Basiccalculus q3 mod10 chainrule final PDF

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Basic Calculus Quarter 3 – Module 10: Chain Rule

Basic Calculus – Grade 11 Alternative Delivery Mode Quarter 3 – Module 10: Chain Rule First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

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Basic Calculus Quarter 3 – Module 10: Chain Rule

Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson. Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you. Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these. In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning. Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task. If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Thank you.

What I Need to Know

One of the main reasons why this module was created is to ensure that it will assist you to understand the concept and know how to use the chain rule on differentiating certain functions.

When you finish this module, you will be able to: 1. illustrate Chain Rule of Differentiation (STEM_BC11DIIIh-2); and 2. solve problems involving Chain Rule of Differentiation (STEM_BC11DIIIh-i-1).

1

What I Know Determine the derivative of the following functions. Write the letter of the correct answer on a separate sheet of paper. (Use calculator whenever necessary). 1. 𝑦 = 𝑥 4 A. 4𝑥 B. 4𝑥2

C. 4𝑥 3 D. 4𝑥 2 + 1

2. 𝑦 = 𝑥 3 + 1 A. 3𝑥 2 + 2 B. 3𝑥3

C. 3𝑥 + 1 D. 3𝑥 2

3. 𝑦 = 𝑥 1 ⁄2 A. B.

1

C.

𝑥 1 ⁄2 2

D.

𝑥 1 ⁄2

1

2𝑥 1 ⁄2 2

2𝑥 1 ⁄2

4. 𝑦 = sin(2𝑥) A. 2 sin(2𝑥) B. 2cos(2𝑥)

C. 2 sin(𝑥) D. 2 cos(𝑥)

5. 𝑦 = cos(3𝑥) A. −3 sin(3𝑥) B. 3 sin(3𝑥)

C. −3 sin(𝑥) D. 3 sin(𝑥)

6. 𝑦 = (3𝑥 + 5)5 A. 15(3𝑥 + 5)4 B. 5(3𝑥 + 5)4

C. −5(3𝑥 + 5)4 D. −15(3𝑥 + 5)4

7. 𝑦 = (𝑥 2 + 𝑥 − 7)6 A. (12𝑥 − 6)(𝑥 2 + 𝑥 − 7)5 B. (12𝑥 + 6)(𝑥 2 − 𝑥 + 7)5

C. (12𝑥 − 6)(𝑥 2 − 𝑥 + 7)5 D. (12𝑥 + 6)(𝑥 2 + 𝑥 − 7)5

8. 𝑦 = √(𝑥 2 + 3) A. B.

1

2(𝑥 2+3)1 ⁄2 𝑥 (𝑥 2+3)1 ⁄2

2

1

C.

(𝑥 2+3)1 ⁄2

D.

2(𝑥 2+3)1 ⁄2

𝑥

9. 𝑦 = tan(3𝑥 2 − 4) A. 6𝑥csc 2 (3𝑥2 − 4)

C. 3𝑥csc2 (3𝑥 2 − 4)

B. 6𝑥sec 2 (3𝑥2 − 4)

D. 3𝑥sec 2 (3𝑥2 − 4)

10. 𝑦 = (2𝑥)(𝑥 + 1)2 A. 4𝑥 2 + 8𝑥 + 2

C. 4𝑥 2 − 8𝑥 − 2

B. 6𝑥2 − 8𝑥 + 2

D. 6𝑥 2 + 8𝑥 + 2

Match the corresponding Column B derivatives to its Column A functions. Write the letter of the correct answer on a separate sheet of paper. (Use calculator whenever necessary).

Column A 11. 𝑦 = (4𝑥2 − 3𝑥 − 2)3

Column B A. −12 sin(4𝑥 − 5)

12. 𝑦 = 3cos (4𝑥 − 5)

B. 12𝑥 2 + 24𝑥 + 9

13. 𝑦 = √6𝑥 − 3

C. (24𝑥 − 9)(4𝑥 2 − 3𝑥 − 2)2

14. 𝑦 = 𝑥(2𝑥 + 3)2

D. 4𝑥cos(2𝑥 2 − 3)

15. 𝑦 = sin(2𝑥2 − 3)

E.

3

3

(6𝑥−3)1⁄2

Lesson

1

The Chain Rule: Derivative of Composite Function

The bicycle’s chain plays an important accessory of its two-wheel mechanism. It links the large and small sprocket to help it move to further distance. On this lesson, a complex situation can be solved through a certain process called Chain Rule of Differentiation. As you go on with this module, this process will be presented to you in a simple and clear manner.

What’s In

Find the derivative of the following items below by making use of the Power Rule of differentiation. Write your answer on a separate sheet of paper. 1. 𝑓(𝑥) = 𝑥 3

2.

𝑦 = 𝑥 10

3.

𝑓 (𝑥) = 𝑥 275

4.

𝑦 = 𝑥 500

5.

𝑓 (𝑥) = 𝑥 −10

4

What’s New Consider differentiating the function, 𝒚 = (𝟐𝒙 + 𝟐)𝟐. The process in solving such given item is shown below.

Explanation

Computation

The function can be written in another 𝑦 = (2𝑥 + 2)(2𝑥 + 2) form. FOIL method was applied to 𝑦 = 4𝑥 2 + 4𝑥 + 4𝑥 + 4 multiply both terms.

𝑦 = 4𝑥 2 + 8𝑥 + 4

Like terms were simplified.

Application of different differentiation 𝑦 ′ = 4(2)(𝑥)2−1 + 8(1)𝑥 1−1 + 0 rules per term (constant multiple rule and constant rule). Show the simplified result differentiation rule application.

after 𝑦′ = 8𝑥 + 8

The method presented above is one way of getting the derivative of a function. What if you are given a function like 𝑦 = (2𝑥 + 2)10, is it still convenient to take its derivative? Obviously, it is time consuming, due to the exponent 10. Problem arises from this kind of functions when their exponent is raised to a fractional or higher power. To fix such concern, you must understand and practice this lesson to help you solve complex functions like the ones mentioned above. Therefore, if there is a function inside a parenthesis raised to a power, the Chain rule can be used to get the derivative of that certain function.

5

What is It

•

Chain Rule is the process of differentiating a composite function.

Recall: Composite functions are two functions combined to make a single one. For example, the combination of functions 𝑓 and 𝑔:

(𝑓 𝑜 𝑔)(𝑥) = 𝑓(𝑔(𝑥 )) Note: To apply the Chain Rule on composite functions, you must take the derivative of its outside function and then multiply it to the derivative of its inside function. In symbols,

Remember:

𝑑 [𝑓 (𝑔(𝑥))] = 𝑓′(𝑔(𝑥)) ∙ 𝑔′ (𝑥) 𝑑𝑥 Derivative of the outside function

𝑑 𝑛 [𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢′ 𝑑𝑥

Derivative of the inside function

Example 1 Solve for the derivative of 𝒇(𝒈(𝒙)) = (𝒙 + 𝟒)𝟓 . Below are the steps and solutions to get the answer for the equation given above.

Explanation

Computation

Since there is no direct differentiation Let 𝑢 = 𝑥 + 4 rule applicable, the equation inside the parenthesis was represented into single variable 𝒖 resulting into a simpler 𝑓(𝑢 ) = (𝑢 )5 equation raised to an exponent. This equation is the outside function. On the other hand, the actual equation inside the parenthesis is the inside 𝑔(𝑥 ) = 𝑥 + 4 function. Application of chain rule: derivative of 𝑓′(𝑔(𝑥 )) = 5(𝑢 )5−1 ∙ (1) the outside function multiplied by the derivative of the inside function, 4

𝑓′(𝑔(𝑥 )) = 5(𝑢 )

𝑑 𝑛 [𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢 ′ 𝑑𝑥

Return the original equation 𝒙 + 𝟒 and substitute to the variable 𝒖 to get the 𝑓′(𝑔(𝑥 )) = 5(𝑥 + 4)4 answer. The derivative of 𝑓(𝑔(𝑥)) = (𝑥 + 4)5 is equal to 5(𝑥 + 4)4 . 6

Example 2 Differentiate 𝒚 = √𝒙 − 𝟑 . The table below will show the steps and solution that will give you your desired answer.

Explanation

Computation

Again, there is no direct differentiation Let 𝑢 = 𝑥 − 3 rule applicable on this item. Therefore, the equation inside the parenthesis was 𝑓(𝑢 ) = √𝑢 represented into single variable 𝒖 resulting into a simpler equation raised 1⁄ 2 to an exponent. This equation is the 𝑓(𝑢 ) = 𝑢 outside function. On the other hand, the actual equation 𝑔(𝑥 ) = 𝑥 − 3 inside the parenthesis is the inside function. 1 1 Application of chain rule: derivative of −1 ( ) ( ) 2 𝑢 𝑓′(𝑔 𝑥 ) = ∙ ( 1) the outside function multiplied by the 2 derivative of the inside function, 1 1

𝑓′(𝑔(𝑥 )) = (𝑢 )−2 2

𝑑 𝑛 [𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢 ′ 𝑑𝑥

1 To make the exponent positive, by 𝑓′(𝑔(𝑥 )) = 2(𝑢 )1⁄ 2 applying laws of exponent, simply bring down its base and exponent on its denominator. 1 Return the original equation 𝒙 − 𝟑 and ′ substitute to the variable 𝒖 to get the 𝑦 = 𝑓′(𝑔(𝑥 )) = 2(𝑥 − 3)1⁄ 2 answer. The derivative of 𝒚 = √𝒙 − 𝟑 is equal to

𝟏

𝟐(𝒙−𝟑)𝟏⁄𝟐

.

Example 3 Evaluate the derivative of 𝑦 = sin(3𝑥) . Using the table below, it will show you the steps and solution that you need in order to get the final answer on the equation given above.

Explanation

Computation

The equation inside the parenthesis was represented into single variable 𝒖 resulting into much simpler equation. This equation is the outside function. 𝑦 = sin(𝑢 ) (Recall that (𝑥) = 𝑦 .)

Let 𝑢 = 3𝑥

On the other hand, the actual equation inside the parenthesis is the inside 𝑔(𝑥 ) = 3𝑥 function. Application of chain rule: derivative of 𝑦′ = [cos(𝑢 )] ∙ [3(1)𝑥 1−1] the outside function multiplied by the derivative of the inside function, 7

𝑑 𝑛 [𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢′ 𝑑𝑥

𝑦′ = [cos(𝑢 )] ∙ (3)

(Note: 𝒚′ is the symbol for the derivative of 𝒚.) It is proper to put the constant in front of the function and return the original 𝑦′ = 3cos (3𝑥 ) equation 𝟑𝒙 in place of variable 𝒖 to get the answer. The derivative of 𝑦 = sin(3𝑥) is equal to 3cos(3𝑥). Example 4 Find the derivative of 𝑦 = 3𝑥(𝑥 + 1)2 . The table below will show the steps and solution that you need to find out the answer for the equation provided.

Explanation

Computation

For this item, the product rule best suited the situation. Recall: 𝑑 [𝑓(𝑥) ∙ 𝑔(𝑥)] 𝑑𝑥 𝑑 [𝑔(𝑥)] + 𝑔(𝑥) = 𝑓(𝑥) ∙ 𝑑𝑥 𝑑 [𝑓(𝑥)] ∙ 𝑑𝑥

ℎ(𝑥 ) = 3𝑥 𝑗(𝑥 ) = (𝑥 + 1)2

Represent the first function as ℎ(𝑥) = 3𝑥 and the second function as 𝑗(𝑥) = (𝑥 + 1)2 . Product Rule Application: Derivative of the first function Take the first function multiplied by the ℎ′(𝑥) = 3(1)(𝑥)1−1 derivative of the second function. Then, add to the product of the second function ℎ′(𝑥) = 3 and the derivative of the first function. In solving for the derivative of second function, 𝑗(𝑥) = (𝑥 + 1)2 Let, 𝑢 = 𝑥 + 1 To get the derivatives of both functions, the constant multiple rule can be applied in 𝑓(𝑢) = (𝑢)2 taking the derivative of the first function while chain rule can be used in taking the 𝑔(𝑥) = 𝑥 + 1 derivative of the second function. 𝑓′(𝑔(𝑥)) = 2(𝑢)2−1 ∙ (1) 𝑓′(𝑔(𝑥)) = 2(𝑥 + 1) 𝑓 ′(𝑔(𝑥)) = 2𝑥 + 2 𝑗 ′(𝑥) = 𝑓 ′(𝑔(𝑥)) = 2𝑥 + 2

8

Now that the derivatives of both functions 𝑦 ′ = [(3𝑥)(2𝑥 + 2)] + [(𝑥 + 1)2 (3)] are complete, the product rule can be applied. Perform the indicated operation, 𝑦 ′ = 6𝑥 2 + 6𝑥 + (𝑥 2 + 2𝑥 + 1)(3) combine like terms and simplify.

𝑦 ′ = 6𝑥 2 +6𝑥 + 3𝑥 2 + 6𝑥 + 3

Follow the simplification process to get the answer.

𝑦 ′ = 9𝑥 2 + 12𝑥 + 3

The derivative of 𝑦 = 3𝑥(𝑥 + 1)2 is equal to 9𝑥 2 + 12𝑥 + 3 . Example 5 Solve for the derivative of 𝑓(𝑔(𝑥)) = (2𝑥 2 + 3𝑥 − 5)7 Solution and steps are shown in the table below.

Explanation

Computation Let 𝑢 = 2𝑥 + 3𝑥 − 5 2

Since there is no direct differentiation rule applicable, the equation inside the parenthesis was represented into single variable 𝑢 resulting into a simpler equation raised to an exponent. This equation is the outside function. On the other hand, the actual equation inside the parenthesis is the inside function. Application of chain rule: Derivative of the outside function multiplied by the derivative of the inside function,

𝑓 (𝑢 ) = (𝑢 )7 𝑔(𝑥 ) = 2𝑥 2 + 3𝑥 − 5

𝑓′(𝑔(𝑥 )) = 7(𝑢 )7−1 ∙ (4𝑥 + 3)

𝑑 𝑛 [𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢 ′ 𝑑𝑥

𝑓′(𝑔(𝑥 )) = 7(4𝑥 + 3)(𝑢 )6

Simplify the terms that needs to be simplified. Return the original equation 𝒙 + 𝟒 and 𝑓′(𝑔(𝑥 )) = (28𝑥 + 21)(2𝑥 2 + 3𝑥 − 5)6 substitute to the variable 𝒖 to get the answer. The derivative of 𝑓(𝑔(𝑥)) = (2𝑥 2 + 3𝑥 − 5)7 is equal to (28𝑥 + 21)(2𝑥 2 + 3𝑥 − 5)6 .

9

What’s More Find the derivative of the following functions. Write your answer on a separate sheet of paper. 1.

𝑦 = (3𝑥 − 1)25

2.

𝑓(𝑥) = √𝑥 − 1

3.

𝑦 = cos(5𝑥 )

What I Have Learned Express what you have learned in the lesson by answering the questions below. Write your answer on a separate sheet of paper. 1. On what instance does the Chain Rule of differentiation applicable? Explain briefly.

2. How does the Chain Rule help you solve the derivatives of composite functions? Elaborate your answer.

10

What I Can Do

Solve the given item below. Write your answer on a separate sheet of paper. A chemical factory tank stores an amount of liquid. After it has started to drain, it has a given equation of 𝐿 = 100(𝑡 − 20)2 in liters and 𝒕 refers to the time in minutes. At what rate is the liquid running out at the end of 6 minutes?

Assessment Differentiate each function by applying chain rule. Write your answer on a separate sheet of paper.

1. 𝑦 = (4𝑥 − 10)4 A. 16(4𝑥 − 10)2 B. 16(4𝑥 − 10)3

C. 16(4𝑥 + 10)2 D. 16(4𝑥 + 10)3

2. 𝑓(𝑥 ) = (3𝑥 2 − 2𝑥 + 1)6 A. (36𝑥 − 12)(3𝑥 2 − 2𝑥 + 1)5 B. (36𝑥 − 12)(3𝑥 3 − 2𝑥 + 1)5

C. (36𝑥 − 12)( 3𝑥 2 + 2𝑥 − 1)4 D. (36𝑥 − 12)(3𝑥 2 − 2𝑥 − 1)4

3. 𝑦 = 𝑥(𝑥 2 + 3)2 A. 5𝑥 4 + 18𝑥 2 + 9 B. 5𝑥 4 − 18𝑥 2 + 9

C. 5𝑥 4 + 18𝑥 2 − 9 D. 5𝑥 2 + 18𝑥 4 + 9

4. 𝑓(𝑥 ) = 4 sin(2𝑥 − 7) A. 7𝑐𝑜𝑠(2𝑥 − 8) B. 8𝑐𝑜𝑠(2𝑥 + 7)

C. 7𝑐𝑜𝑠(2𝑥 + 8) D. 8𝑐𝑜𝑠(2𝑥 − 7)

5. 𝑦 = tan (𝑥 2 + 3𝑥) A. (2𝑥 + 3)𝑠𝑒𝑐 2 (𝑥 3 + 3𝑥) B. (2𝑥 − 3)𝑠𝑒𝑐 2 (𝑥 2 + 3𝑥)

C. (2𝑥 + 3)𝑠𝑒𝑐 2 (𝑥 2 + 3𝑥) D. (2𝑥 − 3)𝑠𝑒𝑐 2 (𝑥 2 + 2𝑥)

11

Match Column A with Column B, where A is the collection of functions and B is the collection of derivatives. Write the letter of the correct answer on a separate sheet of paper. (Use calculator whenever necessary).

Column A 6. 𝑦 = (𝑥 2 − 5𝑥 − 2)2 7. 𝑦 = sin(2𝑥 − 5)

8. 𝑦 = √𝑥 + 8 9. 𝑦 = 4𝑥 (𝑥 − 2)2 10. 𝑦 = cos (5𝑥 2 − 3)

1 A. 2(𝑥+8)1⁄2

Column B

B. 12𝑥 2 − 32𝑥 + 16 C. −10𝑥 sin(5𝑥 2 − 3) D. 2cos(2𝑥 − 5)

E. (4𝑥 − 10)(𝑥2 − 5𝑥 − 2)

Write true if the statement is correct and false if the statement is incorrect. Write your answer on a separate sheet of paper. 11. Given the function y = √3𝑥 + 2, the derivative of this function is 3 y’ = . 1 ⁄2 (3𝑥+2)

12. If y = tan(4𝑥 + 1), then its derivative is y’ = 4𝑠𝑒𝑐2 (4𝑥 + 1). 13. When y = (2𝑥 − 3)1 ⁄3 , the derivative of this function is y’ =

2

(2𝑥−3)2 ⁄3

.

14. For instance, the given function is y = 2sec (3𝑥), then its derivative is 𝑦 ′ = 6 sec(3𝑥) tan (3𝑥). 15. The derivative of the function y = √6𝑥 + 1 is y’ =

3 (6𝑥+1)1 ⁄2

.

Additional Activities Evaluate the following items below. Write your answer on a separate sheet of paper.

4 . √2𝑥+3

1. Find the derivative of the function

𝑦=

2. Differentiate the function 𝑓[𝑔(𝑥 )]

= cos ( 2 ).

1

𝑥

12

13 What’s More

1. 𝑦 ′ = 75(3𝑥 − 1)24

1 (2𝑥−1) ൗ2

What’s In 1. 2. 3. 4. 5.

1

3x2 10x9 275x274 500x499 -10x-11

What I Know 1. C 2. D 3. C 4. B 5. A 6. A 7. D 8. B 9. B 10. D 11. C 12. A 13. E 14. B 15. D

Answer Key

2. 𝑦 ′ =

14 Additional Activities 1.

y’ = −

2.

y’ =

4

3

(2𝑥+3) 2

2 sin( 𝑥3

1

𝑥2

)

Assessment 1. B 2. A 3. A 4. D 5. C 6. E 7. D 8. A 9. B 10. C 11. False 12. True 13. False 14. True 15. True

What I Can Do

What I have learned

𝑙𝑖𝑡𝑒𝑟𝑠 𝑑𝐿 = −2,800 𝑑𝑡 𝑚𝑖𝑛𝑢𝑡𝑒

Student’s answer may vary.

References DepEd. 2013. Basic Calculus. Teachers Guide. Lim, Yvette F., Nocon, Rizaldi C., Nocon, Ederlina G., and Ruivivar, Leonar A. 2016. Math for Engagement Learning Grade 11 Basic Calculus. Sibs Publishing House, Inc. Mercado, Jesus P., and Orines, Fernando B. 2016. Next Century Mathematics 11 Basic Calculus. Phoenix Publishing House, Inc.

15

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