˆš๐‘ฅ=โˆ’โˆ’9๐‘ฅโˆ’2๐‘ฅโˆ’172โˆš๐‘ฅ.๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ

Certainly, there is a simpler way to differentiate this function. We can begin by simplifying the expression for the function to ๐‘ฆ=โˆ’3๐‘ฅโˆ’2๐‘ฅ+17๐‘ฅ.๏Žข๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Šฑ

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 ๐‘ข๐‘ฃ as the product of two functions ๐‘ข and 1๐‘ฃ. We can now apply the product rule as follows:

dddddd๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ข๐‘ฅ1๐‘ฃ+๐‘ข๐‘ฅ๏€ผ1๐‘ฃ๏ˆ.(1)

Therefore, we only need to know how to evaluate dd๐‘ฅ๏€ผ1๐‘ฃ๏ˆ and then we will have a general formula for derivatives of quotients. Let us consider the effect of a small change in ๐‘ฅ on the value of 1๐‘ฃ. If ๐‘ฅ changes by a small amount given by ฮ”๐‘ฅ, there will be a corresponding change in ๐‘ฃ which we denote ฮ”๐‘ฃ. Therefore, the change in 1๐‘ฃ can be expressed as ฮ”๏€ผ1๐‘ฃ๏ˆ=1๐‘ฃ+ฮ”๐‘ฃโˆ’1๐‘ฃ.

Expressing this as a single fraction, we get ฮ”๏€ผ1๐‘ฃ๏ˆ=๐‘ฃโˆ’(๐‘ฃ+ฮ”๐‘ฃ)๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ)=โˆ’ฮ”๐‘ฃ๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ).

Dividing through by ฮ”๐‘ฅ yields ฮ”๏€ป๏‡ฮ”๐‘ฅ=โˆ’๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ).๏Šง๏“๏‹ฒ๏“๏‹ฒ๏—

Taking the limit as ฮ”๐‘ฅโ†’0, we get the derivative of 1๐‘ฃ: ddlimlim๐‘ฅ๏€ผ1๐‘ฃ๏ˆ=ฮ”๏€ป๏‡ฮ”๐‘ฅ=๏ƒโˆ’๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ)๏.๏‹ฒ๏—โ†’๏Šฆ๏Šง๏“๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏“๏‹ฒ๏—

Using the properties of finite limits on continuous functions, we can rewrite this as ddlimlimlimlim๐‘ฅ๏€ผ1๐‘ฃ๏ˆ=โˆ’ฮ”๐‘ฃฮ”๐‘ฅ๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ)=โˆ’ฮ”๐‘ฃฮ”๐‘ฅ๐‘ฃ+๐‘ฃฮ”๐‘ฃ.๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ๏Šจ๏‹ฒ๏—โ†’๏Šฆ

Since limdd๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ฃฮ”๐‘ฅ=๐‘ฃ๐‘ฅ and lim๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ฃ=0, we have dddd๐‘ฅ๏€ผ1๐‘ฃ๏ˆ=โˆ’1๐‘ฃ๐‘ฃ๐‘ฅ.๏Šจ

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

dddddd๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ข๐‘ฅ1๐‘ฃโˆ’๐‘ข๐‘ฃ๐‘ฃ๐‘ฅ.๏Šจ(2)

Geometrically, we can visualize this ๐‘ข๐‘ฃ as representing the area of a rectangle whose sides are of lengths ๐‘ข and 1๐‘ฃ as shown in the figure.

By changing ๐‘ฅ by a small value ฮ”๐‘ฅ, the area of the rectangle will change. Assuming that both ๐‘ข and ๐‘ฃ are increasing, the change in ๐‘ข will increase the area of the rectangle by 1๐‘ฃฮ”๐‘ข, whereas the change in 1๐‘ฃ will decrease the area of the rectangle by ๐‘ขฮ”๏€ผ1๐‘ฃ๏ˆ. We will also need to subtract ฮ”๐‘ขฮ”๏€ผ1๐‘ฃ๏ˆ from the increase due to ๐‘ข in order to find the area of the new rectangle.

Often, the quotient rule (equation (2)) is expressed as a single fraction as follows: dd๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃโˆ’๐‘ข๐‘ฃ.dddd๏‘๏—๏“๏—๏Šจ

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.

Rule: Quotient Rule

Given two differentiable functions ๐‘ข and ๐‘ฃ, the derivative of their quotient is given by dd๐‘ฅ=๏€ฝ๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ)๏‰=๐‘ฃ(๐‘ฅ)(๐‘ข(๐‘ฅ))โˆ’๐‘ข(๐‘ฅ)(๐‘ฃ(๐‘ฅ))(๐‘ฃ(๐‘ฅ)).dddd๏—๏—๏Šจ

This can be written succinctly using prime notation as follows: ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ.๏Ž˜๏Ž˜๏Ž˜๏Šจ

There is an alternative way to derive the quotient rule without appealing to the product rule. We can consider the changes in ๐‘ข and ๐‘ฃ are ฮ”๐‘ข and ฮ”๐‘ฃ, as a result of a small change in ฮ”๐‘ฅ. Then, the corresponding change in ๐‘ข๐‘ฃ is given by ฮ”๏€ป๐‘ข๐‘ฃ๏‡=๐‘ข+ฮ”๐‘ข๐‘ฃ+ฮ”๐‘ฃโˆ’๐‘ข๐‘ฃ.

Expressing this as a single fraction, we have ฮ”๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ(๐‘ข+ฮ”๐‘ข)โˆ’๐‘ข(๐‘ฃ+ฮ”๐‘ฃ)๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ)=๐‘ฃฮ”๐‘ขโˆ’๐‘ขฮ”๐‘ฃ๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ).

Dividing by ฮ”๐‘ฅ, we have ฮ”๏€ป๏‡ฮ”๐‘ฅ=๐‘ฃโˆ’๐‘ข๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ).๏‘๏“๏‹ฒ๏‘๏‹ฒ๏—๏‹ฒ๏“๏‹ฒ๏—

Taking the limit as ฮ”๐‘ฅโ†’0, we get the derivative of ๐‘ข๐‘ฃ as follows: ddlimlim๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=ฮ”๏€ป๏‡ฮ”๐‘ฅ=๏ƒ๐‘ฃโˆ’๐‘ข๐‘ฃ(๐‘ฃ+ฮ”๐‘ฃ)๏.๏‹ฒ๏—โ†’๏Šฆ๏‘๏“๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏‘๏‹ฒ๏—๏‹ฒ๏“๏‹ฒ๏—

Using the rules of finite limits for continuous functions, we have ddlimlimlim๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=โŽ›โŽœโŽœโŽ๐‘ฃฮ”๐‘ขฮ”๐‘ฅโˆ’๐‘ขฮ”๐‘ฃฮ”๐‘ฅ๐‘ฃ+๐‘ฃฮ”๐‘ฃโŽžโŽŸโŽŸโŽ .๏‹ฒ๏—โ†’๏Šฆ๏‹ฒ๏—โ†’๏Šฆ๏Šจ๏‹ฒ๏—โ†’๏Šฆ

Since limdd๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ขฮ”๐‘ฅ=๐‘ข๐‘ฅ, limdd๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ฃฮ”๐‘ฅ=๐‘ฃ๐‘ฅ, and lim๏‹ฒ๏—โ†’๏Šฆฮ”๐‘ฃ=0, we have dd๐‘ฅ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃโˆ’๐‘ข๐‘ฃ.dddd๏‘๏—๏“๏—๏Šจ

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 ๐‘ฆ=8๐‘ฅ+53๐‘ฅ+22.

Answer

To find the derivative of ๐‘ฆ, we will apply the quotient rule: ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ.๏Ž˜๏Ž˜๏Ž˜๏Šจ

We set ๐‘ข=8๐‘ฅ+5 and ๐‘ฃ=3๐‘ฅ+22. Before we can apply the quotient rule, we need to calculate the derivatives of ๐‘ข and ๐‘ฃ. Using the power rule of differentiation, we have ๐‘ข=8,๐‘ฃ=3.๏Ž˜๏Ž˜

Substituting these expressions into the quotient rule, we have dd๐‘ฆ๐‘ฅ=8(3๐‘ฅ+22)โˆ’3(8๐‘ฅ+5)(3๐‘ฅ+22).๏Šจ

Expanding the parentheses in the numerator, we have dd๐‘ฆ๐‘ฅ=24๐‘ฅ+176โˆ’24๐‘ฅโˆ’15(3๐‘ฅ+22)=161(3๐‘ฅ+22).๏Šจ๏Šจ

Letโ€™s consider another example for the quotient rule.

Example 3: Differentiating Quotient Functions

Differentiate ๐‘“(๐‘ฅ)=4๐‘ฅโˆ’5๐‘ฅ+83๐‘ฅโˆ’4๏Šจ.

Answer

We will apply the quotient rule, ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ,๏Ž˜๏Ž˜๏Ž˜๏Šจ to find the derivative of ๐‘“. We begin by setting ๐‘ข=4๐‘ฅโˆ’5๐‘ฅ+8๏Šจ and ๐‘ฃ=3๐‘ฅโˆ’4. Now we need to find the derivatives of ๐‘ข and ๐‘ฃ. We can do this using the power rule as follows: ๐‘ข=8๐‘ฅโˆ’5,๐‘ฃ=3.๏Ž˜๏Ž˜

Substituting these expressions into the quotient rule, we have ๐‘“(๐‘ฅ)=(8๐‘ฅโˆ’5)(3๐‘ฅโˆ’4)โˆ’3๏€น4๐‘ฅโˆ’5๐‘ฅ+8๏…(3๐‘ฅโˆ’4).๏Ž˜๏Šจ๏Šจ

We now expand the parentheses in the numerator and simplify as follows: ๐‘“(๐‘ฅ)=24๐‘ฅโˆ’47๐‘ฅ+20โˆ’12๐‘ฅ+15๐‘ฅโˆ’24(3๐‘ฅโˆ’4)=12๐‘ฅโˆ’32๐‘ฅโˆ’4(3๐‘ฅโˆ’4).๏Ž˜๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

In the next example, we will find an unknown constant in a fractional expression when the value of the derivative of the expression at a point is given.

Example 4: Using the Quotient Rule

Suppose ๐‘“(๐‘ฅ)=๐‘ฅ+๐‘Ž๐‘ฅโˆ’๐‘Ž and ๐‘“(2)=โˆ’2๏Ž˜. Determine ๐‘Ž.

Answer

Since we have been given the value of the derivative of ๐‘“ at a particular point, we first need an expression for the derivative of ๐‘“. We can find such an expression using the quotient rule: ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ.๏Ž˜๏Ž˜๏Ž˜๏Šจ

Setting ๐‘ข=๐‘ฅ+๐‘Ž and ๐‘ฃ=๐‘ฅโˆ’๐‘Ž, we have ๐‘ข=1,๐‘ฃ=1.๏Ž˜๏Ž˜

Substituting these expressions into the quotient rule, we have ๐‘“(๐‘ฅ)=(๐‘ฅโˆ’๐‘Ž)โˆ’(๐‘ฅ+๐‘Ž)(๐‘ฅโˆ’๐‘Ž)=โˆ’2๐‘Ž(๐‘ฅโˆ’๐‘Ž).๏Ž˜๏Šจ๏Šจ

To find the value of ๐‘Ž, we will use the fact that ๐‘“(2)=โˆ’2๏Ž˜. Substituting ๐‘ฅ=2 into the expression for our derivative, we have ๐‘“(2)=โˆ’2๐‘Ž(2โˆ’๐‘Ž).๏Ž˜๏Šจ

Hence, โˆ’2=โˆ’2๐‘Ž(2โˆ’๐‘Ž).๏Šจ

Multiplying both sides of the equation by (2โˆ’๐‘Ž)๏Šจ and dividing by โˆ’2 yield (2โˆ’๐‘Ž)=๐‘Ž.๏Šจ

We can now expand the parentheses as follows: 4โˆ’4๐‘Ž+๐‘Ž=๐‘Ž.๏Šจ

Hence, by subtracting ๐‘Ž from both sides of the equation, we get ๐‘Žโˆ’5๐‘Ž+4=0.๏Šจ

This expression can be factored by inspection, which gives (๐‘Žโˆ’4)(๐‘Žโˆ’1)=0; alternatively, we could have used the quadratic formula or another method. Whatever approach taken, we get either solution of ๐‘Ž=4 and ๐‘Ž=1.

In the next example, we will apply the quotient rule when the algebraic expressions for functions are not provided.

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

Let ๐‘”(๐‘ฅ)=๐‘“(๐‘ฅ)โˆ’4โ„Ž(๐‘ฅ)โˆ’5. Given that ๐‘“(โˆ’2)=โˆ’1, ๐‘“(โˆ’2)=โˆ’8๏Ž˜, โ„Ž(โˆ’2)=โˆ’2, and โ„Ž(โˆ’2)=5๏Ž˜, find ๐‘”(โˆ’2)๏Ž˜.

Answer

We begin by using the quotient rule, ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ,๏Ž˜๏Ž˜๏Ž˜๏Šจ to find an expression for the derivative of ๐‘”. Let ๐‘ข=๐‘“(๐‘ฅ) and ๐‘ฃ=โˆ’4โ„Ž(๐‘ฅ)โˆ’5. We begin by finding the derivatives of ๐‘ข and ๐‘ฃ as follows: ๐‘ข=๐‘“(๐‘ฅ),๐‘ฃ=โˆ’4โ„Ž(๐‘ฅ).๏Ž˜๏Ž˜๏Ž˜๏Ž˜

Substituting these into the quotient rule, we have ๐‘”(๐‘ฅ)=(โˆ’4โ„Ž(๐‘ฅ)โˆ’5)๐‘“(๐‘ฅ)โˆ’(โˆ’4โ„Ž(๐‘ฅ))๐‘“(๐‘ฅ)(โˆ’4โ„Ž(๐‘ฅ)โˆ’5)=4๐‘“(๐‘ฅ)โ„Ž(๐‘ฅ)โˆ’(4โ„Ž(๐‘ฅ)+5)๐‘“(๐‘ฅ)(4โ„Ž(๐‘ฅ)+5).๏Ž˜๏Ž˜๏Ž˜๏Šจ๏Ž˜๏Ž˜๏Šจ

Setting ๐‘ฅ=โˆ’2, we have ๐‘”(โˆ’2)=4๐‘“(โˆ’2)โ„Ž(โˆ’2)โˆ’(4โ„Ž(โˆ’2)+5)๐‘“(โˆ’2)(4โ„Ž(โˆ’2)+5).๏Ž˜๏Ž˜๏Ž˜๏Šจ

Since ๐‘“(โˆ’2)=โˆ’1, ๐‘“(โˆ’2)=โˆ’8๏Ž˜, โ„Ž(โˆ’2)=โˆ’2, and โ„Ž(โˆ’2)=5๏Ž˜, we have ๐‘”(โˆ’2)=4(โˆ’1)(5)โˆ’(4(โˆ’2)+5)(โˆ’8)(4(โˆ’2)+5)=โˆ’449.๏Ž˜๏Šจ

Before applying the quotient rule, it is worth checking whether the given expression can be simplified. This is particularly important when the given expression involves a sum or a difference of quotients. Letโ€™s consider an example where we simplify the given fractional expression before applying the quotient rule.

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

If ๐‘ฆ=๐‘ฅ+5๐‘ฅโˆ’5โˆ’๐‘ฅโˆ’5๐‘ฅ+5, find dd๐‘ฅ๐‘ฆ.

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 ๐‘ฆ as a single fraction: ๐‘ฆ=(๐‘ฅ+5)โˆ’(๐‘ฅโˆ’5)(๐‘ฅโˆ’5)(๐‘ฅ+5).๏Šจ๏Šจ

We can now expand the parentheses in the numerator and denominator and simplify: ๐‘ฆ=๐‘ฅ+10๐‘ฅ+25โˆ’๏€น๐‘ฅโˆ’10๐‘ฅ+25๏…๐‘ฅโˆ’25=20๐‘ฅ๐‘ฅโˆ’25.๏Šจ๏Šจ๏Šจ๏Šจ We can now differentiate ๐‘ฆ using the quotient rule, ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ,๏Ž˜๏Ž˜๏Ž˜๏Šจ by setting ๐‘ข=20๐‘ฅ and ๐‘ฃ=๐‘ฅโˆ’25๏Šจ. We begin by finding the derivatives of ๐‘ข and ๐‘ฃ as follows: ๐‘ข=20,๐‘ฃ=2๐‘ฅ.๏Ž˜๏Ž˜

We now substitute these expressions into the quotient rule to get dd๐‘ฅ๐‘ฆ=20๏€น๐‘ฅโˆ’25๏…โˆ’2๐‘ฅ(20๐‘ฅ)(๐‘ฅโˆ’25)=โˆ’20๐‘ฅโˆ’500(๐‘ฅโˆ’25).๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Letโ€™s recap a few important concepts from this explainer.

Key Points

  • To find the derivative of the quotient of two differentiable functions ๐‘ข(๐‘ฅ) and ๐‘ฃ(๐‘ฅ), we can use the quotient rule which states that dd๐‘ฅ๏€ฝ๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ)๏‰=๐‘ฃ(๐‘ฅ)(๐‘ข(๐‘ฅ))โˆ’๐‘ข(๐‘ฅ)(๐‘ฃ(๐‘ฅ))(๐‘ฃ(๐‘ฅ)).dddd๏—๏—๏Šจ This is often written more succinctly using prime notation as follows: ๏€ป๐‘ข๐‘ฃ๏‡=๐‘ฃ๐‘ขโˆ’๐‘ฃ๐‘ข๐‘ฃ.๏Ž˜๏Ž˜๏Ž˜๏Šจ
  • Before applying the quotient rule, it is worth checking whether it is possible to simplify the expression for the function.

ุงู†ุถู… ุฅู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ

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