Basic calculus about Chain rule, Lecture notes of Mathematics

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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

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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-by-

step 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).

9. 𝑦 = tan( 3 𝑥

2

A. 6 𝑥csc

2

2

B. 6 𝑥sec

2

2

C. 3 𝑥csc

2

2

D. 3 𝑥sec

2

2

2

A. 4 𝑥

2

B. 6 𝑥

2

C. 4 𝑥

2

D. 6 𝑥

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 Column B

2

3

A. − 12 sin

12. 𝑦 = 3 cos ( 4 𝑥 − 5 ) B. 12 𝑥

2

C.

2

2

2

D. 4 𝑥cos

2

15. 𝑦 = sin( 2 𝑥

2

E.

3

( 6 𝑥− 3 )

1 ⁄ 2

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.

3

10

275

500

− 10

Lesson

The Chain Rule: Derivative

of Composite Function

[𝑓(𝑔(𝑥))] = 𝑓′(𝑔(𝑥)) ∙ 𝑔

′

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:

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,

Example 1

Solve for the derivative of 𝒇(𝒈

𝟓

Below are the steps and solutions to get the answer for the equation given above.

Explanation Computation

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.

Let 𝑢 = 𝑥 + 4

5

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,

[𝑢

𝑛

] = 𝑛(𝑢)

𝑛− 1

′

5 − 1

4

Return the original equation 𝒙 + 𝟒 and

substitute to the variable 𝒖 to get the

answer.

4

The derivative of 𝑓(𝑔(𝑥)) = (𝑥 + 4 )

5

is equal to 5 (𝑥 + 4 )

4

Derivative of

the outside

function

Derivative of

the inside

function

Remember:

[𝑢

𝑛

] = 𝑛(𝑢)

𝑛− 1

′

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

rule applicable on this item. Therefore,

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.

Let 𝑢 = 𝑥 − 3

1 2

⁄

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,

[𝑢

𝑛

] = 𝑛(𝑢)

𝑛− 1

′

To make the exponent positive, by

applying laws of exponent, simply bring

down its base and exponent on its

denominator.

1

2

− 1

−

1

2

1 ⁄ 2

Return the original equation 𝒙 − 𝟑 and

substitute to the variable 𝒖 to get the

answer.

′

1 ⁄ 2

The derivative of 𝒚 = √𝒙 − 𝟑 is equal to

𝟏

𝟐

( 𝒙−𝟑

)

𝟏 ⁄𝟐

Example 3

Evaluate the derivative of 𝑦 = sin( 3 𝑥).

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.

(Recall that (𝑥) = 𝑦 .)

Let 𝑢 = 3 𝑥

𝑦 = sin

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,

𝑦′ = [cos

] ∙ [ 3 ( 1 )𝑥

1 − 1

]

Now that the derivatives of both functions

are complete, the product rule can be

applied. Perform the indicated operation,

combine like terms and simplify.

Follow the simplification process to get the

answer.

′

= [( 3 𝑥)( 2 𝑥 + 2 )] + [(𝑥 + 1 )

2

( 3 )]

′

2

2

′

2

2

′

2

The derivative of 𝑦 = 3 𝑥

2

is equal to 9 𝑥

2

Example 5

Solve for the derivative of 𝑓(𝑔(𝑥)) = ( 2 𝑥

2

7

Solution and steps are shown in the table below.

Explanation Computation

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.

Let 𝑢 = 2 𝑥

2

7

On the other hand, the actual equation

inside the parenthesis is the inside

function.

2

Application of chain rule: Derivative of

the outside function multiplied by the

derivative of the inside function,

[𝑢

𝑛

] = 𝑛(𝑢)

𝑛− 1

′

Simplify the terms that needs to be

simplified.

7 − 1

6

Return the original equation 𝒙 + 𝟒 and

substitute to the variable 𝒖 to get the

answer.

2

6

The derivative of 𝑓(𝑔(𝑥)) = ( 2 𝑥

2

7

is equal to ( 28 𝑥 + 21 )( 2 𝑥

2

6

Find the derivative of the following functions. Write your answer on a separate sheet

of paper.

25

3. 𝑦 = cos

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.

What’s More

What I Have Learned

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 Column B

2

2

A.

( 𝑥+ 8

)

1 ⁄ 2

7. 𝑦 = sin

B. 12 𝑥

2

C. − 10 𝑥 sin

2

2

D. 2cos

10. 𝑦 = cos

2

E.

(

)(

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 = √

3 𝑥 + 2 , the derivative of this function is

y’ =

3

( 3 𝑥+ 2 )

1 ⁄ 2

12. If y = tan( 4 𝑥 + 1 ), then its derivative is y’ = 4 𝑠𝑒𝑐

2

13_._ When y = ( 2 𝑥 − 3 )

1 ⁄ 3

, the derivative of this function is y’ =

2

( 2 𝑥− 3 )

2 ⁄ 3

14. For instance, the given function is y = 2 sec ( 3 𝑥), then its derivative is

′

= 6 sec( 3 𝑥) tan ( 3 𝑥).

15. The derivative of the function y = √

6 𝑥 + 1 is y’ =

3

( 6 𝑥+ 1 )

1 ⁄ 2

Additional Activities

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

1. Find the derivative of the function 𝑦 =

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

1

𝑥

2

ow What I Kn

C 1.

D 2.

C 3.

B 4.

A 5.

A 6.

D 7.

B 8.

B 9.

D 10.

C 11.

A 12.

E 13.

B 14.

D 15.

’s In What

3x 1.

2

10x 2.

9

275x 3.

274

500x 4.

499

10x- 5.

11 -

’s More What

′

24

′

1

(

1 −𝑥 2 )

1

2

ൗ

Answer Key

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.

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