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Basic Calculus
Quarter 3 – Module 10:
Chain Rule
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Basic Calculus – Grade 11
Alternative Delivery Mode
Quarter 3 – Module 10: Chain Rule
First Edition, 2020
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Basic Calculus
Quarter 3 – Module 10:
Chain Rule
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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-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.
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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).
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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. 4𝑥 2
C. 4𝑥 3
D. 4𝑥 2 + 1
2. 𝑦 = 𝑥 3 + 1
A. 3𝑥 2 + 2
B. 3𝑥 3
C. 3𝑥 + 1
D. 3𝑥 2
3. 𝑦 = 𝑥 1⁄2
A.
B.
1
𝑥 1⁄2
2
𝑥 1⁄2
C.
D.
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
C.
1
(𝑥 2 +3)1⁄2
D.
𝑥
2(𝑥 2 +3)1⁄2
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9. 𝑦 = tan(3𝑥 2 − 4)
A. 6𝑥csc 2 (3𝑥 2 − 4)
C. 3𝑥csc 2 (3𝑥 2 − 4)
B. 6𝑥sec 2 (3𝑥 2 − 4)
D. 3𝑥sec 2 (3𝑥 2 − 4)
10. 𝑦 = (2𝑥)(𝑥 + 1)2
A. 4𝑥 2 + 8𝑥 + 2
C. 4𝑥 2 − 8𝑥 − 2
B. 6𝑥 2 − 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).
11. 𝑦 =
(4𝑥 2
Column A
− 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(2𝑥 2 − 3)
E.
3
3
(6𝑥−3)1⁄2
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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
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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.
Like terms were simplified.
𝑦 = 4𝑥 2 + 8𝑥 + 4
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.
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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
rule applicable, the equation inside the
Let 𝑢 = 𝑥 + 4
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 .
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Example 2
Differentiate 𝒚 = √𝒙 − 𝟑 .
The table below will show the steps and solution that will give you your desired
answer.
Explanation
Again, there is no direct differentiation
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.
Computation
Let 𝑢 = 𝑥 − 3
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
(
)
(
)
𝑓′(𝑔
𝑥
)
=
𝑢
∙ (1)
the outside function multiplied by the
2
derivative of the inside function,
1
1
𝑓′(𝑔(𝑥 )) =
𝑑 𝑛
[𝑢 ] = 𝑛(𝑢)𝑛−1 ∙ 𝑢′
𝑑𝑥
2
(𝑢)−2
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
′
(
)
𝑦
=
𝑓′(𝑔
𝑥
)
=
substitute to the variable 𝒖 to get the
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,
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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
ℎ(𝑥 ) = 3𝑥
𝑑
[𝑔(𝑥)] + 𝑔(𝑥)
= 𝑓(𝑥) ∙
𝑑𝑥
𝑗(𝑥 ) = (𝑥 + 1)2
𝑑
[𝑓(𝑥)]
∙
𝑑𝑥
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
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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
′
2
Follow the simplification process to get the 𝑦 = 9𝑥 + 12𝑥 + 3
answer.
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
Since there is no direct differentiation Let 𝑢 = 2𝑥 + 3𝑥 − 5
2
rule applicable, the equation inside the
parenthesis was represented into single 𝑓 (𝑢) = (𝑢)7
variable 𝑢 resulting into a simpler
equation raised to an exponent. This
equation is the outside function.
On the other hand, the actual equation 𝑔(𝑥 ) = 2𝑥 2 + 3𝑥 − 5
inside the parenthesis is the inside
function.
Application of chain rule: Derivative of
the outside function multiplied by the 𝑓′(𝑔(𝑥 )) = 7(𝑢)7−1 ∙ (4𝑥 + 3)
derivative of the inside function,
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 .
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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.
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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)
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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
2
(
6. 𝑦 = 𝑥 − 5𝑥 − 2)2
7. 𝑦 = sin(2𝑥 − 5)
8. 𝑦 = √𝑥 + 8
9. 𝑦 = 4𝑥 (𝑥 − 2)2
10. 𝑦 = cos (5𝑥 2 − 3)
Column B
1
A. (
2 𝑥+8)1⁄2
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
y’ =
3
(3𝑥+2)1⁄2
.
12. If y = tan(4𝑥 + 1), then its derivative is y’ = 4𝑠𝑒𝑐 2 (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).
3
15. The derivative of the function y = √6𝑥 + 1 is y’ =
(6𝑥+1)1⁄2
.
Additional Activities
Evaluate the following items below. Write your answer on a separate sheet of paper.
4
1. Find the derivative of the function
𝑦=
2. Differentiate the function 𝑓[𝑔(𝑥 )]
= cos (𝑥 2 ).
1
12
.
√2𝑥+3
13
What’s More
1. 𝑦 ′ = 75(3𝑥 − 1)24
2. 𝑦 ′ =
1
1
(2𝑥−1) ൗ2
What’s In
1.
2.
3.
4.
5.
What I Know
3x2
10x9
275x274
500x499
-10x-11
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
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14
Additional Activities
y’ =
2.
y’ = −
1.
Assessment
4
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
3
(2𝑥+3)2
1
2 sin( 2 )
𝑥
𝑥3
What I have learned
What I Can Do
𝑑𝐿
𝑙𝑖𝑡𝑒𝑟𝑠
= −2,800
𝑑𝑡
𝑚𝑖𝑛𝑢𝑡𝑒
Student’s answer may vary.
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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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For inquiries or feedback, please write or call:
Department of Education – Region III,
Schools Division of Bataan - Curriculum Implementation Division
Learning Resources Management and Development Section (LRMDS)
Provincial Capitol Compound, Balanga City, Bataan
Telefax: (047) 237-2102
Email Address: bataan@deped.gov.ph