This learning packet or any part thereof is solely for the exclusive use of Xavier
University Senior High School students.
Subject: Basic Calculus
Department: MATH/ICT
Duration: 9 weeks
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2
Grading System for Specialized Subject
Semestral Grade = Midterm Grade (50%)
Final Grade (50%)
Written Works
25%
Written Works
25%
Performance Task
45%
Performance Task
45%
Quarterly Assessment 30%
Quarterly Assessment 30%
Written Works
Written works constitute long tests, post-tests, quizzes, seat works, assignments, or any
written activity. These will be given depending on the need and will usually be scheduled.
Quiz questions are usually exercises or applications to practice the concepts learned but will
sometimes be “objective.” They may be held before or after the lecture or activity, so students
are held responsible for studying and preparing. Assignments or homework are given to
encourage self-study among students to review their lecture notes and pursue skills
development not readily acquired during class hours or learning time.
Performance Tasks
Performance tasks (PT) are given or announced to students at the start of the quarter. Ample
time and appropriate scaffold tasks are provided so that students can prepare well.
Quarterly Assessment
There are two major exams for quarterly assessment: Midterm and Final. A delayed exam
will be given within one week after the scheduled examination. There will be no removal
examinations.
Reading Assignments
Students are expected to read all required material(s) given by the teachers. One must refer
to the topics in the course outline.
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Non-regular Activity
During the cautious return to on-campus learning and depending on facilities' availability,
some class periods may be spent in the Audio-Visual Room for other instructional modes like
video-showing, PowerPoint presentations, or multi-media activities.
Study/Activity group and Activity Partner
During the class orientation on the first day, a study/activity group and partner are created.
It is designed to allow students to interact through group activities or with a partner.
Class Decorum
For virtual classes, students are expected to observe netiquette. Taking down notes is an
integrated function of every student as well as studying and participating in class. Students
who can’t join the virtual classes will have to do the tasks on their own, following the
prescribed Learning Time.
GENERAL OBJECTIVES
At the end of the semester, the students must know how to determine the limit of a
function, differentiate algebraic and trigonometric functions in one variable, and formulate
and solve problems involving continuity, extreme values, and instantaneous rate of change.
Duration: 9 weeks (4th Quarter)
WEEK
1
ESSENTIAL TOPICS
Assessments
Limits (Part 1)
Written Works
Limits (Part 2)
Written Works
Continuity
Mini Task 1
Written Works
Review Week
Written Works
March 14-18
2
March 21-25
3
March 28April 1
4
April 4-8
5
4
Midterm Week
Midterm Exam
Derivatives (Part 1)
Written Works
Derivatives (Part 2)
Written Works
Performance Task Week
Performance Task
Finals Week
Final Exam
April 11-13
6
April 18-22
7
April 25-29
8
May 2-6
9
May 10-13
*Schedule may be subject to change.
Tips on how to pass this subject
1. Read diligently this learning packet and understand
each topic.
2. Since this is a self-paced learning, always manage
your time efficiently. Self-discipline is the key to
passing this subject.
3. Do not hesitate to ask for help from your teacher or
any family member who is knowledgeable of this
subject.
4. Practice.
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BASIC CALCULUS
GRADE LEVEL AND STRAND: Grade 12 (STEM)
SY 2021-2022
LESSONS OVERVIEW
Welcome to Basic Calculus!
In this learning packet, you will take the following topics, and you are expected to
demonstrate the following as evidence of learning:
Essential Topics
1. Limits
2. Continuity
3. Derivatives
Learning Objectives
At the end of this learning packet, students will be able to
1. illustrate the limit of a function using a table of values
and the graph of the function;
2. distinguish between lim 𝑓(𝑥) 𝑎𝑛𝑑 𝑓(𝑎) ;
𝑥→𝑎
3. illustrate the limit laws and the definition of
0
indeterminate form of type 0;
4. apply the limit laws and the definition of indeterminate
0
form of type 0 in evaluating the limit of algebraic
functions (polynomial, rational, and radical);
5. distinguish the difference between one-sided limits,
infinite limits, and limits at infinity;
6. evaluate one-sided limits, infinite limits, limits at
infinity, and the limit of trigonometric functions with
sin 𝑡
1−cos 𝑡
expressions 𝑡 and 𝑡 ;
7. illustrate continuity of a function at a number;
8. determine whether a function is continuous at a
number or not;
9. illustrate continuity of a function on an interval;
10. illustrate different types of discontinuity
(hole/removable, jump/essential, asymptotic/infinite);
11. solve problems involving continuity of a function;
12. illustrate the tangent line to the graph of a function at
a given point;
13. applies the definition of the derivative of a function at
a given number;
14. relate the derivative of a function to the slope of the
tangent line;
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15. determine the relationship between differentiability
and continuity of a function;
16. apply the differentiation rules in computing the
derivative of algebraic, exponential, and trigonometric
functions;
17. solve situational problems involving instantaneous
rate of change;
18. illustrate the Chain Rule of differentiation;
19. illustrate the Extreme Value Theorem;
20. compute the higher-order derivatives of algebraic
functions; and
21. solve optimization problems that yield polynomial
functions.
Evidence of Learning
1. Task Completion (includes activities, MT, and PT)
2. Time Management
3. Self-regulated Learning
4. Discipline
Self-regulated Strategies
1. Goal Setting
2. Resource Management
By the end of the quarter, you are expected to show the following as Performance Task:
The BASCAL Company is one of the popular packaging industries in the city. You
are working in that company as a paper designer where your job is to come up with
different designs and sizes of boxes. One client contracted your company to come up
with a box design for a certain product. The condition of your manager is to maximize
the material to be used per box so that no material will be wasted as well as to
identify the maximum volume of the box which it can hold. As a paper designer,
you are tasked to come up with the dimensions of the box, its maximum volume, and
the proposed price per box which you will submit to your manager. Your written report
will be evaluated according to the organization of your output, use of mathematical
concepts, accuracy of computation, promptness of submission, and completeness of
necessary information.
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To manage your time well, observe the Learning Time below:
Lesson
Worksheet No.
Limits (Part 1)
Limits (Part 2)
Continuity
Derivatives (Part 1)
Derivatives (Part 2)
1
2
3
4
5&6
Mini Task 1
Performance Task
Target Date of
Completion
March 18, 2022
March 25, 2022
April 1, 2022
April 22, 2022
April 29, 2022
April 8, 2022
May 6, 2022
My Thoughts About
This Subject
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----PRE-TEST---Let us find out how much you already know about the topics in this learning packet. Encircle
the letter that you think best answers the question. Please answer all items. After taking this
short test, you will see the answers immediately through the given link(s). Take note of the items
that you were not able to correctly answer and review those items as you go through this
learning packet.
1. What is the limit of a constant k as x approaches the number a?
a. 𝑘 + 4
b. k
c. 0
d. does not exist
2. Evaluate
2
a. − 7
b.
−1
1
7
2
c. 7
d. 7
c. 3𝜋
d. −3𝜋
c. +∞
d. −∞
3. Evaluate
a. 3
b. −3
4. Evaluate
a.
1
2
b.
−1
2
5. Given the function
. What is the value of a so that the
1
function f is continuous at 𝑥 = 2 ?
a. −1
6. The function
a. −1 and 2
b. 0
c. 1
d. 2
is continuous everywhere except when 𝑥 is/are ______.
b. −2 and 1
c. 2
d. −1
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7. Which of the following is equal to the
a.
−3
b.
√2
?
3
c.
√2
−2
8. Which of the following is equal to the
a.
2
b.
3
9. Evaluate
−1
c. −∞
5
c. 1
if
a. 0
b. 2
12. If
c. 3
, then what is
a. 8𝑥 7
14. What is
15. What is
a. 8x + 3
?
d. 5𝜋
c. 𝑥 8
d. x
c. 3
d. x
b. 4x − 3
c. 8𝑥 + 1
d. 8x − 3
?
b. −80x + 17
c. 4 − 5x
d. −40
, then what is ?
b. −5
a. 4𝑥 2 − 3x
c. 𝜋
d. 4
?
b. 𝑥 7
if
d. 2
.
11. Which of the following is the derivative of
a. 0
b. −100
a. 3x
d. +∞
.
b. 0
10. Evaluate
2
√3
?
if
a. −1
13. If
d.
√3
?
10
16. What is
a.
−47
(8𝑥+3)2
?
4−5𝑥
8𝑥
b. (8𝑥+3)2
−5𝑥
c. (8𝑥+3)2
d. (8𝑥+3)2
17. Which of the following is the derivative of
a. cos 𝑥 + sin 𝑥
b. cos 𝑥
c. sin 𝑥
?
18. Which of the following is the derivative of
a. x(sec 𝑥)2
b. tan 𝑥
?
c. 𝑥(sec 𝑥) + tan 𝑥
, then what is
a.
47
121
−47
c. 144
21. Which of the following are the critical numbers of
a. −1 and 2
b. −2 and 2
c. −1 and 2
22. Find the absolute minimum of
a. (−2,1)
b. (−1,12)
c. (2, −15)
23. Which of the following is the relative maximum of
a. (−2,6)
b. (0,11)
c. (2, −6)
24. If
a. 48x − 24
b. 48
at
.
1
d. 4
?
57
b. 121
d. (csc 𝑥)2
2
19. Find the slope of the tangent line to the curve
a. 0
b. 1
c. −1
20. If
d. sin 2𝑥
, then what is
?
c. 24𝑥 2 − 24𝑥 − 10
−57
d. 144
d. 1 and −4
on the interval
d. (4,37)
?
.
?
d. (0,10)
d. 8𝑥 3 − 12𝑥 2 − 10𝑥
25. You have 40 linear feet of fencing with which to enclose a rectangular space for a garden.
Find the largest area that can be enclosed with this much fencing and the dimensions of the
corresponding garden.
a. 1,000 𝑓𝑡 2
b. 100 𝑓𝑡 2
c. 20 𝑓𝑡 2
d. 400 𝑓𝑡 2
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CONTEXT
Change is the only thing that will not change in this world. Every day in our lives, we
observe the constant change of everything, physically or emotionally. From our first
heartbeat until our last breath, we are always part of this never-ending process of change.
We are living in this ever-changing world. Observing the kind of life in the past and
comparing it to the sort of a life today, we can see that there are a lot of changes that have
happened, good or bad. So, as we reflect, we can say that we are just part of the process. The
moment we leave this world, change is still there, and the world will continue to move
forward.
The idea of “change” has a significant impact on the world of mathematics. This idea
of change paved the way to the birth of Calculus. It is an area of mathematics that deals with
change.
In this learning packet, as an introduction to Calculus, we will observe the behavior
of a function as we evaluate its limit and determine its continuity. To simply put this one, we
will examine the values (range) of a function as we keep on changing the values of the domain
that approach a particular value. Also, we will discuss derivatives of functions where
different rules of derivatives and their application (focusing on instantaneous rate of change
and optimization) will be highlighted.
As a pre-requisite in this subject, you must have a better understanding of functions
discussed in General Mathematics and Pre-Calculus. By this, you will have a good grasp of
the topics discussed in this subject. You can always review your past lessons if you find
yourself breaking your back. In the end, mathematics is not concerned about how brilliant
you are but how you persevere in the learning process.
In this learning packet, you will learn and expected to show the following as evidence of
learning:
1. illustrate the limit of a function using a table of values and the graph of the function;
2. distinguish between lim 𝑓(𝑥) 𝑎𝑛𝑑 𝑓(𝑎);
𝑥→𝑎
0
3. illustrate the limit laws and the definition of indeterminate form of type 0;
0
4. apply the limit laws and the definition of indeterminate form of type 0 in evaluating
the limit of algebraic functions (polynomial, rational, and radical);
5. distinguish the difference between one-sided limits, infinite limits, and limits at
infinity;
6. evaluate one-sided limits, infinite limits, limits at infinity, and the limit of
sin 𝑡
1−cos 𝑡
trigonometric functions with expressions 𝑡 and 𝑡 ;
7. illustrate continuity of a function at a number;
8. determine whether a function is continuous at a number or not;
12
9. illustrate continuity of a function on an interval;
10. illustrate different types of discontinuity (hole/removable, jump/essential,
asymptotic/infinite);
11. solve problems involving continuity of a function;
12. illustrate the tangent line to the graph of a function at a given point;
13. applies the definition of the derivative of a function at a given number;
14. relate the derivative of a function to the slope of the tangent line;
15. determine the relationship between differentiability and continuity of a function;
16. apply the differentiation rules in computing the derivative of algebraic, exponential,
and trigonometric functions;
17. solve situational problems involving instantaneous rate of change;
18. illustrate the Chain Rule of differentiation;
19. illustrate the Extreme Value Theorem;
20. compute the higher-order derivatives of algebraic functions; and
21. solve optimization problems that yield polynomial functions.
By the end of the lesson, you are expected to show the following scaffold to the
Performance Task:
MINI-TASK 1: Solving Problems Involving Limits and Continuity Through Video
Presentation
You will be given problem sets involving Limits and Continuity. You will answer
these problem sets and create a pre-recorded video on how you solved each problem in
the background. You will also explain the solution process by emphasizing essential
topics discussed in class. The video presentation may include concept sharing, solving
techniques, applications, etc. You may use applications like MS PowerPoint, online
whiteboard, physical whiteboard, manila paper, etc., depending on the availability of
resources you have at home.
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Give your initial ideas and what you want to know about limits and continuity of a
function and differentiation and its application by answering the K and W column of
the KWL chart.
K
(What I know)
W
(What I want to know)
L
(What I learned)
Let us see how these ideas may be affirmed or revised as we go along this learning packet.
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15
Unit 1: Limits and
Continuity
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EXPERIENCE
Lesson 1.1 Definition and Illustration of Limits
Limits are the cornerstone of calculus, called the Mathematics of Change. The study
of limits is fundamental in studying change in detail. The evaluation of a particular limit
supports the formulation of the derivative and the integral of a function (Arceo et al., 2016).
The concept of limits can be described as the best prediction of a point that cannot be
observed, and the best way of predicting it is through examining its neighboring points
cautiously. This process is called an estimate. Since Mathematics has “black hole” scenarios
such as zero dividing by zero, going to infinity, and the likes yield indeterminate results, the
concept of limits is significant because it can give us an estimate when the result cannot be
computed directly (Azad et al., 2022).
In other words, limits refer to the limits of single-variable functions. Studying the
function's behavior as its variable approaches a particular value (a constant) shows that it
can only take values closest to the constant. Still, it cannot equal the constant itself. However,
the limit can clearly describe what is happening to the function near the constant (Arceo et
al., 2016).
Definition of Limits
As mentioned, the concept of limits can be described as the best prediction of a point
that cannot be observed, and the best way to predict it is through an estimate. Try to consider
the given below:
Solve for 𝑓(𝑥) =
𝑥 2 −1
𝑥−1
when 𝑥 = 1
Solution:
𝑓(1) =
12 −1
1−1
𝟎
𝑓(1) = 𝟎
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In Mathematics, the expression
𝟎
𝟎
yields an indeterminate result. It means that the
value of the expression does not exist. If the value of 𝑓(𝑥) when 𝑥 = 1 does not exist, estimate
(when we use the neighboring points of 1) is the best option to predict the value of the
function. To put it differently, instead of using 𝑥 = 1, let us try approaching it closer and
closer. Try to examine tables 1 and 2 below:
Approaching the value
of 𝒙 = 𝟏 from the left
𝒙
Approaching the value
of 𝒙 = 𝟏 from the right
𝑥2 − 1
𝑓(𝑥) =
𝑥−1
𝒙
𝑥2 − 1
𝑓(𝑥) =
𝑥−1
0.5
1.50000
1.5
2.50000
0.9
1.90000
1.1
2.10000
0.99
1.99000
1.01
2.01000
0.999
1.99900
1.001
2.00100
0.9999
1.99990
1.0001
2.00010
0.99999
1.99999
1.00001
2.00001
Table 1
Table 2
Initially, the function yields an indeterminate result when 𝑥 = 1. But, the tables
clearly show that as we use the values closest to 1, it can be predicted that the value of the
function is 2. In other words, as the value of 𝑥 approaches 1, the value of the function is going
to be 2. However, we cannot say that it should be 2; that is why mathematicians devise the
term “limit.” The limit of 𝑓(𝑥) =
𝑥 2 −1
𝑥−1
as 𝑥 approaches 1 is 2. It can be written in a
mathematical statement as:
. Thus,
it is a unique way of saying, “ignoring what
happens when we get there, but as we get closer
and closer, the answer gets closer and closer to
2”. Moreover, let us examine the graph of the
function described in illustration 1. It clearly
shows that we cannot say what the value at 𝑥 =
1 is, but we can say that as we approach 1, the
limit is 2.
Illustration 1
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See table 3 for the formal definition of limits. To deepen your understanding of its
definition, you may visit this link: https://www.mathsisfun.com/calculus/limitsformal.html.
Let 𝒇 be a function defined at every number in some open interval
containing 𝒂, except possibly at the number 𝒂 itself. The limit of 𝒇(𝒙)
as 𝒙 approaches 𝒂 is 𝑳, written as
𝒍𝒊𝒎 𝒇(𝒙) = 𝑳
𝒙→𝒂
if the following statement is true:
Given any 𝜺 > 𝟎, however small, there exists a 𝜹 > 𝟎 such that
if 𝟎 < |𝒙 − 𝒂| < 𝜹 then |𝒇(𝒙) − 𝑳| < 𝜺
Table 3
Illustrating the Limit of a Function
As observed, the limit of a function can be illustrated in two (2) ways: (1) Table of
Values and (2) Graph of a Function. In the following examples, see how constructing a table
of values and looking at the graph of 𝑦 = 𝑓(𝑥) helps in explaining the limit of the functions.
Example 1.1.1.
Investigate lim (𝑥 2 + 1)
x→−1
A. Table of Values
It clearly shows that 𝑎 = −1 and 𝑓(𝑥) = 𝑥 2 + 1.
We start by approaching −1 from the left.
𝑥
𝑓(𝑥)
−1.5
3.25
−1.2
2.44
−1.01
2.0201
−1.0001
2.00020001
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Now approach −1 from the right.
𝑥
𝑓(𝑥)
−0.5
1.25
−0.8
1.64
−0.99
1.9801
−0.9999
1.99980001
The tables show that as 𝑥 approaches −1, 𝑓(𝑥) approaches 2. In symbols, lim (𝑥 2 + 1) = 2.
x→−1
B. Graph of 𝑓(𝑥) = 𝑥 2 + 1
Its graph is given by
illustration 2. It shows a
parabola with a y-intercept
(0,1). Look at the graph in the
vicinity of 𝑥 = −1. You can
somehow see the points from the
table of values approaching the
level where 𝑦 = 2. Hence, the
graph confirms that lim (𝑥 2 +
x→−1
1) = 2.
Illustration 2
Example 1.1.2.
Investigate lim|𝑥|
𝑥→0
A. Table of Values
It clearly shows that 𝑎 = 0 and 𝑓(𝑥) = |𝑥|.
20
We start by approaching 0 from the left.
𝑥
𝑓(𝑥)
−0.3
0.3
−0.01
0.01
−0.00009
0.00009
−0.00000001
0.00000001
Now approach 0 from the right.
𝑥
𝑓(𝑥)
0.3
0.3
0.01
0.01
0.00009
0.00009
0.00000001
0.00000001
The tables show that as 𝑥 approaches 0, 𝑓(𝑥) approaches 0. In symbols, lim|𝑥| = 0.
𝑥→0
B. Graph of 𝑓(𝑥) = |𝑥|
Its graph is given by
illustration 3. It is clear that
lim|𝑥| = 0, that is, the two sides of
𝑥→0
the graph both move down to the
origin (0,0) as 𝑥 approaches 0.
Illustration 3
21
Example 1.1.3.
Investigate lim
𝑥 2 −5𝑥+4
𝑥→1
𝑥−1
A. Table of Values
It clearly shows that 𝑎 = 1 and 𝑓(𝑥) =
𝑥 2 −5𝑥+4
𝑥−1
. Note that 1 is not in the domain of 𝑓,
but this is not a problem. In evaluating a limit, remember that we only need to go very close
to 1; we will not go to 1 itself. We now start by approaching 1 from the left.
𝑥
𝑓(𝑥)
0.5
−3.5
0.88
−3.12
0.996
−3.004
0.9999
−3.0001
Now approach 1 from the right.
𝑥
𝑓(𝑥)
1.5
−2.5
1.17
−2.83
1.003
−2.997
1.0001
−2.999
The tables show that as 𝑥 approaches 1, 𝑓(𝑥) approaches −3. In symbols, lim
𝑥→1
B. Graph of 𝑓(𝑥) =
𝑥 2 −5𝑥+4
𝑥−1
= −3.
𝑥 2 −5𝑥+4
𝑥−1
Its graph is given by illustration 4. Take note that
𝑓(𝑥) =
𝑥 2 −5𝑥+4
𝑥−1
=
(𝑥−4)(𝑥−1)
𝑥−1
= 𝑥 − 4, provided 𝑥 ≠
1. Hence, the graph of 𝑓(𝑥) =
𝑥 2 −5𝑥+4
𝑥−1
is also the
graph of 𝑓(𝑥) = 𝑥 − 1, excluding the point where
𝑥 = 1.
Look at the graph in the vicinity of 𝑥 = 1. You
can somehow see the points from the table of values
approaching the level where 𝑦 = −3. Hence, the
graph confirms that lim
𝑥→1
𝑥 2 −5𝑥+4
𝑥−1
= −3.
Illustration 4
22
Example 1.1.4.
Investigate lim 𝑓(𝑥) if 𝑓(𝑥) = {
𝑥→4
𝑖𝑓 𝑥 < 4
.
𝑖𝑓 𝑥 ≥ 4
𝑥+1
(𝑥 − 4)2 + 3
A. Table of Values
This looks slightly different, but the logic and procedure are the same. We still
approach the constant 4 from the left and the right, but we should evaluate the appropriate
corresponding functional expression. In this case, when 𝑥 approaches 4 from the left, the
taken values should be substituted in 𝑓(𝑥) = 𝑥 + 1. Indeed, this part of the function accepts
values less than 4. So,
𝑥
𝑓(𝑥)
3.7
4.7
3.85
4.85
3.995
4.995
3.99999
4.99999
On the other hand, when 𝑥 approaches 4 from the right, the values should be substituted in
𝑓(𝑥) = (𝑥 − 4)2 + 3. Thus,
𝑥
𝑓(𝑥)
4.3
3.09
4.1
3.01
4.001
3.000001
4.00001
3.0000000001
Observe that the values that 𝑓(𝑥) approaches are not equal, namely, 𝑓(𝑥) approaches 5 from
the left while it approaches 3 from the right. In such a case, we say that the limit of the given
function does not exist (DNE). In symbols, lim 𝑓(𝑥) DNE. There are two (2) concepts that
need to be emphasized here:
𝑥→4
(1) We do not say that lim 𝑓(𝑥) “equals DNE,” nor do we write “lim 𝑓(𝑥) = 𝐷𝑁𝐸, because
𝑥→4
𝑥→4
“DNE” is not a value. In the example, “DNE” indicated that the function moves in different
directions as to its variable approaches 𝑎 from the left and the right. In other cases, the limit
1
fails to exist because it is undefined, such as for lim 𝑥 which leads to the division of 1 by zero.
𝑥→0
(2) We have been specifying whether 𝑥 will approach a value 𝑎 from the left, through values
less than 𝑎, or from the right, through values greater than 𝑎. This direction may be specified
in the limit notation, lim 𝑓(𝑥) by adding certain symbols.
𝑥→𝑎
23
•
•
If 𝑥 approaches 𝑎 from the left, or through values less than 𝑎, then we write
lim− 𝑓(𝑥).
𝑥→𝑎
If 𝑥 approaches 𝑎 from the right, or through values greater than 𝑎, then we
write lim+ 𝑓(𝑥).
𝑥→𝑎
Furthermore, we say lim 𝑓(𝑥) = 𝐿 if and only if lim− 𝑓(𝑥) = 𝐿 and lim+ 𝑓(𝑥) = 𝐿. In other
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
words, for a limit L to exist, the limits from the left and the right must both exist and are equal
to L. Therefore, lim 𝑓(𝑥) 𝐷𝑁𝐸 whenever lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥). These limits, lim− 𝑓(𝑥) and
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
lim+ 𝑓(𝑥) are also called one-sided limits since you only consider values on one side of 𝑎
𝑥→𝑎
(Arceo et al., 2016).
B. Graph of 𝑓(𝑥) if 𝑓(𝑥) = {
𝑖𝑓 𝑥 < 4
𝑖𝑓 𝑥 ≥ 4
𝑥+1
(𝑥 − 4)2 + 3
Its graph is given by illustration 5.
We can see from the graph that 𝑓(𝑥) has
no limit as 𝑥 approaches 𝟒. The two
separate parts of the function move
toward different 𝑦-levels (𝑦 = 5 from the
left, 𝑦 = 3 from the right) in the vicinity
𝑎 = 4.
So, in general, if we have the
graph of a function as shown in
illustration 6, determining limits can be
done much faster and easier by
inspection. For instance, it can be seen
from the graph in illustration 6 that:
Illustration 5
a. lim 𝑓(𝑥) = 1
𝑥→−2
b. lim 𝑓(𝑥) = 3. Here, it does not matter that 𝑓(0)
𝑥→0
does not exist (that is, it is undefined, or 𝑥 = 0 is not
in the domain of 𝑓). Always remember that what
matters is the behavior of the function close to 𝑎 =
0 and not precisely at 𝑎 = 0. Even if 𝑓(0) were
defined and equal to any other constant (not equal
to 3), like 100 or −5000, this would still have no
bearing on the limit. In cases like this, lim 𝑓(𝑥) = 3
𝑥→0
prevails regardless of the value of 𝑓(0) if any.
Illustration 6
24
c. lim 𝑓(𝑥) 𝐷𝑁𝐸. As shown in illustration 6, the two parts of the graph near 𝑎 = 3 do not
𝑥→3
move towards a common 𝑦-level as 𝑥 approaches 𝑎 = 3.
Practice #1
Try this! ☺
Answer what is asked on each question:
1. Investigate 𝑙𝑖𝑚 𝑥 2 − 2𝑥 + 4 through table of values.
𝑥→1
2. Consider the function 𝑓(𝑥) whose graph is shown below. Determine the following:
a) 𝑙𝑖𝑚 𝑓(𝑥)
𝑥→−3
b) 𝑙𝑖𝑚 𝑓(𝑥)
𝑥→1
c) 𝑙𝑖𝑚 𝑓(𝑥)
𝑥→3
d) 𝑙𝑖𝑚 𝑓(𝑥)
𝑥→−1
e) 𝑙𝑖𝑚 𝑓(𝑥)
𝑥→5
Process Questions
1. How to determine the limit of a function through table of values? graphing?
2. How to determine that the limit of a function does not exist through tables of
values? graphing?
3. Why is it incorrect to say that say that the lim 𝑓(𝑥) “equals DNE”, nor do we write
𝑥→𝑎
“lim 𝑓(𝑥) = 𝐷𝑁𝐸?
𝑥→𝑎
4. Is it correct to say that the lim 𝑓(𝑥) = 𝑓(𝑎) since 𝑎 is in the domain of 𝑓? Why? Why
not?
𝑥→𝑎
Student’s responses:
Please write your answers on page 158.
25
My Working Space
(write your solutions to the practice problems here)
26
Lesson 1.2 Limit Theorems
One might ask if it is required to construct a table of values or graph the function when
asked to identify the limit, apparently not. The following theorems might help us directly
evaluate the limit of a function without doing the tedious task of constructing a table of
values or graphing the function.
In the following statements, 𝑎 is a constant, and f and g are functions that may or may
not have 𝑎 in their domains.
1. The limit of a constant is itself. If k is any constant, then
lim 𝑘 = 𝑘
𝑥→𝑎
2. The limit of x as x approaches 𝑎 is equal to 𝑎.
lim 𝑥 = 𝑎
𝑥→𝑎
For the following theorems, we will assume that the limits of f and g exist as x
approaches 𝑎 and that these limits are as follows:
and
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
lim 𝑔(𝑥) = 𝑀
𝑥→𝑎
3. Constant Multiple Theorem. The limit of a multiple of a function is simply that
multiple of the limit of the function.
lim 𝑘 ∙ 𝑓(𝑥) = 𝑘 ∙ 𝐿
𝑥→𝑎
4. Addition Theorem. The limit of a sum of functions is the sum of the limits of the
individual functions, and the limit of the difference of functions is the difference of
their limits.
lim (𝑓(𝑥) + 𝑔(𝑥)) = lim 𝑓(𝑥) + lim 𝑔(𝑥) = 𝐿 + 𝑀
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
lim (𝑓(𝑥) − 𝑔(𝑥)) = lim 𝑓(𝑥) − lim 𝑔(𝑥) = 𝐿 − 𝑀
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
5. Multiplication Theorem. The limit of a product of functions is equal to the product
of the limits.
lim (𝑓(𝑥) • 𝑔(𝑥)) = lim 𝑓(𝑥) • lim 𝑔(𝑥) = 𝐿 • 𝑀
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
27
6. Division Theorem. The limit of a quotient of functions is equal to the quotient of
the limits of the individual functions, provided the denominator limit is not equal to
0.
lim 𝑓(𝑥)
𝑓(𝑥) 𝑥→𝑎
𝐿
lim
=
=
, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑀 ≠ 0
𝑥→𝑎 𝑔(𝑥)
lim 𝑔(𝑥) 𝑀
𝑥→𝑎
7. Power Theorem. The limit of an integer power p of a function is just that power of
the limit of the function.
𝑝
𝑝
lim (𝑓(𝑥)) = (lim 𝑓(𝑥)) = 𝐿𝑝
𝑥→𝑎
𝑥→𝑎
8. Radical/Root Theorem. If n is a positive integer, the limit of the nth root of a
function is the nth root of the limit of the function, provided that the nth root of the
limit is a real number. If n is even, then L must be positive.
𝑛
𝑛
lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐿
𝑥→𝑎
𝑥→𝑎
Note: The addition and multiplication theorems may be applied to more than two
functions' sums, differences, and products.
Examples:
1.2.1
1.2.2
1.2.3
lim 5 = 5
𝑥→2
lim 𝑥 = 2
𝑥→2
lim(3𝑥 − 4) = lim 3𝑥 − lim 4
𝑥→1
𝑥→1
𝑥→1
= 3lim 𝑥 − lim 4
𝑥→1
𝑥→1
(Addition Theorem)
(Constant Multiple Theorem)
= 3(1) −4
= −1
1.2.4
(Addition Theorem)
lim (2𝑥 4 − 5𝑥 + 3) = lim 2𝑥 4 − lim 5𝑥 + lim 3
𝑥→−2
𝑥→−2
𝑥→−2
𝑥→−2
= 2 lim 𝑥 4 −5lim 𝑥 + lim 3
𝑥→−2
𝑥→−2
4
𝑥→−2
(Constant Multiple Theorem)
= 2 ( lim 𝑥) −5 lim 𝑥 + lim 3
𝑥→−2
4
𝑥→−2
𝑥→−2
(Power Theorem)
= 2(−2) − 5(−2) + 3
= 2(16) + 10 + 3
= 45
28
Examples:
1.2.5.
𝑥+3
lim 𝑥 2 −1 =
𝑥→2
lim 𝑥+lim 3
𝑥→2
(Addition Theorem)
𝑥→2
lim 𝑥 2 −lim 1
𝑥→2
𝑥→2
lim 𝑥+lim 3
𝑥→2
=
(Power theorem)
𝑥→2
2
(lim 𝑥) −lim 1
𝑥→2
=
=
1.2.6.
𝑥→2
2+3
(2)2 −1
5
3
3
lim √2𝑥 + 3 = 3√ lim (2𝑥 + 3)
𝑥→−5
𝑥→−5
= 3√ lim 2𝑥 + lim 3
𝑥→−5
𝑥→−5
= 3√2 lim 𝑥 + lim 3
𝑥→−5
𝑥→−5
3
= √2(−5) + 3
3
= √−7
3
= − √7
1.2.7.
lim
𝑥 2 −9
𝑥→4 𝑥−4
The limit of the numerator is 7, while the limit of the denominator is 0. This
will result to
7
0
which is undefined. Thus, lim
𝑥 2 −9
𝑥→4 𝑥−4
does not exist (DNE).
Practice #2
Try this! ☺
Evaluate the following limits.
1. lim (−𝑥 3 + 3𝑥 − 6)
𝑥→−3
4. lim5
𝑥→
2
√2𝑥−5
𝑥−4
5−𝑥
2. lim (𝑥 2−3𝑥−10)
𝑥→4
𝑥 2 −9
3. lim 𝑥−6
𝑥→3
5
5. lim 2𝑥+6
𝑥→−3
29
Process Questions
1. What could be the possible cases that the limit of a function does not exist?
2. In which real-life situation can you apply the practical (may not be mathematical)
concept of limit?
3. How does limit describes the behavior of a function?
4. What concepts in algebra are needed in evaluating limits of a function?
Student’s responses:
Please write your answers on page 158.
My Working Space
(write your solutions to the practice problems here)
30
Lesson 1.3 Indeterminate Forms
There are functions whose limits cannot be determined immediately using the Limit
Theorems introduced previously. In these cases, the functions must be manipulated so that
the limit, if it exists, can be calculated. We call such limit expressions indeterminate forms.
0
In this lesson, we will only be discussing a particular indeterminate form the " 0 ". See
its definition indicated below.
𝟎
Definition of Indeterminate Form of Type " 𝟎 "
𝑓(𝑥)
If lim 𝑓(𝑥) = 0 and lim 𝑔(𝑥) = 0, then lim 𝑔(𝑥) is called an indeterminate form
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
0
of type " 0 ".
0
Note: If the limit is indeterminate of type " 0 ", one should find an expression equivalent to
the original to find the actual value. This can be done through factoring or rationalizing.
Hopefully, the expression that will emerge after factoring or rationalizing will have a
computable limit (Arceo et al., 2016). See the following examples.
Example 1.3.1.
Evaluate lim
𝑥 2 +2𝑥+1
𝑥→−1
𝑥+1
Solution:
The limit of both numerator and denominator as 𝑥 approaches −1 is 0. Thus, the limit
0
is indeterminate of type " ". However, observe that (𝑥 + 1) is a factor common to the
0
numerator and the denominator, and
lim
𝑥→−1
𝑥 2 +2𝑥+1
𝑥+1
𝑥 2 +2𝑥+1
𝑥+1
=
(𝑥+1)2
𝑥+1
= 𝑥 + 1, when 𝑥 ≠ −1. Therefore,
= lim (𝑥 + 1) = 0.
𝑥→−1
Example 1.3.2.
Evaluate lim
𝑥 2 −1
𝑥→1 √𝑥−1
Solution:
Since lim 𝑥 2 − 1 = 0 and lim √𝑥 − 1 = 0, then lim
𝑥→1
𝑥 2 −1
0
type " 0 ". To find the limit, observe that if 𝑥 ≠ 1, then
(𝑥 + 1)(√𝑥 + 1). So, we have lim
𝑥 2 −1
𝑥→1 √𝑥−1
is an indeterminate form of
𝑥→1 √𝑥−1
𝑥 2 −1
𝑥→1
√
∙
𝑥−1
√𝑥+1
√𝑥+1
=
(𝑥−1)(𝑥+1)(√𝑥+1)
𝑥−1
=
= lim(𝑥 + 1)(√𝑥 + 1) = 4.
𝑥→1
31
Practice #3
Try this! ☺
Evaluate the following limits.
1. lim
𝑡 2 −1
2𝑧−𝑧 2
2. lim ( 𝑧 2−4 )
𝑡→−1 𝑡 2 +4𝑡+3
3
𝑧→2
𝑥2 −𝑥−2
3. lim 𝑥3−6𝑥2−7𝑥
𝑥→−1
4. lim
𝑥→−1
√𝑥2 +3−2
𝑥2 −1
Process Questions
1. What are the strategies that need to be mindful of in solving the actual limit of the
0
function when its initial limit is indeterminate of type " 0 ".
2. How important are the concepts of factoring or rationalizing when the limit of the
0
function is indeterminate of type " 0 "?
Student’s responses:
Please write your answers on page 159.
Worksheet #1: (See page 152)
My Working Space
(write your solutions to the practice problems here)
32
33
34
Lesson 1.4
One-sided Limits, Infinite Limits, Limits at
Infinity, and Infinite Limits at Infinity
One-sided Limits
This section considers limits at a point a on the real line by letting x approach a.
Because x can approach a from the left-side or from the right-side, we have a left-hand limit
and right-hand limit. They are called one-sided limits.
The limit exists if the resulting function values from both sides approach the same
number L. Also, it is crucial to consider that the limits are not defined for some functions
when evaluated as to their variable approaches a particular number a.
Example 1.4.1.
Let 𝑓(𝑥) = √𝑥 .
The domain of f is [0, ∞) and so f is defined
on the right-side of 0. Note that if x is close to and
greater than 0, then 𝑓(𝑥) is close to 0, as shown
in illustration 7.
This means that:
lim √𝑥 = 0.
𝑥→0+
Illustration 7
Therefore, the right-hand limit exists.
As x approaches a from the right, we use values greater than a but sufficiently close
to the value of a. We call this right-hand limit of 𝒇 as x approaches a.
Right-hand Limit
Definition:
Let f be a function defined at every number in some open interval (a, b). Then the
limit of f(x) as x approaches a from the right is L written as
lim 𝑓(𝑥) = 𝐿.
𝑥→𝑎+
Take note that, in example 1.4.1, the left-hand limit, lim− √𝑥 , does not exist (DNE).
𝑥→0
This leads us to consider separate evaluation of limit for cases when x approaches a through
numbers greater than a and through numbers less than a.
35
On the other hand, as x approaches a from the left, we use values that are less than a
but sufficiently close to the value of a. This is called the left-hand limit of 𝒇 as x approaches
a.
Left-hand Limit
Definition:
Let 𝑓 be a function defined at every number in some open interval (𝑎, 𝑏). Then the
limit of 𝑓(𝑥) as x approaches a from the left is L written as
lim 𝑓(𝑥) = 𝐿.
𝑥→𝑎−
Example 1.4.2.
Suppose we want to find the limit of √𝑥 − 4 as x approaches 4 from
the right.
The domain of 𝑓 is [4, ∞) and 𝑓 is defined on the
right-side of 4. Note that if x is close to and greater than
4, then 𝑓(𝑥) is close to 0 as shown in illustration 8.
This means that:
lim √𝑥 − 4 = 0.
𝑥→4+
Therefore, the right-hand limit exists.
Take note that the left-hand limit, lim− √𝑥 − 4 , does not
𝑥→4
exist (DNE).
Example 1.4.3.
Illustration 8
lim √3 − 𝑥
𝑥→3−
The domain of 𝑓 is (−∞, 3] and 𝑓 is defined
on the left-side of 3. Note that if x is close to and less
than 3, then 𝑓(𝑥) is close to 0 as shown in
illustration 9.
This means that:
lim √3 − 𝑥 = 0.
𝑥→3−
Therefore, the left-hand limit exists.
Take note that the right-hand limit, lim+ √3 − x , does not exist (DNE).
Illustration 9
𝑥→3
36
Existence of Limit Theorem:
The limit of f as x approaches a exists and is equal to the number L, if and only if
lim+ 𝑓(𝑥) and lim− 𝑓(𝑥) both exist and are equal to L. That is:
𝑥→𝑎
𝑥→𝑎
lim 𝑓(𝑥) = lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝐿 .
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥+1
Example 1.4.4.
Consider the function defined by 𝑓(𝑥) = {−√1 − 𝑥 2
5
a. Sketch the graph of 𝑓(𝑥)
b. Evaluate the following limits (if they exist):
i.
ii.
lim 𝑓(𝑥),
lim 𝑓(𝑥),
𝑖𝑓 − 1 ≤ 𝑥 < 1
𝑖𝑓 𝑥 ≥ 1
lim 𝑓(𝑥), 𝑎𝑛𝑑 lim 𝑓(𝑥)
𝑥→−1−
𝑥→1−
𝑖𝑓 𝑥 < −1
𝑥→−1+
𝑥→−1
lim 𝑓(𝑥), 𝑎𝑛𝑑 lim 𝑓(𝑥)
𝑥→1+
𝑥→1
Solutions:
a. Graph
b. The limits are evaluated as follows:
i.
lim 𝑓(𝑥) =
𝑥→−1−
lim 𝑓(𝑥) =
𝑥→−1+
lim 𝑓(𝑥) =
𝑥→−1
lim (𝑥 + 1) = 0,
𝑥→−1−
lim − √1 − 𝑥 2 = 0,
𝑥→−1+
𝑓(𝑥) = 𝑥 + 1 since 𝑥 < −1
𝑓(𝑥) = −√1 − 𝑥 2 since 𝑥 > −1
lim = lim+ = 0
𝑥→−1−
𝑥→−1
37
ii.
lim 𝑓(𝑥) =
𝑥→1−
lim 𝑓(𝑥) =
𝑥→1+
lim − √1 − 𝑥 2 = 0,
𝑥→1−
𝑥→1
Example 1.4.5.
𝑓(𝑥) = 5 since 𝑥 > 1
lim 5 = 5,
𝑥→1+
Since lim− 𝑓(𝑥) ≠
𝑓(𝑥) = −√1 − 𝑥 2 since 𝑥 < 1
lim 𝑓(𝑥), lim 𝑓(𝑥) does not exist
𝑥→1+
𝑥→1
Let g be defined by 𝑔(𝑥) = {
a. Sketch the graph of g
b. Find the lim 𝑔(𝑥) if it exists
⎸𝑥⎹ 𝑖𝑓 𝑥 ≠ 0
2 𝑖𝑓 𝑥 = 0
𝑥→0
Solutions:
a. Graph
b. The limits are evaluated as follows:
lim 𝑔(𝑥) = lim−(−𝑥) = 0 and
𝑥→0−
𝑥→0
lim 𝑔(𝑥) = lim+(𝑥) = 0
𝑥→0+
𝑥→0
In example 1.4.5, lim− 𝑔(𝑥) = lim+ 𝑔(𝑥). Because the left-hand limit and the right𝑥→0
𝑥→0
hand limits are equal, we say that the two-sided limits, lim 𝑔(𝑥) = 0, exists. Note that
𝑔(0) = 2, which has no effect on lim 𝑔(𝑥).
𝑥→0
𝑥→0
38
Infinite Limits
This section will consider the asymptotes, lines where the graph begins to get closer
and closer. This time we will consider vertical asymptotes.
We previously evaluated the limit of a rational function that leads to
𝑝(𝑥)
0
lim 𝑓(𝑥) = lim 𝑞(𝑥) = 0. We simplified rational functions by factoring and canceling common
𝑥→𝑎
𝑥→𝑎
factors in order to evaluate the limit.
𝑝(𝑥)
Now, consider the case where lim 𝑓(𝑥) = lim 𝑞(𝑥) gives a non-zero constant for 𝑝(𝑎)
𝑥→𝑎
while 𝑞(0) = 0.
Example 1.4.6.
𝑥→𝑎
1
Consider the graph of function 𝑓(𝑥) = 𝑥.
Based on the graph as shown in
illustration 10, we can observe that as x
𝟏
approaches 0 from the left or 𝐥𝐢𝐦− 𝒙 , the graph
𝒙→𝟎
decreases infinitely, or 𝑓 is said to decrease
without bound to the negative infinity
(symbol: − ∞).
Similarly, as x approaches from the right
𝟏
or 𝐥𝐢𝐦+ 𝒙 , the graph increases infinitely, or 𝑓 is
𝒙→𝟎
said to increase without bound to the positive
infinity (symbol: +∞ or simply ∞).
Illustration 10
Table 4
𝐥𝐢𝐦−
𝒙→𝟎
x
𝟏
𝒙
𝐥𝐢𝐦+
𝒙→𝟎
𝑓(𝑥) =
-1
-0.9
-0.5
-0.1
-0.01
-0.001
-0.0001
1
𝑥
x
-1
-1.11
-2
-10
-100
-1,000
-10,000
𝟏
1
0.9
0.5
0.1
0.01
0.001
0.0001
𝟏
𝟏
𝟏
𝒙
𝑓(𝑥) =
1
1.11
2
10
100
1,000
10,000
Since the 𝐥𝐢𝐦− 𝒙 ≠ 𝐥𝐢𝐦+ 𝒙 , 𝐥𝐢𝐦 𝒙 does not exist.
𝒙→𝟎
𝒙→𝟎
𝒙→𝟎
1
𝑥
Our observation is also
verified by the table of values as
shown in table 4, as x approaches 0
from the left, 𝑓 gets negatively large,
and as x approaches 0 from the
right, 𝑓 gets positively large.
𝐥𝐢𝐦
𝟏
𝒙→𝟎− 𝒙
Therefore, we write the
𝟏
= −∞ and 𝐥𝐢𝐦+ 𝒙 = +∞.
𝒙→𝟎
[By the Existence of Limit Theorem]
39
1
Example 1.4.7.
We investigate the function 𝑓(𝑥) = 𝑥 2 .
𝐥𝐢𝐦−
𝒙→𝟎
x
-1
-0.9
-0.5
-0.1
-0.01
-0.001
-0.0001
𝟏
𝒙𝟐
𝐥𝐢𝐦+
𝒙→𝟎
𝑓(𝑥) =
1
x
𝑥2
1
0.9
0.5
0.1
0.01
0.001
0.0001
1
1.23
4
100
10,000
1,000,000
100,000,000
𝟏
𝒙𝟐
𝑓(𝑥) =
1
𝑥2
1
1.23
4
100
10,000
1,000,000
100,000,000
Table 5
Illustration 11
As shown in illustration 11 and table 5, we can observe that as x approaches 0 from
both the left and right sides, the function 𝑓 increases infinitely or increases without bound
to the positive infinity.
1
1
Therefore, we write lim− 𝑥 2 = +∞ and lim+ 𝑥 2 = + ∞.
𝑥→0
𝟏
𝑥→0
𝟏
Since 𝐥𝐢𝐦− 𝒙𝟐 = 𝐥𝐢𝐦+ 𝒙𝟐 = +∞, the limit exists.
𝒙→𝟎
𝒙→𝟎
𝟏
[By the Existence of Limit Theorem]
1
Both 𝑓(𝑥) = 𝒙 and 𝑥 2 resulted in an infinite decrease or increase in values as x
approaches 0. These kinds of limits are called infinite limits. These allow us to obtain the
vertical asymptote of the graph. As the graph gets closer to the vertical asymptote, the
function values increase or decrease infinitely.
This will conclude that:
lim 𝑓(𝑥) = +∞, lim− 𝑓(𝑥) = −∞, lim+ 𝑓(𝑥) = +∞, 𝑜𝑟 lim+ 𝑓(𝑥) = −∞,
𝑥→𝑎−
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
then 𝑥 = 𝑎 is a vertical asymptote of 𝑦 = 𝑓(𝑥).
𝑝(𝑥)
Thus, a vertical asymptote 𝑥 = 𝑎 of a function 𝑓(𝑥) = 𝑞(𝑥) is when 𝑞(𝑎) = 0 and
𝑝(𝑎) ≠ 0.
The tabular and graphical approaches to evaluating infinite limits can show if the
function values increase or decrease infinitely. We only assign very close values to the
limiting number a to show how the function behaves, and we only get either −∞ or +∞.
40
However, there are other ways of determining infinite limits. We only need to remember
that infinite limits occur when the function's denominator is 0 at a and the numerator is
not.
The following theorem and its properties will help us quickly evaluate infinite limits.
Theorem:
Let r be a positive integer.
1. If r is even, then
a.
1
1
b. lim+ 𝑟 𝑥 = + ∞
lim− 𝑥 𝑟 = + ∞
𝑥→ 0
√
𝑥→0
2. If r is odd, then
1
1
a. lim− 𝑥 𝑟 = − ∞
c. lim+ 𝑥 𝑟 = + ∞
1
d. lim+ 𝑟 𝑥 = + ∞
𝑥→0
𝑥→0
1
b. lim− 𝑟 𝑥 = − ∞
𝑥→0
√
𝑥→0
√
Let us evaluate the following limits:
Example 1.4.8.
1
lim − (𝑥+4)3
𝑥→ − 4
Solution:
The limit is infinite because the denominator becomes 0 when 𝑥 = −4. We just need
to determine the sign of infinity. Note that the function value becomes negative when we
approach −4 from the left, say −4.1. Thus:
1
lim
𝑥→−4− (𝑥+4)3
Example 1.4.9.
= −∞
1
lim+ 𝑥 2 −9
𝑥→3
Solution:
1
1
It is essential to rewrite in factored form and simplify as: 𝑥 2 −9 = (𝑥−3)(𝑥+3) where the
denominator is equal to 0 at 𝑥 = 3. We can observe that the function value is positive when
we take a number from the right side of 3, say 3.01. Hence,
lim
1
𝑥→3+ 𝑥 2 −9
1
= lim+ (𝑥−3)(𝑥+3) = +∞
𝑥→3
41
𝑥−2
Example 1.4.10.
lim− 𝑥 2 −4𝑥+4
𝑥→2
Solution:
0
We can observe that when 𝑥 = 2, the function value is 0 which is indeterminate and
unacceptable. This can be remedied by factoring and canceling the same factors such as:
lim
𝑥−2
𝑥→2− 𝑥 2 −4𝑥+4
𝑥−2
1
= lim− (𝑥−2)(𝑥−2) = lim− 𝑥−2
𝑥→2
𝑥→2
Taking a number from the left side of 2, say 1.99, the sign of the function value
becomes negative. Thus
𝑥−2
𝑥−2
1
lim− 𝑥 2 −4𝑥+4 = lim− (𝑥−2)(𝑥−2) = lim− 𝑥−2 = −∞
𝑥→2
𝑥→2
𝑥→2
The following properties of limits will help us evaluate infinite limits involving the
operations on addition, multiplication, and division.
Properties of Infinite Limits
Let a and L be real numbers:
a. Sum
a. 𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝐿, then lim [𝑓(𝑥) + 𝑔(𝑥)] = +∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
b. 𝐼𝑓 lim 𝑓(𝑥) = −∞, and lim 𝑔(𝑥) = 𝐿, then lim [𝑓(𝑥) + 𝑔(𝑥)] = −∞.
𝑥→𝑎
c.
𝑥→𝑎
𝑥→𝑎
𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim ℎ(𝑥) = −∞, then lim [𝑓(𝑥) − ℎ(𝑥)] = +∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
d. If lim 𝑓(𝑥) = +∞, and lim ℎ(𝑥) = −∞, then lim [ℎ(𝑥) − 𝑓(𝑥)] = −∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Note that +∞ − (+∞) and −∞ − (−∞) are said to be in indeterminate forms.
b. Product
a. 𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝐿 > 0, then lim 𝑓(𝑥) ∙ 𝑔(𝑥) = +∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
b. 𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝐿 < 0, then lim 𝑓(𝑥) ∙ 𝑔(𝑥) = −∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
c. 𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim ℎ(𝑥) = +∞, then lim 𝑓(𝑥) ∙ ℎ(𝑥) = +∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
d. 𝐼𝑓 lim 𝑓(𝑥) = +∞, and lim ℎ(𝑥) = −∞, then lim 𝑓(𝑥) ∙ ℎ(𝑥) = −∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
42
e. If lim 𝑓(𝑥) = −∞, and lim ℎ(𝑥) = −∞, then lim 𝑓(𝑥) ∙ ℎ(𝑥) = +∞.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Note that (0)(+∞) and (0)(−∞) are said to be in indeterminate forms.
c. Quotient
𝑔(𝑥)
a. lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝐿, then lim 𝑓(𝑥) = 0.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑔(𝑥)
b. lim ℎ(𝑥) = −∞, and lim 𝑔(𝑥) = 𝐿, then lim ℎ(𝑥) = 0.
𝑥→𝑎
𝑥→𝑎
+∞ +∞ −∞
Note that +∞ , −∞ , +∞ , and
𝑥→𝑎
−∞
−∞
are said to be in indeterminate forms.
Evaluate the following using the properties of infinite limits:
Example 1.4.11.
4
lim (𝑥 − 𝑥−1)
𝑥→1−
Solution:
4
lim− (𝑥 − 𝑥−1) = lim−𝑥 − 4 lim−
𝑥→1
𝑥→1
Example 1.4.12.
𝑥→1
3
1
𝑥−1
= 1 − 4(−∞) = +∞
1
lim ( ) (𝑥 − 2)
𝑥→0+ 𝑥 2
Solution:
3
1
3
1
1
lim+ (𝑥 2 ) (𝑥 − 2) = lim+ (𝑥 2 ) lim+ (𝑥 − 2) = +∞ (− 2) = −∞
𝑥→0
𝑥→0
Example 1.4.13.
lim − (𝑥 2 −16)
𝑥→0
𝑥−3
𝑥→−4
Solution:
𝑥−3
1
lim − (𝑥 2 −16) = lim −(𝑥 − 3) ∙ lim − (𝑥 2 −16) = (−7)(+∞) = −∞
𝑥→−4
𝑥→−4
𝑥→−4
Note to Self:
This is a safe space. You can write
anything here.
43
Limits at Infinity
1
Consider the function 𝑓(𝑥) = 𝑥
We will evaluate the function using numbers that become extremely small and
1
extremely large. The following illustration and table show the computed values of 𝑓(𝑥) = 𝑥.
𝟏
𝒙→−∞ 𝒙
𝟏
𝒙→+∞ 𝒙
𝐥𝐢𝐦
𝐥𝐢𝐦
1
𝑥
𝑥 → −∞
𝑓(𝑥) =
-1
-10
-100
-1,000
-10,000
-1
-0.1
-0.01
-0.001
-0.0001
-0.00001
-0.000001
-100,000
-1,000,000
𝑥 → +∞
1
10
100
1,000
10,000
100,000
1,000,000
𝑓(𝑥) =
1
0.1
0.01
0.001
0.0001
0.00001
0.000001
Table 6
Illustration 12
Table 6 shows that when x gets extremely small, the function values become closer
and closer to 0. Similarly, when x gets extremely large, the function values become closer and
closer to 0. We can therefore conclude that the limit of
of
1
𝑥
as 𝑥 → −∞ is 0 and the limit
as 𝑥 → +∞ is also 0.
Illustration 12 also shows that as x approaches negative infinity, the graph of 𝑓(𝑥) =
gets closer and closer to the x-axis where y = 0. In like manner, as x approaches positive
1
infinity, the graph of 𝑓(𝑥) = 𝑥 still gets closer and closer to the x-axis. Hence,
lim
1
𝑥→−∞ 𝑥
= 0 and lim
1
𝑥→+∞ 𝑥
=0
This type of limit is called limits at infinity. This leads us to the following theorem.
Theorem:
Let k be any real number and r be any positive rational number, then
1.
2.
lim
𝑘
𝑥→+∞ 𝑥 𝑟
lim
𝑘
𝑥→−∞ 𝑥 𝑟
1
𝑥
= 0.
= 0, provided 𝑥 𝑟 is defined when using 𝑥 < 0.
44
Evaluate the following limits:
Example 1.4.14.
lim
4𝑥 2 +2𝑥−1
𝑥→+∞ 2𝑥 2 +3𝑥+1
Solution:
To apply the theorem, we must rewrite the terms of the polynomial in the numerator
𝑘
and denominator in the form 𝑥 𝑟 . To do this, we will divide each term, both in the numerator
and denominator by 𝑥 2 . (Trick: Divide all terms by the highest degree of x in the denominator).
Applying the limits at infinity,
Therefore,
Illustration 13
Note that we will get the same result of 2 when we change 𝑥 → +∞ to 𝑥 → −∞. When
x increases or decreases without bound, the function values get closer and closer to 2. This
is also evident in illustration 13.
As shown in illustration 13, the graph is asymptotic to the line 𝑦 = 2 which is the
horizontal asymptote. This clearly shows that we can associate limits at infinity with
horizontal asymptotes. So, the line 𝑦 = 𝐿 is a horizontal asymptote of the graph 𝑦 = 𝑓(𝑥)
whenever lim 𝑓(𝑥) = 𝐿 or lim 𝑓(𝑥) = 𝐿.
𝑥→−∞
x→+∞
4𝑥 2 +2𝑥−1
We also say that 𝑦 = 2 is the horizontal asymptote of the function 𝑓(𝑥) = 2𝑥 2 +3𝑥+1.
45
Example 1.4.15.
lim
𝑥→−∞
𝑥 2 −𝑥−2
𝑥 3 +𝑥 2 +𝑥−4
Solution:
Applying the theorem on limits at infinity,
Therefore,
. We also say that 𝑦 = 0 is a horizontal asymptote
𝑥 2 −𝑥−2
of 𝑓(𝑥) = 𝑥 3+𝑥 2 +𝑥−4.
Example 1.4.16.
lim
𝑥−2
𝑥→∞ √2𝑥 2 −𝑥+1
Solution:
Therefore,
.
46
Infinite Limits at Infinity
This section deals with functions that increase or decrease without bound when x
approaches the positive or negative infinity. The functions below illustrate this behavior.
Example 1.4.16.
Consider the function 𝑓(𝑥) = 𝑥 2 + 1, a parabola that opens upward.
Determine the following: (a) lim (𝑥 2 + 1) and (b) lim (𝑥 2 + 1) .
𝑥→−∞
𝑥→+∞
From illustration 14, we noticed that as
the values of x get extremely small (𝑥 → −∞),
the values of 𝑦 increase without bound to the
positive infinity.
The same observation shows that as the
values of x get extremely large (𝑥 → +∞), the
values of 𝑦 also increase without bound to
the positive infinity.
Now, we can say that for the function
𝑓(𝑥) = 𝑥2 + 1, as 𝑥 → −∞, 𝑦 → +∞, and as
𝑥 → +∞, 𝑦 → +∞. Thus, we can say that:
a)
lim (𝑥 2 + 1) = +∞
and
𝑥→−∞
Illustration 14
b) lim (𝑥 2 + 1) = −∞.
𝑥→+∞
Example 1.4.17.
Given the function 𝑓(𝑥) = 𝑥 3 . Determine the following: (a) lim 𝑥 3
𝑥→−∞
and (b) lim 𝑥 3 .
𝑥→+∞
From illustration 15, we observed that
as x increases without bound, the values of f
also increase without bound. Similarly, as x
decreases without bound, the values of f also
decrease without bound. Thus, we say that:
a)
b)
lim 𝑥 3 = −∞
𝑥→ −∞
and
lim 𝑥 3 = +∞.
𝑥→ +∞
All the above examples illustrate what we
call infinite limits at infinity. To summarize,
we have the following properties.
Illustration 15
47
Properties of Infinite Limits at Infinity
If the values of 𝑓(𝑥) become extremely large when x approaches positive or negative
infinity, then we write
lim 𝑓(𝑥) = +∞ or lim 𝑓(𝑥) = +∞, respectively.
𝑥→ +∞
𝑥→ −∞
If the values of 𝑓(𝑥) become extremely small when x approaches positive or negative
infinity, then we write
lim 𝑓(𝑥) = −∞ or lim 𝑓(𝑥) = −∞, respectively.
𝑥→ +∞
𝑥→ −∞
Example 1.4.18.
lim
𝑥 2 − 3𝑥
𝑥→ −∞ 1 + 2𝑥
Solution:
Divide both the numerator and denominator by x, we get:
Thus, the
.
The result shows the limit is infinity. It can be verified by assigning a very small
number to x, and we get the negative infinity limit.
Note to Self:
This is a safe space. You can write
anything here.
48
Practice #4
Try this! ☺
A. Use the graph to determine the following limits.
a.
lim 𝑓(𝑥)
g.
lim 𝑓(𝑥)
h.
lim 𝑓(𝑥)
i.
d. lim+ 𝑓(𝑥)
j.
e.
k. 𝑓(1)
b.
c.
𝑥→−3−
𝑥→−3+
𝑥→−3
𝑥→1
f.
lim 𝑓(𝑥)
lim 𝑓(𝑥)
𝑥→3−
lim 𝑓(𝑥)
𝑥→3+
lim 𝑓(𝑥)
𝑥→3
𝑓(−3)
𝑥→1−
lim 𝑓(𝑥)
l.
𝑓(3)
𝑥→1
B. Evaluate the following:
1.
2.
3.
lim+
𝑥 2 − 4𝑥 + 3
𝑥→2
lim
𝑥 2 −3𝑥+2
5𝑥 3 −2𝑥 2 −1
𝑥→∞ 𝑥 3 −𝑥 + 1
lim (4𝑥 2 + 𝑥 + 1)
𝑥→ +∞
Process Questions
1. When we conclude that the limit of a certain function is −∞ or +∞, does the limit
really exist? Explain.
0 ∞
2. Is it meaningful or possible to operate infinities such as: 0, ∞, or ∞ − ∞? Explain
your answer for each of the three situations mentioned.
Student’s responses:
Please write your answers on page 159.
49
My Working Space
(write your solutions to the practice problems here)
50
Lesson 1.5 Limits of Trigonometric Functions
Trigonometric functions are as important as algebraic functions. Like in the previous
sections, we can also use graphical and tabular methods to find the limits of trigonometric
functions. However, we focus our attention on using easier but more accurate ways of
finding those limits, that is, by using established results or theorems. We also take advantage
of the fact that the basic forms of trigonometric functions are continuous at every point of
their respective domains (Egarguin et al., 2017).
Let us first consider evaluating the limit of a trigonometric function by using a table
of values.
Example 1.5.1.
Evaluate lim sin 𝑥
𝑥→0
Solution:
We will construct the table of values for 𝑓(𝑥) = sin 𝑥. We first approach 0 from the
left or through the values less than but close to 0.
𝑥
𝑓(𝑥)
−1
−0.8414709848
−0.5
−0.4794255386
−0.1
−0.09983341665
−0.01
−0.00999983333
−0.001
−0.00099999983
−0.0001
−0.00009999999
−0.00001
−0.00000999999
Now we consider approaching 0 from its right or through values greater than but
close to 0.
𝑥
𝑓(𝑥)
1
0.8414709848
0.5
0.4794255386
0.1
0.09983341665
0.01
0.00999983333
0.001
0.00099999983
0.0001
0.00009999999
0.00001
0.00000999999
51
As the values of x get closer and closer to 0, the values of 𝑓(𝑥) get closer and closer to
0. In symbols, 𝐥𝐢𝐦 𝐬𝐢𝐧 𝒙 = 𝟎.
𝒙→𝟎
We can also find lim sin 𝑥 by using the graph of the sine function. Consider the graph
𝑥→0
of 𝑓(𝑥) = sin 𝑥 as shown in illustration 16.
Illustration 16
The graph validates our observation in example 1.5.1 that lim sin 𝑥 = 0. Also, using
𝑥→0
the graph, we have the following:
(𝑎) lim𝜋 sin 𝑥 = 1
𝑥→
(𝑐) lim𝜋 sin 𝑥 = −1
𝑥→−
2
(𝑏) lim sin 𝑥 = 0
2
(𝑑) lim sin 𝑥 = 0
𝑥→𝜋
𝑥→− 𝜋
Now recall the six basic trigonometric functions and their respective domains.
Definition:
The following are the six basic trigonometric functions and their domains:
(𝑖) 𝑠𝑖𝑛𝑒:
𝑓(𝑥) = sin 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ
(𝑖𝑖) 𝑐𝑜𝑠𝑖𝑛𝑒:
𝑓(𝑥) = cos 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ
(𝑖𝑖𝑖) 𝑐𝑜𝑠𝑒𝑐𝑎𝑛𝑡 ∶
𝑓(𝑥) = csc 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ − {𝑘𝜋|𝑘 ∈ ℤ}
(2𝑘+1)𝜋
(𝑖𝑣) 𝑠𝑒𝑐𝑎𝑛𝑡 ∶
𝑓(𝑥) = sec 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ − { 2 |𝑘 ∈ ℤ}
(𝑣) 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 ∶
(𝑣𝑖) 𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 ∶
(2𝑘+1)𝜋
𝑓(𝑥) = tan 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ − { 2 |𝑘 ∈ ℤ}
𝑓(𝑥) = cot 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ∈ ℝ − {𝑘𝜋|𝑘 ∈ ℤ}
Note: ℝ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑛𝑑
ℤ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠
52
Observe that the six trigonometric functions are continuous everywhere in their
respective domains. This result is formally stated in the following theorem.
Theorem:
Sine, cosine, secant, cosecant, tangent, and cotangent functions are continuous at
every point in their respective domains.
The previous theorem has a significant impact on finding the limits of trigonometric
functions. This implication is stated in the following theorem.
Theorem:
Let a be contained in the domain of the function whose limit is being evaluated. Then:
(𝑖)
(𝑖𝑖)
lim sin 𝑥 = sin 𝑎
𝑥→𝑎
lim cos 𝑥 = cos 𝑎
𝑥→𝑎
(𝑖𝑖𝑖) lim sec 𝑥 = sec 𝑎
𝑥→𝑎
(𝑖𝑣) lim csc 𝑥 = csc 𝑎
(𝑣)
𝑥→𝑎
lim tan 𝑥 = tan 𝑎
𝑥→𝑎
(𝑣𝑖) lim cot 𝑥 = cot 𝑎
𝑥→𝑎
Example 1.5.2.
Evaluate the limit of the following if it exists.
(𝑎) lim sin 𝑥
𝑥→0
𝑥→
(𝑏) lim𝜋 cos 𝑥
𝑥→−
(𝑔) lim5𝜋 tan 𝑥
4
2
(ℎ) lim𝜋
𝑥→
6
2 sin 𝑥
√3 sec 𝑥−𝑐𝑜𝑡 2 𝑥
(𝑐) lim tan 𝑥
𝑥→ 𝜋
(𝑑) lim𝜋 cot 𝑥
𝑥→−
3
(𝑒) lim sec 𝑥
𝑥→2𝜋
(𝑓)
lim 𝜋 csc 𝑥
𝑥→ −
6
53
Solution:
(a) lim sin 𝑥 = sin 0 = 0
𝑥→0
𝜋
(𝑏) lim𝜋 cos 𝑥 = cos (− ) =
4
𝑥→−
√2
2
4
sin 𝜋
0
(𝑐) lim tan 𝑥 = tan 𝜋 =
= −1 = 0
cos 𝜋
𝑥→ 𝜋
𝜋
(𝑑) lim𝜋 cot 𝑥 = cot (− ) =
3
𝑥→−
3
𝜋
3
𝜋
sin(− )
3
cos(− )
1
1
2
√3
−
2
=
=−
1
√3
=−
√3
3
1
(𝑒) lim sec 𝑥 = sec 2𝜋 =
=1=1
cos 2𝜋
𝑥→2𝜋
(𝑓)
𝜋
lim 𝜋 csc 𝑥 = csc (− 6 ) =
𝑥→ −
6
(𝑔) lim5𝜋 tan 𝑥 = tan
𝑥→
2
2
(ℎ) lim𝜋
𝑥→
5𝜋
6
=
5𝜋
)
2
5𝜋
cos( )
2
sin(
√3
𝜋
6
=
1
−
1
2
= −2
1
= 0 , 𝐷𝑜𝑒𝑠 𝑁𝑜𝑡 𝐸𝑥𝑖𝑠𝑡 (𝐷𝑁𝐸) 𝑜𝑟 𝑁𝑜 𝐿𝑖𝑚𝑖𝑡
𝜋
6
(2)(sin )
2 sin 𝑥
=
sec 𝑥−𝑐𝑜𝑡 2 𝑥
1
sin(− )
𝜋
𝜋 2
(√3)(sec )−(cot )
6
6
=
1
2
(2)( )
(√3)(
2
2
)−(√3)
√3
1
= 2−3 = −1
Evaluating the limits of trigonometric functions is simply finding the trigonometric
function values at 𝑥 = 𝑎. But there are two essential limits involving trigonometric functions
that cannot be evaluated using the same technique. These are:
𝐬𝐢𝐧 𝒙
𝟏 − 𝐜𝐨𝐬 𝒙
𝒂𝒏𝒅 𝐥𝐢𝐦
𝒙→𝟎 𝒙
𝒙→𝟎
𝒙
𝐥𝐢𝐦
Example 1.5.3.
Evaluate lim
sin 𝑥
𝑥
𝑥→0
Solution:
sin 𝑥
We will construct the table of values for 𝑓(𝑥) = 𝑥 . We first approach the number
0 from the left or through values less than but close to 0.
𝑥
𝑓(𝑥)
−1
0.8414709848
−0.5
0.9588510772
−0.1
0.9983341665
54
−0.01
0.9999833334
−0.001
0.9999998333
−0.0001
0.9999999983
Now we consider approaching 0 from the right or through values greater than but
close to 0.
Since lim−
𝑥→0
Example 1.5.4.
sin 𝑥
𝑥
𝑥
𝑓(𝑥)
1
0.8414709848
0.5
0.9588510772
0.1
0.9983341665
0.01
0.9999833334
0.001
0.9999998333
0.0001
0.9999999983
and lim+
𝑥→0
sin 𝑥
𝑥
Evaluate lim
are both equal to 1, we conclude that lim
𝑥→0
sin 𝑥
𝑥
= 1.
1−cos 𝑥
𝑥→0
𝑥
Solution:
1−cos 𝑥
We will construct the table of values for 𝑔(𝑥) = 𝑥 . We first approach the number
0 from the left or through the values less than but close to 0.
𝑥
𝑓(𝑥)
−1
−0.4596976941
−0.5
−0.2448348762
−0.1
−0.04995834722
−0.01
−0.0049999583
−0.001
−0.0004999999
−0.0001
−0.00005
Now we consider approaching 0 from the right or through values greater than but
close to 0.
𝑥
𝑓(𝑥)
−1
0.4596976941
55
Since lim−
1−cos 𝑥
𝑥
𝑥→ 0
−0.5
0.2448348762
−0.1
0.04995834722
−0.01
0.0049999583
−0.001
0.0004999999
−0.0001
0.00005
= 0 and lim+
1−cos 𝑥
𝑥
𝑥→ 0
= 0, we conclude that lim
𝑥→ 0
1−cos 𝑥
𝑥
= 0.
Based on those mentioned earlier, it will lead us to the following theorem.
Theorem:
sin 𝑥
=1
𝑥→0 𝑥
(𝑖) lim
1 − cos 𝑥
=0
𝑥→0
𝑥
(𝑖𝑖) lim
Example 1.5.5.
Evaluate the limits of the following if it exists.
3 sin 2𝑥
𝑥→0
𝑥
(𝑎) lim
sin 3𝑥
(𝑏) lim
sin 5𝑥
𝑥→0
(𝑐) lim
cos 𝑥−1
𝑥→0 2 sin 𝑥
(𝑑) lim cot 𝑥 sin 2𝑥
𝑥→0
(𝑒) lim
2𝑡𝑎𝑛2 𝑥
𝑥→0
𝑥2
Solution:
(𝑎) Let 𝑦 = 2𝑥. 𝑇ℎ𝑒𝑛 𝑦 → 0 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 → 0. 𝑆𝑜 𝑤𝑒 ℎ𝑎𝑣𝑒:
3 sin 2𝑥
3 sin 𝑦
6 sin 𝑦
sin 𝑦
) = (6)(1) = 6
= lim 𝑦 = lim
= (6) (lim
𝑥→0
𝑦→0
𝑦→0
𝑦→0 𝑦
𝑥
𝑦
2
lim
Thus, lim
𝑥→0
3 sin 2𝑥
𝑥
= 6.
56
sin 3𝑥
(𝑏) lim
=
sin 5𝑥
𝑥→0
sin 3𝑥
lim sin 5𝑥 =
𝑥→0
sin 3𝑥
sin 3𝑥
)
3𝑥
sin 5𝑥
lim 5(
)
5𝑥
𝑥→0
lim 3(
𝑥→0
. 𝐴𝑠 𝑥 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 𝑧𝑒𝑟𝑜, 𝑠𝑜 𝑑𝑜 3𝑥 𝑎𝑛𝑑 5𝑥. 𝐻𝑒𝑛𝑐𝑒,
sin 3𝑥
)
3𝑥
sin 5𝑥
5 lim (
)
𝑥→0 5𝑥
3 lim (
𝑥→0
(3)(1)
3
= (5)(1) = 5
3
Thus, lim sin 5𝑥 = 5.
𝑥→0
(𝑐) lim
cos 𝑥−1
𝑥→0 2 sin 𝑥
Thus, lim
= lim
−(1−cos 𝑥)
2 sin 𝑥
𝑥→0
cos 𝑥−1
𝑥→0 2 sin 𝑥
= − lim
𝑥→0
1−cos 𝑥
𝑥
2 sin 𝑥
𝑥
lim
1−cos 𝑥
𝑥
= − 𝑥→0 2 sin
𝑥 = −
lim
𝑥→0
𝑥
0
0
sin 𝑥
2 lim
𝑥→0 𝑥
= − 2(1) = 0
= 0.
(𝑑) lim cot 𝑥 sin 2𝑥 = lim
sin 2𝑥
𝑥→0 tan 𝑥
𝑥→0
=
= lim
sin 2𝑥
sin 𝑥
cos 𝑥
𝑥→0
sin 2𝑥
lim
𝑥→0 2𝑥
sin 𝑥
1
(lim
)(lim
)
𝑥→0 2𝑥
𝑥→0cos 𝑥
=
= lim
sin 2𝑥
1
𝑥→0 (sin 𝑥)(cos 𝑥)
1
1
sin 𝑥
1
( lim
)(
)
2 𝑥→0 𝑥
cos 0
=
= lim
𝑥→0
1
1
1
( )(1)( )
2
1
sin 2𝑥
2𝑥
sin 𝑥
1
(
)(
)
2𝑥
cos 𝑥
=
1
1
2
=2
Thus, lim cot 𝑥 sin 2𝑥 = 2.
𝑥→0
(𝑒) lim
2𝑡𝑎𝑛2 𝑥
𝑥→0
𝑥2
2(sin 𝑥)2
= lim 𝑥 2 (cos 𝑥)2 = lim (
𝑥→0
𝑥→0
2
sin 𝑥 2
𝑥
2
) ((cos 𝑥)2) = (lim
𝑥→0
sin 𝑥 2
𝑥
2
) (lim (cos 𝑥)2)
𝑥→0
2
= (1)2 ((cos 0)2) = (1) (12 ) = (1)(2) = 2
Thus, lim
𝑥→0
2𝑡𝑎𝑛2 𝑥
𝑥2
= 2.
Note to Self:
This is a safe space. You can write
anything here.
57
Practice #5
Try this! ☺
Evaluate the limit of the following if it exists.
1. lim𝜋 sin 𝑥
𝑥→
6
2. lim𝜋 cot 𝑥
𝑥→
4
3. lim𝜋 𝑠𝑒𝑐 2 𝑥
𝑥→
3
4. lim𝜋(sin 𝑥 + cos 𝑥)
𝑥→
3
5. lim𝜋(sec 𝑥 + tan 𝑥)
𝑥→
2
6. lim
sin 𝑥
𝑥→0 3𝑥
7. lim
2 cos 𝑥−2
𝑥
𝑥→0
8. lim
cos 𝑥
𝑥→0
𝑥
9. lim 2𝑥 cot 2𝑥
𝑥→0
10. lim
𝑥→0
tan 𝑥
𝑥
Process Questions
1. Is there a limit involving trigonometric functions that will not exist? Cite an
example.
2. Is there any case where the limit of trigonometric functions can be determined by
observing some values of the function or its graph? Cite an example.
Student’s responses:
Please write your answers on page 160.
Worksheet #2: (See page 153)
58
My Working Space
(write your solutions to the practice problems here)
59
60
Lesson 1.6 Continuity at a Point
What comes to your mind when you hear or read the word “continuous”? Perhaps you
would think that it does not stop, or in the context of machines, it is an activity or operation
that is not interrupted. In the language of mathematics, it means the same thing. We can say
that a given function is continuous at a point or an interval if we can trace the graph of that
function that passes through a given point without lifting our pencil from the paper.
Examine the following:
lim 𝑓(𝑥) exists and
𝑓(𝑎) is not defined.
The graph has a
hole at 𝑥 = 𝑎. The
graph is
discontinuous.
𝑓(𝑎) is defined but
lim 𝑓(𝑥) does not
𝑥→𝑎
𝑓(𝑎) is defined but
lim 𝑓(𝑥) ≠ 𝑓(𝑎). The
𝑥→𝑎
exist. The graph is
discontinuous.
𝑥→𝑎
graph has a hole at
𝑥 = 𝑎. The graph is
discontinuous.
lim 𝑓(𝑥) does not
𝑥→𝑎
exist. The graph is
discontinuous.
lim 𝑓(𝑥) exists and
𝑥→𝑎
𝑓(𝑎) is defined;
lim 𝑓(𝑥) = 𝑓(𝑎). The
𝑥→𝑎
graph is continuous.
Note. From Definition of Continuity, 2022, Math Warehouse
(https://www.mathwarehouse.com/calculus/continuity/continuity-definitions.php).
Copyright Math Warehouse.
The illustrated graphs provide the definition of a continuous function.
61
Definition
A function 𝑓 is said to be continuous at 𝑥 = 𝑎 when all the following conditions are met:
(i) 𝑓(𝑎) is defined
(ii) lim 𝑓(𝑥) exists
(iii) lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
𝑥→𝑎
Otherwise, function 𝑓 is discontinuous at 𝑥 = 𝑎.
Types of Discontinuity
a) Essential (Nonremovable)
This type of discontinuity occurs when function 𝑓 is discontinuous at 𝑎 and condition (ii)
is not satisfied.
b) Removable
This type of discontinuity occurs when function 𝑓 satisfies condition (ii) but fails to satisfy
either (i) or (iii). Given this type of discontinuity, we can remove the discontinuity by
redefining function 𝑓.
Example 1.6.1.
Determine whether 𝑓(𝑥) = 𝑥 2 + 𝑥 + 1 is continuous at 𝑎 = 1. If it is
discontinuous, determine the type of discontinuity.
Solution:
First, we need to evaluate 𝑓 at 𝑎 = 1. That is,
𝑓(1) = 12 + 1 + 1
𝒇(𝟏) = 𝟑
Therefore, 𝑓(𝑎) exists.
Now, we will determine if lim 𝑓(𝑥) exists.
𝑥→𝑎
lim 𝑓(𝑥) = 12 + 1 + 1
𝑥→1
𝐥𝐢𝐦 𝒇(𝒙) = 𝟑
𝒙→𝟏
After doing the previous two procedures, we will compare whether lim 𝑓(𝑥) = 𝑓(𝑎).
𝑥→𝑎
Since both yield the same values, which is 3, we can say that 𝒇(𝒙) = 𝒙𝟐 + 𝒙 + 𝟏 is
continuous at 𝒂 = 𝟏.
62
2−𝑥
Example 1.6.2.
Determine whether 𝑔(𝑥) = 𝑥+1 is continuous at 𝑎 = 2. If it is
discontinuous, determine the type of discontinuity.
Solution:
Evaluate 𝑔 at 𝑎 = 2. That is,
𝑔(2) =
2−2
2+1
𝒈(𝟐) = 𝟎
Therefore, 𝑔(𝑎) exists.
Determine if lim 𝑔(𝑥) exists.
𝑥→𝑎
lim 𝑔(𝑥) =
𝑥→2
2−2
2+1
𝐥𝐢𝐦 𝒈(𝒙) = 𝟎
𝒙→𝟏
Since lim 𝑔(𝑥) = 𝑔(𝑎) we can say that 𝒈(𝒙) =
𝑥→𝑎
𝟐−𝒙
𝒙+𝟏
is continuous at 𝒂 = 𝟐.
𝑥 2 +2𝑥−3
Example 1.6.3.
Determine whether ℎ(𝑥) =
discontinuous, determine the type of discontinuity.
𝑥−1
is continuous at 𝑎 = 1. If it is
Solution:
Clearly, at 𝑎 = 1, function ℎ is not defined, but we can reconstruct our function to
evaluate its limit later. Now,
ℎ(𝑥) =
ℎ(𝑥) =
𝑥 2 + 2𝑥 − 3
𝑥−1
(𝑥 − 1)(𝑥 + 3)
𝑥−1
ℎ(𝑥) = 𝑥 + 3
At 𝑎 = 1, we have ℎ(1) = 1 + 3 = 4. Evaluating its limit, we have lim ℎ(𝑥) = 1 + 3 =
𝑥→1
4. So, we can say that the given function ℎ has a removable discontinuity. Therefore, we
can redefine the function to
𝑥 2 + 2𝑥 − 3
,
𝐻(𝑥) = { 𝑥 − 1
4
,
𝑥≠1
𝑥=1
63
1
Example 1.6.4.
Determine whether 𝑘(𝑥) = 𝑥 2 is continuous at 𝑎 = 0. If it is
discontinuous, determine the type of discontinuity.
Solution:
Clearly, at 𝑎 = 0, function 𝑘 is not defined. If we are going to evaluate lim 𝑘(𝑥), we can
𝑥→0
observe that the limit does not exist, and there is no such way to redefine function 𝑘 to make
1
it continuous. Therefore, we can say that 𝑘(𝑥) = 𝑥 2 has an essential discontinuity.
Practice #6
Try this! ☺
Determine if each of the given functions is continuous at 𝑎. If it is discontinuous, state
which of the three conditions is/are not met and determine what type of discontinuity is
involved.
2𝑥
1. 𝑓(𝑥) = 1−𝑥 2 ; 𝑎 = −1
2. 𝑔(𝑥) = −𝑥 2 + 3𝑥 − 4; 𝑎 = 4
Process Questions
1. By just looking at the graph of any function, can you really tell whether continuity
or discontinuity exists? Explain.
2. Do you think all quadratic functions are continuous? Why?
Student’s responses:
Please write your answers on page 160.
Note to Self:
This is a safe space. You can write
anything here.
64
My Working Space
(write your solutions to the practice problems here)
65
Lesson 1.7 Continuity on an Interval
Since we have already discussed the continuity of a function at a given point, we will
now expand what we have learned so far in the previous lesson to discuss the continuity of
a function on a given interval.
Consider the graph shown in illustration 17.
Illustration 17
The given graph above is the graph of 𝑓(𝑥) = √4 − 𝑥 2 . Clearly, we can say that 𝑓 is
defined at −2 < 𝑥 < 2; and from this, 𝑓 is said to be continuous for all 𝑥 on the open interval
(−2,2) for which 4 − 𝑥 2 > 0. We will also check whether 𝑓 is still continuous at its endpoints
−2 and 2 to eventually show that it is continuous on the closed interval [−2,2].
i) lim + 𝑓(−2) = √4 − 𝑥 2 = 0
𝑥→−2
ii) lim− 𝑓(2) = √4 − 𝑥 2 = 0
𝑥→2
Since the limit exists for both lim + 𝑓(−2) and lim − 𝑓(2), we can say that 𝑓 is
𝑥→−2
continuous on the closed interval [−2,2].
𝑥→−2
66
Definition
If 𝑓 is continuous at every point on an open interval (𝑎, 𝑏), we say that 𝒇 is continuous on
(𝒂, 𝒃). On the other hand, we say that 𝒇 is continuous on the closed interval [𝒂, 𝒃], if 𝑓 is
continuous on the open interval (𝑎, 𝑏) and
lim 𝑓(𝑥) = 𝑓(𝑎) and
𝑥→𝑎+
lim 𝑓(𝑥) = 𝑓(𝑏)
𝑥→𝑏 −
Note: If 𝑓 is continuous on any values of 𝑥 on the given interval (−∞, ∞), we just say that 𝑓 is
continuous.
Example 1.7.1.
Determine whether 𝑓(𝑥) = 3𝑥+1 is continuous at [−1,3].
Solution:
By looking at the given function, we can say that 𝑓 is defined on whatever values of 𝑥
on the given closed interval [−1,3]. Therefore, we can say that 𝒇 is continuous at [−𝟏, 𝟑].
Example 1.7.2.
Determine whether 𝑔(𝑥) =
1
𝑥−2
is continuous at [1,4].
Solution:
Function 𝑔 is a rational function that is not defined at 𝑥 = 2. Since 2 is included on the
interval [1,4], it implies that 𝑔 is not continuous on the interval (1,4) as well as [1,4].
Therefore, we can say that 𝒈 is not continuous at [𝟏, 𝟒].
Example 1.7.3.
1
Determine whether ℎ(𝑥) = 𝑥−2 is continuous at [1,2).
Solution:
The given function ℎ is not defined at 𝑥 = 2. Now, we will check whether it is
continuous on the open interval (1,2). Clearly, we can observe that ℎ is continuous on
whatever values of 𝑥 on the open interval (1,2).
Based on the definition of the continuity of a function on an interval, we are going to
show that ℎ(1) = lim+ ℎ(𝑥).
𝑥→1
1
i) ℎ(1) = 1−2 = −1
1
ii) lim+ ℎ(𝑥) = 1−2 = −1
𝑥→1
Since ℎ(1) = lim+ ℎ(𝑥), we can say that 𝒉 is continuous at [𝟏, 𝟐).
𝑥→1
67
Example 1.7.4.
4 − 𝑥2 , 𝑥 < 1
, 𝑥 = 1 is continuous at (−2,5].
Determine whether 𝑘(𝑥) = { 2
√𝑥 − 1 , 𝑥 > 1
Solution:
We need to investigate whether 𝑘 is continuous on the open interval (−2,5). To do
this, using our previous knowledge on determining the continuity of a function at a given
point, let us investigate first whether 𝑘 is continuous at a given value of 𝑎 (which in this case
is 𝑥 = 1).
Recall that we need to satisfy the following:
(i) 𝑘(𝑎) is defined
(ii) lim 𝑘(𝑥) exists
(iii) lim 𝑘(𝑥) = 𝑘(𝑎)
𝑥→𝑎
𝑥→𝑎
Now, given 𝑘(𝑎) = 𝑘(1) = 2, we can say that 𝑘(𝑎) is defined. The next thing we need
to do is evaluate whether the limit exists from both sides of 𝑎 = 1.
lim 𝑘(𝑥) = 4 − 𝑥 2 = 4 − 12 = 3
𝑥→1−
lim 𝑘(𝑥) = √1 − 1 = 0
𝑥→1+
From the solution above, we can observe that 𝑘(𝑎) ≠ lim 𝑘(𝑥) because the limit does
𝑥→1
not exist at 𝑥 = 1. Since 1 is included in the interval (−2,5], therefore, we can conclude that
𝒌 is not continuous at (−𝟐, 𝟓].
Example 1.7.5.
Determine whether 𝑀(𝑥) = {
𝑥 + 3, 𝑥 < 2
is continuous at [1,2].
𝑥2, 𝑥 ≥ 2
Solution:
We need to investigate whether 𝑀 is continuous on the open interval (1,2). Since the
given function has polynomial sub-functions, we can say that it is defined on the open
interval (1,2). Now, we are going to check whether lim+ 𝑀(𝑥) = 𝑀(𝑎) and lim− 𝑀(𝑥) =
𝑥→1
𝑥→2
𝑀(𝑏).
𝑀(𝑎) = 𝑀(1) = 1 + 3 = 4
lim 𝑀(𝑥) = 𝑥 + 3 = 1 + 3 = 4
𝑥→1+
Therefore, lim+ 𝑀(𝑥) = 𝑀(𝑎)
𝑥→1
𝑀(𝑏) = 𝑀(2) = 22 = 4
lim 𝑀(𝑥) = 2 + 3 = 5
𝑥→2−
Therefore, lim+ 𝑀(𝑥) ≠ 𝑀(𝑏)
𝑥→1
68
Since we have found out that lim+ 𝑀(𝑥) ≠ 𝑀(𝑏), we can say that 𝑴(𝒙) =
𝑥→1
𝒙 + 𝟑, 𝒙 < 𝟐
{
is not continuous at [𝟏, 𝟐].
𝒙𝟐 , 𝒙 ≥ 𝟐
Example 1.7.6.
Determine the intervals on which 𝑓(𝑥) = √𝑥 + 3 is continuous.
Solution:
The given function will be continuous if and only if 𝑥 + 3 ≥ 0. Now, let us solve the
inequality and determine the interval that will make 𝑓 continuous.
𝑥+3≥0
𝑥 ≥ −3
Therefore, we can say that 𝒇(𝒙) = √𝒙 + 𝟑 is continuous on the interval [−𝟑, ∞).
Practice #7
Try this! ☺
Determine if the given function is continuous on each of the given intervals.
3𝑥 − 1
,
𝑥 ≤ −1
2
,
−1 < 𝑥 < 1
𝑓(𝑥) = {𝑥 + 5𝑥
3
3𝑥
,
𝑥≥1
1. (−∞, −1]
2. [−1,0]
3. [−1,2)
4. [−1,1]
5. [1, ∞)
Process Questions
1. What do you need to be mindful of when determining the continuity of a rational
function on a given interval? Why?
2. What do you need to be mindful of when determining the continuity of a piecewise
function on a given interval? Why?
Student’s responses:
Please write your answers on page 161.
Worksheet #3: (See page 154)
69
My Working Space
(write your solutions to the practice problems here)
70
Unit 2: Derivatives
71
72
Lesson 2.1 The Tangent Line and the Derivative
You have heard of the words like secant line and tangent line during your discussion
in Geometry. As shown in illustration 18, a secant line to a curve is determined by two points
on the curve (the blue line), while a tangent line (red line) is a line that touches the curve
once.
Illustration 18
Recalling what you have learned about slope (𝑚) of the line containing the points
(𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ), the slope of the line is obtained by
𝑚=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
Some call it the “change in y over change in x” or the “rise over run.”
Now, consider the graph containing the points 𝑃 (𝑥, 𝑓(𝑥)) and 𝑄 (𝑥 + ℎ, 𝑓(𝑥 + ℎ))
as shown in illustration 19.
(𝑥, 𝑓(𝑥))
Illustration 19
The slope of the line 𝑃𝑄 can be solved as
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒚
𝒈
𝒎𝑷𝑸 = 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒙 = 𝒉 =
𝒇(𝒙+𝒉)−𝒇(𝒙)
(𝒙+𝒉)−𝒙
=
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝒉
.
73
As point 𝑄 moves closer and closer to point 𝑃, as shown in illustration 20, the value
of ℎ gets smaller and smaller.
Illustration 20
If you move Q all the way to touch P, then the secant line will become a tangent line,
and you can now compute for the slope of the tangent line.
It brings us to the following definition:
Limit Definition of the slope of a Tangent Line to a Given Curve
Given a point 𝑃(𝑎, 𝑓(𝑎)) on the graph of the function with the equation 𝑦 = 𝑓(𝑥) , the
slope of the line tangent to the graph of 𝒇 at point 𝑷 is given by
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
ℎ→0
ℎ
𝑚(𝑎) = lim
if the limit exists.
Note: Some books use ∆𝑥 (delta 𝑥) instead of ℎ.
Example 2.1.1.
Consider the function 𝑓(𝑥) = 𝑥 2 + 2𝑥. Determine the slope of the
tangent line to the graph of 𝑓 at the point (1, 3).
Solution:
Using the limit definition,
𝑚(1) = lim
ℎ→0
= lim
ℎ→0
𝑓(1+ℎ)−𝑓(1)
ℎ
[(1+ℎ)2 +2(1+ℎ)]−[(1)2 +2(1)]
ℎ
[1 + 2ℎ + ℎ2 + 2 + 2ℎ] − [3]
= lim
ℎ→0
ℎ
74
= lim
[ℎ2 +4ℎ+3]−[3]
ℎ
ℎ→0
= lim
ℎ2 +4ℎ
, factor out ℎ
ℎ
ℎ→0
= lim (ℎ + 4)
ℎ→0
= 0+4
𝒎(𝟏) = 𝟒
Example 2.1.2.
Given the function 𝑔(𝑥) = 𝑥 3 + 1, find the equation of the tangent line
to the graph of 𝑔 at 𝑥 = −1.
Solution:
Using the limit definition,
𝑔(−1 + ℎ) − 𝑔(−1)
ℎ→0
ℎ
𝑚(−1) = lim
= lim
[(−1+ℎ)3 +1]−[(−1)3 +1]
ℎ
ℎ→0
= lim
[(−1+3ℎ−3ℎ2 +ℎ3 )+1]−[0]
ℎ
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
ℎ3 −3ℎ2 +3ℎ
ℎ
ℎ(ℎ2 −3ℎ+3)
ℎ
= lim (ℎ2 − 3ℎ + 3)
ℎ→0
= 02 − 3(0) + 3
𝒎(−𝟏) = 𝟑
To determine the equation of the tangent line, recall the point-slope form of the
equation of the line:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ).
At 𝑥 = −1, 𝑔(−1) = 0, thus the corresponding point is (−1, 0) with a slope of 𝑚 = 3.
Substituting these values to the equation,
𝑦 − 0 = 3(𝑥 − (−1))
𝑦 = 3(𝑥 + 1)
𝒚 = 𝟑𝒙 + 𝟑
75
Example 2.1.3.
Find the equation of the tangent line to the curve defined by
3𝑥+1
𝐹(𝑥) = 𝑥−1 at the point where 𝑥 = 2.
Solution:
Using the limit definition,
𝐹(2 + ℎ) − 𝐹(2)
ℎ→0
ℎ
𝑚(2) = lim
= lim [
3(2+ℎ)+1 3(2)+1
−
(2+ℎ)−1
2−1
ℎ
ℎ→0
= lim [
6+3ℎ+1
−7
2+ℎ−1
= lim [
3ℎ+7
−7
ℎ+1
3ℎ+7−7(ℎ+1)
ℎ+1
ℎ
ℎ→0
= lim [
3ℎ+7−7ℎ−7
ℎ+1
ℎ
ℎ→0
= lim [
ℎ→0
= lim [
]
ℎ
ℎ→0
= lim [
]
ℎ
ℎ→0
ℎ→0
]
−4ℎ
ℎ+1
ℎ
]
]
]
−4ℎ 1
∙ ]
ℎ+1 ℎ
−4
= lim [ ℎ+1]
ℎ→0
−4
= 0+1
𝒎(𝟐) = −𝟒
With slope 𝑚 = −4, and point (2,7), the equation of the tangent line will be
𝑦 − 7 = −4(𝑥 − 2)
𝑦 − 7 = −4𝑥 + 8
𝑦 = −4𝑥 + 8 + 7
𝒚 = −𝟒𝒙 + 𝟏𝟓
76
Remarks: Possibly, the limit does not exist; that is, (1) the limit itself is not defined or (2)
the limit is infinity. The first case implies no tangent line to the curve at the given point. It
usually happens to the functions whose graphs have sharp turning points such as the vertex
of the graph of the absolute value function 𝑦 = 𝑓(𝑥) = |𝑥|. The second case leads to a
“vertical tangent line,” which leads to the equation 𝑥 = 𝑎.
Example 2.1.4.
Find the tangent line, if any, at the vertex of the graph of the function
𝑦 = 𝑓(𝑥) = |𝑥|.
Solution:
The vertex of the graph is at (0,0). Using the limit definition,
𝑓|0 + ℎ| − 𝑓(0)
ℎ→0
ℎ
𝑚(0) = lim
= lim
|0+ℎ|−𝑓(0)
ℎ
ℎ→0
= lim
|ℎ|−0
ℎ
ℎ→0
= lim
|ℎ|
ℎ→0 ℎ
Evaluating the limit will result in an indeterminate form. To eliminate the absolute
value bar, you must consider the following definition of absolute value:
|𝑥| = {
𝑥
−𝑥
Applying the limit definition to lim
|ℎ|
ℎ→0 ℎ
𝑖𝑓 𝑥 ≥ 0
𝑖𝑓 𝑥 < 0
means breaking the limit into left- and right-
hand limits. The right-hand limit is given by lim+
ℎ→0
limit is given by lim−
ℎ→0
|ℎ|
ℎ
= lim−
ℎ→0
−ℎ
ℎ
|ℎ|
ℎ
ℎ
= lim+ ℎ = lim+1 = 1. The left-hand
ℎ→0
ℎ→0
= lim− (−1) = −1. Since the right-hand and the left-
hand limits are not equal, then the lim
ℎ→0
|ℎ|
ℎ→0 ℎ
does not exist.
Thus, there is no such tangent line to the graph of 𝒇 at (0,0).
77
Example 2.1.5.
3
Find the tangent line, if any, to the curve 𝑔(𝑥) = √𝑥 at the origin.
Solution:
The slope of the tangent line at 𝑥 = 0 is
𝑔(0 + ℎ) − 𝑔(0)
ℎ→0
ℎ
𝑚(0) = lim
3
3
√0+ℎ− √0
ℎ
ℎ→0
= lim
3
√ℎ−0
ℎ→0 ℎ
= lim
3
√ℎ
ℎ→0 ℎ
= lim
Rationalizing the numerator will result to
3
3
𝑚(0) = lim [
ℎ→0
= lim
√ℎ
ℎ
∙
√ℎ2
3
√ℎ2
]
ℎ
3
ℎ→0 ℎ √ℎ2
= lim 3
1
ℎ→0 √ℎ2
Evaluating this infinite limit will lead to
1
𝑚(0) = ( + )
0
= +∞
Since the limit is infinity, the tangent line is vertical, and the equation of the tangent
line is 𝑥 = 0 since the point is at the origin.
Note to Self:
This is a safe space. You
can write anything here.
78
Limit Definition of the Derivative
The derivative of a function is the function 𝑓′ defined by
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉→𝟎
𝒉
𝒇′(𝒙) = 𝒍𝒊𝒎
if this limit exists.
Let us now relate the slope of a tangent line to a curve to the limit definition of the
derivative.
Note: The process of obtaining the derivative of a function is called differentiation. To
differentiate a function means to get its derivative with respect to the independent variable.
Other notations commonly used to denote derivatives of the function defined by 𝑦 = 𝑓(𝑥)
are the following:
𝒅𝒚
𝒚′ , 𝒇′ (𝒙), 𝑫𝒙 𝒚 ,
𝒂𝒏𝒅 𝑫𝒙 [𝒇(𝒙)]
𝒅𝒙
When getting the derivative of a function at a number C, the corresponding
notations are used:
Example 2.1.6.
Solve for the derivative of 𝑓(𝑥) = 7𝑥 + 4.
Solution:
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑓′(𝑥) = lim
= lim
[7(𝑥+ℎ)+4]−(7𝑥+4)
ℎ
ℎ→0
= lim
[7𝑥+7ℎ+4]−7𝑥−4
ℎ
ℎ→0
= lim
7ℎ
ℎ→0 ℎ
= lim 7
ℎ→0
𝒇′(𝒙) = 𝟕
79
Example 2.1.7.
3
Solve for the derivative of 𝑦 = 𝑥 2 , then find 𝑦 ′ (3).
Solution:
Using the limit definition of the derivative,
3
3
−
(𝑥 + ℎ)2 𝑥 2
𝑦′ = lim [
]
ℎ→0
ℎ
3
3
−
(𝑥+ℎ)2 𝑥2
= lim [
ℎ
ℎ→0
= lim [
3𝑥2 −3(𝑥+ℎ)2
𝑥2 (𝑥+ℎ)2
ℎ
ℎ→0
= lim [
]
3𝑥 2 −3(𝑥 2 +2ℎ𝑥+ℎ2 )
𝑥 2 (𝑥+ℎ)2
ℎ→0
= lim [
ℎ→0
]
3𝑥 2 −3𝑥 2 −6ℎ𝑥−3ℎ2
ℎ𝑥 2 (𝑥+ℎ)2
1
∙ ℎ]
]
−6ℎ𝑥−3ℎ2
= lim [ℎ𝑥 2 (𝑥+ℎ)2 ]
ℎ→0
ℎ(−6𝑥−3ℎ)
= lim [ ℎ𝑥 2 (𝑥+ℎ)2 ]
ℎ→0
(−6𝑥−3ℎ)
= lim [ 𝑥 2 (𝑥+ℎ)2 ]
ℎ→0
=
=
(−6𝑥−3(0))
𝑥 2 (𝑥+0)2
−6𝑥
𝑥4
𝟔
𝒚′ = − 𝒙𝟑
Then compute for the specific derivative 𝑦 ′ (3) by replacing x in the final answer with 3.
Thus,
𝟔
𝟐
𝒚′ (𝟑) = − 𝟑𝟑 = − 𝟗.
80
Practice #8
Try this! ☺
Answer what is asked on each question:
1. Determine the equation of the tangent line to the graph at the specified point.
a. 𝑓(𝑥) = 𝑥 2 − 4𝑥; P(1, -3)
b. g(x)=
2𝑥−2
𝑥−4
; P(6,5)
2. Determine the derivative of each function using the limit definition
a. 𝑦 ′ if 𝑦 = 3𝑥 2 − 2
5
b. 𝑓 ′ (𝑥) if 𝑓(𝑥) = 𝑥 3
𝑑𝑦
𝑥+3
c. 𝑑𝑥 if 𝑦 = 3𝑥−2
Process Questions
1. How to describe the relationship between the slope of the tangent line to a curve
and the limit definition of the derivative?
2. What is the difference between derivative and differentiation?
Student’s responses:
Please write your answers on page 161.
Note to Self:
This is a safe space. You
can write anything here.
81
My Working Space
(write your solutions to the practice problems here)
82
Lesson 2.2
Differentiability and
Continuity of Functions
From the topic about continuity, we knew that a function 𝑓 is continuous at a number
𝑎 if it satisfies the following conditions: (1) 𝑓(𝑎) is defined, (2) lim 𝑓(𝑥) exists, and (3)
𝑥→𝑎
lim 𝑓(𝑥) = 𝑓(𝑎). The following theorem relates continuity with differentiability, which is
𝑥→𝑎
the state when a function has a derivative or when its derivative exists:
Theorem 1(Differentiability Implies Continuity)
If a function is differentiable at a number, then it is also continuous at that number.
Logic tells us that since Theorem 1 is an if-then statement, it means that its
contrapositive is also true. This is stated in the following corollary.
Corollary 1
If a function is discontinuous at a number, then it is not differentiable at that number.
Consider examples 2.2.1 and 2.2.2 and see if the given function is continuous at
any particular number included in its domain.
Example 2.2.1.
𝑓(𝑥) = 𝑥 2 − 2𝑥
Solution:
Using the limit definition of the derivative, we get 𝑓′(𝑥) = 2𝑥 − 2. Since the function's
derivative is linear, it is defined for any real number. This means that the original function 𝑓
is differentiable for any real number (or in fact, differentiable in its domain). Thus, it
automatically follows by Theorem 1 that the given function is continuous at any
particular number included in its domain.
Example 2.2.2.
1
𝑓(𝑥) = 2−𝑥
Solution:
Finding the derivative of the given function using the limit definition, we get 𝑓′(𝑥) =
1
. By choosing the value 3 in the domain of 𝑓, we can say that the derivative is defined,
(2−𝑥)2
and we have 𝑓′(3) = 1. Thus, the function is differentiable at 𝑥 = 3. By Theorem 1, the
function is continuous at 3. We should be able to check that for all values in the domain of
the rational function; the function is differentiable. The only real number for which the
83
function is not differentiable is at 𝑥 = 2. Note that 𝑥 = 2 is not part of the domain of 𝑓; the
function is not defined at 2. Hence, 𝒇 is discontinuous at 𝒙 = 𝟐.
Note: The underlined should not be interpreted that the non-differentiability of 𝑓 made it
discontinuous at 2. On the contrary, Corollary 1 assures that the discontinuity at 2 makes the
function not differentiable at 2.
Example 2.2.3.
differentiable at 0.
2 if 𝑥 ≠ 0
Consider the function 𝑓(𝑥) = {𝑥
. Determine if the function is
1 if 𝑥 = 0
Solution:
Using the limit definition of the derivative, we determine if the function is
differentiable at 0.
𝑓(0 + ℎ) − 𝑓(0)
𝑓(ℎ) − 1
𝑓′(0) = lim
= lim
ℎ→0
ℎ→0
ℎ
ℎ
Since the limiting condition requires that ℎ approaches, without necessarily being
equal to, 0, it follows that 𝑓′(0) = lim
𝑓(ℎ)−1
ℎ
ℎ→0
= lim
ℎ→0
Now, if ℎ approaches 0 from the right, lim+
ℎ→0
ℎ2 −1
ℎ
.
ℎ2 −1
ℎ
−1
= (0+ ) = −∞.
On the other hand, if ℎ approaches 0 from the left lim−
ℎ→0
ℎ2 −1
ℎ
The results of one-sided limits lead to the conclusion that lim
Thus, the function is not differentiable at 𝟎.
−1
= ( 0− ) = +∞.
ℎ→0
ℎ2 −1
ℎ
does not exist.
Note: The function fails to be continuous at 0. Although both 𝑓(0) = 1 and lim 𝑓(𝑥) =
ℎ→0
lim 𝑥 2 = 0 exist, their different values mean that 𝑓 is discontinuous at 0; thus, by Corollary
ℎ→0
1, the function is not differentiable at 0.
To examine when a function is differentiable over an interval, especially those
containing one or two endpoints, there is a need to introduce one-sided derivatives. A onesided derivative is either a left-hand or a right-hand derivative.
Definition 1 (Left-hand and Right-hand Derivatives of a Function)
The right-hand derivative of a function 𝒇 at a number 𝑪 is given by
𝑓+ ′(𝐶) = 𝑙𝑖𝑚+
ℎ→0
𝑓(𝐶 + ℎ) − 𝑓(𝐶)
,
ℎ
if this limit exists.
The left-hand derivative of a function 𝒇 at a number 𝑪 is given by
𝑓− ′(𝐶) = 𝑙𝑖𝑚−
ℎ→0
𝑓(𝐶 + ℎ) − 𝑓(𝐶)
,
ℎ
if this limit exists.
84
2 𝑖𝑓 𝑥 ≥ 1
Consider the function given by 𝑓(𝑥) = { 2𝑥
.
4𝑥 − 2 𝑖𝑓 𝑥 < 1
Determine 𝑓+ ′(1) and 𝑓− ′(1).
Example 2.2.4.
Solution:
Using Definition 1, we have:
𝑓+ ′(1) = 𝑙𝑖𝑚+
ℎ→0
𝑓(1+ℎ)−𝑓(1)
ℎ
= 𝑙𝑖𝑚+
ℎ→0
2(1+ℎ)2 −2
ℎ
= 𝑙𝑖𝑚+
4ℎ+2ℎ2
ℎ
ℎ→0
= 𝑙𝑖𝑚+(4 + 2ℎ) = 4
ℎ→0
and
𝑓− ′(1) = 𝑙𝑖𝑚−
ℎ→0
𝑓(1+ℎ)−𝑓(1)
ℎ
= 𝑙𝑖𝑚−
ℎ→0
[4(1+ℎ)−2]−2
ℎ
= 𝑙𝑖𝑚−
4+4ℎ−2−2
ℎ
ℎ→0
= 𝑙𝑖𝑚−
ℎ→0
4ℎ
ℎ
= 4.
The preceding example leads to an introductory remark regarding the existence of
the derivative.
.
Theorem 2 (Existence of the Derivative at a Number)
If for a function 𝑓 it can be shown that 𝑓+′ (𝐶) = 𝑓− ′(𝐶), then 𝑓′(𝐶) exists.
It follows from Theorem 2 that 𝑓 is differentiable at the number 𝐶. The function in
example 2.2.4 is differentiable at 1. Consequently, it makes the function continuous at 1.
3𝑥 − 2 if 𝑥 ≥ 1
Consider the function given by 𝑓(𝑥) = { 2
.
𝑥 + 𝑥 if 𝑥 < 1
Determine 𝑓+ ′(1) and 𝑓− ′(1). Also, determine if 𝑓 is differentiable at 1.
Example 2.2.5.
Solution:
Using the definitions for right-hand and left-hand derivatives at a number, we have:
𝑓+′ (1) = 𝑙𝑖𝑚+
𝑓(1+ℎ)−𝑓(1)
ℎ→0
ℎ
= 𝑙𝑖𝑚+
[3(1+ℎ)−2]−1
ℎ→0
ℎ
= 𝑙𝑖𝑚+
ℎ→0
3ℎ
ℎ
= 𝑙𝑖𝑚+ 3 = 3,
ℎ→0
and
𝑓− ′(1) = 𝑙𝑖𝑚−
ℎ→0
𝑓(1+ℎ)−𝑓(1)
ℎ
= 𝑙𝑖𝑚−
ℎ→0
(1+ℎ)2 +(1+ℎ)−1
ℎ
= 𝑙𝑖𝑚−
ℎ→0
ℎ2 +3ℎ+1
ℎ
1
= (0− ) = −∞.
Clearly, it follows that 𝑓 is not differentiable at 1. However, we cannot conclude
regarding the continuity or discontinuity of 𝑓. With this, the conclusion on the
nondifferentiability of the function follows Corollary 1.
On the other hand, if a function is continuous at a point, it can be differentiable or not
differentiable at that point. See example 2.2.6.
85
Example 2.2.6.
𝑓(𝑥) = | 𝑥 | = {
𝑥 if 𝑥 ≥ 0
−𝑥 if 𝑥 < 0
Solution:
The given function is continuous at 0. However, finding the derivatives 𝑓+′ (0) and
𝑓−′ (0) results to unequal values, making the function nondifferentiable at 0.
Hence, based on examples 2.2.5 and 2.2.6, a function that is nondifferentiable when
the left-hand and right-hand derivatives are not equal can lead to either a continuous or
discontinuous function.
Definition 2
A function is said to be differentiable on an open interval (𝑎, 𝑏) if its derivative exists at
every number in (𝑎, 𝑏).
Note:
2𝑥
a. The function 𝑓(𝑥) = 𝑥−1 is differentiable on the interval (2, 5) since the point of
discontinuity of 𝑓, which is 1, is not contained in the given interval.
b. The same function 𝑓 in item (a) fails to be continuous on (−2, 3) since 1 ∈ (−2, 3). Thus, 𝑓
is not differentiable on the said interval.
To evaluate differentiability on intervals that include at least one endpoint, we will use
Definition 3 and any one-sided derivatives.
Definition 3 (Differentiability on Semi-open, Semi-closed, and Closed Intervals)
A function 𝑓 is said to be differentiable on:
1. the interval [𝑎, 𝑏) if it is differentiable on (𝑎, 𝑏) and 𝑓+ ′(𝑎) exists;
2. the interval [𝑎, +∞) if it is differentiable on (𝑎, +∞) and 𝑓+ ′(𝑎) exists;
3. the interval (𝑎, 𝑏] if it is differentiable on (𝑎, 𝑏) and 𝑓− ′(𝑏) exists;
4. the interval (−∞, 𝑏] if it is differentiable on (−∞, 𝑏) and 𝑓− ′(𝑏) exists;
5. the interval [𝑎, 𝑏] if it is differentiable on (𝑎, 𝑏) and both 𝑓+ ′(𝑎) and 𝑓− ′(𝑏)exist; and
6. the interval (−∞, +∞) or ℝ if it is differentiable at every real number.
Note: In most books, the following are stated as theorems:
a. Any polynomial function is differentiable on (−∞, +∞).
86
A specific example would be 𝑓(𝑥) = 𝑥 2 whose derivative is 𝑓 ′ (𝑥) = 2𝑥. Note
that the derivative is a linear function, which is defined everywhere, implying that the
derivative of 𝑓 exists on ℝ.
b. A rational function is differentiable in its domain.
It means that a function is differentiable at any subset of intervals in the
domain that do not contain the point of discontinuity.
c. The square root function 𝑓(𝑥) = √𝑥 is continuous on [0, +∞). However, it fails to be
differentiable on the same interval since 𝑓− ′(0) does not exist. Thus, the said function
is differentiable for any interval type that is a subset of (0, +∞) only.
𝑥 𝑖𝑓 𝑥 ≥ 0
as example 2.2.6 is
−𝑥 𝑖𝑓 𝑥 < 0
differentiable only on [0, +∞) or its subsets; and (0, +∞) or its subsets.
d. The absolute value function 𝑓(𝑥) = | 𝑥 | = {
Practice #9
Try this! ☺
Consider each given function, number 𝑎, and the interval. Then answer the following
questions and show your solution:
𝑥
a. 𝑓(𝑥) = 𝑥+1
2
𝑖𝑓 𝑥 ≥ 1
b. 𝑓(𝑥) = { 𝑥 + 4
4𝑥 + 1 𝑖𝑓 𝑥 < 1
𝑎=0
[−1, 1]
𝑎=1
(1, 2)
i. Is the function differentiable at the given number?
ii. Is the function continuous at the given number?
iii. Is the function differentiable at the given interval?
iv. Is the function continuous at the given interval?
Process Questions
1. How to describe the relationship between differentiability and continuity of a
function?
2. How to determine if the given function is differentiable at the given point?
3. How to determine if the given function is differentiable at the given interval?
Student’s responses:
Please write your answers on page 162.
Worksheet #4: (See page 155)
87
My Working Space
(write your solutions to the practice problems here)
88
89
Lesson 2.3
Theorems of Differentiation and
Instantaneous Rate of Change
Evaluating the derivative or differentiation using the concept of limit may become
more tedious, especially for some complicated functions. For this purpose, we derive the
theorems or basic rules that we can apply for some functions, more specifically algebraic
functions, exponential functions, and trigonometric functions. The derivative would still use
the limit definition of the derivative.
Note: The process of solving the derivative of a function is called differentiation.
Some notations of differentiation to know to minimize confusion:
Calculus was invented by Newton and Leibniz independently in the 17th century. Thus,
𝑑𝑥
different notations came up with the same meaning. Newton used 𝑥̇ whereas Leibniz 𝑑𝑡 to
denote the derivative of 𝑥 with respect to time (𝑡).
The following are the derivative notations that you may encounter in this lesson.
Notation
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑥
Explanation
𝑦= 𝑓′ = 𝑓′(𝑥)
It is a function or an expression in 𝑥.
𝑑
𝑑𝑥
It is considered an operation.
To find 𝑑𝑥 means to perform the differentiation operation on 𝑦.
𝐷𝑥 and 𝐷𝑥 𝑦
Some authors use the notation 𝐷𝑥 where D stands for the
differentiation operator. Some authors use 𝐷𝑥 𝑦 to emphasize
that the variable is 𝑥.
𝑑𝑦
Table 7
𝑑𝑦
Caution: 𝑑𝑥 is NOT a fraction. It does not mean “divide 𝑑𝑦 by 𝑑𝑥”.
Recall: The limit definition of the derivative of a function:
𝑓′(𝑥) = 𝑙𝑖𝑚
∆𝑥→0
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
or 𝑓′(𝑥) = 𝑙𝑖𝑚
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
90
Derivatives of Algebraic Functions
To determine the derivative of algebraic functions without using the limit definition,
certain rules must be considered.
Note that these rules may also be applied to transcendental functions such as
exponential and trigonometric functions which may be discussed in the next lessons.
Theorem 3 (The Constant Rule of Differentiation)
The derivative of a constant function is 0. That is, if 𝑎 is a real number, then
𝑑
𝑑𝑥
[𝑎] = 0.
The rule clearly shows that the derivative of any constant is always zero.
Derivation of the Constant Rule
Let 𝑓(𝑥) = 𝑎. Using the limit definition of the derivative, we have
𝑑
[𝑎] = lim
𝑑𝑥
𝑎−𝑎
∆𝑥→0 ∆𝑥
Example 2.3.1.
a. 𝑦 = 10
Solution:
a.
𝑑𝑦
𝑑𝑥
=0
= lim 0 = 0.
∆𝑥→0
Find the derivative of the following functions using the constant rule:
2
b. 𝑓(𝑥) = √2
c. 𝑔(𝑥) =
b. 𝑓′(𝑥) = 0
c. 𝑔′(𝑥) = 0
3
Theorem 4 (The Identity Rule of Differentiation)
𝑑
The derivative of an identity function is 1. That is, 𝑑𝑥 [𝑥] = 1.
Derivation of the Identity Rule of Differentiation
𝑑
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
[𝑥] = 𝑙𝑖𝑚
∆𝑥→0
𝑑𝑥
∆𝑥
𝑥 + ∆𝑥 − 𝑥
∆𝑥→0
∆𝑥
= 𝑙𝑖𝑚
= 𝑙𝑖𝑚
∆𝑥
∆𝑥→0 ∆𝑥
= 𝑙𝑖𝑚 1
∆𝑥→0
=1
91
Theorem 5 (The Power Rule of Differentiation)
If 𝑛 is a nonzero constant, then the derivative of the function 𝑓(𝑥) = 𝑥 𝑛 is
𝑑 𝑛
[𝑥 ] = 𝑛𝑥 𝑛−1
𝑑𝑥
The derivative of a power function is a function in which the power on 𝑥 becomes the
coefficient of the term and the power on 𝑥 in the derivative decreases by 1.
Derivation of the Power Rule
Let 𝑓(𝑥) = 𝑥 𝑛 , where 𝑛 is any positive integer besides 1, using the limit definition of
the derivative and binomial expansion,
𝑑
𝑑𝑥
[𝑥 𝑛 ] = 𝑙𝑖𝑚
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
[𝑥 𝑛 +𝑛𝑥 𝑛−1 (∆𝑥)+
= 𝑙𝑖𝑚
(Limit definition of the derivative)
∆𝑥
∆𝑥→0
𝑛(𝑛−1)𝑥𝑛−2
2
(∆𝑥)2 + ...+ (∆𝑥)𝑛 − 𝑥 𝑛 ]
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚 [𝑛𝑥 𝑛−1 +
𝑛(𝑛−1)𝑥𝑛−2
∆𝑥→0
2
(Binomial Expansion)
(∆𝑥) + . . . + (∆𝑥)𝑛−1 ] (Algebraic manipulation to
simplify the resulting expression)
= 𝑛𝑥 𝑛−1 + 0 + ⋯ + 0
(Evaluation of limit)
= 𝑛𝑥 𝑛−1
Therefore,
𝑑
𝑑𝑥
[𝑥 𝑛 ] = 𝑛𝑥 𝑛−1
Example 2.3.2.
Find the derivative of the following functions using the power rule.
a. 𝑦 = 𝑥 3
d. 𝑔(𝑡) = √𝑡
b. 𝑓(𝑥) = 𝑥 6
e. 𝑔(𝑡) =
1
2
𝑡3
c. 𝑓(𝑥) =
Solution:
a.
𝑑𝑦
𝑑𝑥
1
𝑥8
= 3𝑥 3−1 = 3𝑥 2
b. 𝑓 ′ (𝑥) = 6𝑥 6−1 = 6𝑥 5
92
1
c. 𝑓(𝑥) = 𝑥 8 = 𝑥 −8
(Transform the function to the power form)
𝑓′(𝑥) = −8𝑥 −8−1 = −8𝑥 −9 =
−8
𝑥9
1
d. 𝑔(𝑡) = √𝑡 = 𝑡 2
1 1−1
𝑡2
2
𝑔′(𝑡) =
e. 𝑔(𝑡) =
1
2
𝑡3
(Transform the function to the power)
1
=2
√𝑡
.
(Power Rule)
2
= 1 (𝑡 − 3 )
2
2
(Power Rule)
(Transform the function to the power form)
5
2
𝑔′ (𝑡) = − 3 𝑡 −3 − 1 = − 3 𝑡 −3 =
−2
5
(Power Rule)
3𝑡 3
Theorem 6 (The Constant Multiple Rule of Differentiation)
If f is a differentiable function and 𝑎 is a real number, then the derivative of the
function a f is:
𝑑
[𝑎 𝑓(𝑥)] = 𝑎 𝑓′(𝑥).
𝑑𝑥
The derivative of a constant a multiplied by a function f is the same as the constant
multiplied by the derivative.
In the case of powers such as 𝑓(𝑥) = 𝑥 𝑛 , we have:
𝑑
𝑑𝑥
[𝑎𝑓(𝑥)] =
𝑑
𝑑𝑥
[𝑎𝑥 𝑛 ]
𝑑
= 𝑎 𝑑𝑥 [𝑥 𝑛 ]
(Constant Multiple Rule)
= 𝑎(𝑛𝑥 𝑛 − 1 )
(Power Rule)
Derivation of the Constant Multiple Rule
𝑑
𝑑𝑥
[𝑎 𝑓(𝑥) ] = 𝑙𝑖𝑚
𝑎 𝑓(𝑥 + ∆𝑥) − 𝑎 𝑓(𝑥)
∆𝑥→0
= 𝑙𝑖𝑚 𝑎 [
∆𝑥→0
= 𝑎 [ 𝑙𝑖𝑚
∆𝑥→0
(Limit definition of the derivative)
∆𝑥
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
𝑓 (𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
(Algebraic manipulation to simplify the
]
resulting expression)
]
(Evaluation of Limits)
= 𝑎 𝑓′(𝑥)
93
Example 2.3.3.
multiple rule.
Find the derivative of the following functions using the constant
1
a. 𝑦 = 2𝑥
d. 𝑔(𝑥) = 3𝑥 6
b. 𝑓(𝑥) = 5𝑥 3
e. 𝑠 = 5 𝑡 2
4 1
c. 𝑓(𝑥) = 10𝑥 8
Solutions:
a.
𝑑𝑦
𝑑𝑥
𝑑
= 2 𝑑𝑥 (𝑥) = 2(1) = 2
b. 𝑓′(𝑥) = 5(3𝑥 2 ) = 15𝑥 2
c. 𝑓′(𝑥) = 10(8𝑥 7 ) = 80𝑥 7
d. 𝑔(𝑥) =
1
1
3𝑥 6
= 3 (𝑥 −6 )
1
2
𝑔′ (𝑥) = ( ) (−6𝑥 −7 ) = −2𝑥 −7 = − 7
3
𝑥
e.
𝑑𝑠
𝑑𝑡
4 𝑑
=5
1
4
1
1
(𝑡 2 ) = 5 (2 𝑡 −2 ) =
𝑑𝑡
2
1
5𝑡 2
Theorem 7 (The Sum and Difference Rules of Differentiation)
Let f and g be differentiable functions, then the sum (or difference) of f and g is also
differentiable.
𝑑
The derivative of the sum of f and g is given by 𝑑𝑥 [𝑓(𝑥) + 𝑔(𝑥)] = 𝑓 ′(𝑥) + 𝑔 ′(𝑥).
𝑑
The derivative of the difference of f and g is given by 𝑑𝑥 [𝑓(𝑥) − 𝑔(𝑥)] = 𝑓 ′(𝑥) − 𝑔 ′(𝑥).
The derivative of the sum of a function f and a function g is the same as the sum of the
derivative of f and the derivative of g.
The derivative of the difference of a function f and a function g is the same as the
difference between the derivative of f and the derivative of g.
94
Derivation of the Sum Rule
[𝑓(𝑥 + ∆𝑥) + 𝑔(𝑥 + ∆𝑥] − [ 𝑓 (𝑥) + 𝑔 (𝑥) ]
𝑑
[𝑓(x) + 𝑔(x) ] = 𝑙𝑖𝑚
𝑑𝑥
∆𝑥
∆𝑥 →0
= 𝑙𝑖𝑚
𝑓(𝑥 + ∆𝑥) + 𝑔(𝑥 + ∆𝑥) − ( 𝑓(𝑥) − 𝑔(𝑥)
∆𝑥
∆𝑥→ 0
= 𝑙𝑖𝑚
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) + 𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
∆𝑥
∆𝑥→ 0
= 𝑙𝑖𝑚
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
∆𝑥→ 0
+ 𝑙𝑖𝑚
∆𝑥→ 0
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
∆𝑥
= 𝑓′(𝑥) + 𝑔′(𝑥)
The derivation of the difference rule follows the same steps as in the sum rule
derivation except that the operation between f and g is subtraction. Also, these rules may be
applied to a finite number n of functions 𝑓1 , 𝑓2 ,…, 𝑓𝑛 ; that is,
.
Example 2.3.4.
𝑑
Given 𝑓(𝑥) = 4𝑥 5 + 7𝑥 2 , find 𝑑𝑥 [4𝑥 5 + 7𝑥 2 ].
Solution:
𝑑
𝑑𝑥
𝑑
𝑑
The Sum and Difference Rule
(4𝑥 5 + 7𝑥 2 ) = 𝑑𝑥 [4𝑥 5 ] + 𝑑𝑥 [7𝑥 2 ]
𝑑
𝑑
= (4) 𝑑𝑥 [𝑥 5 ] + (7) 𝑑𝑥 [𝑥 2 ]
The Constant Rule
= (4)(5𝑥 4 ) + (7)(2𝑥)
The Power Rule
= 20𝑥 4 + 14𝑥
Hence, the derivative of 𝒇(𝒙) = 𝟒𝒙𝟓 + 𝟕𝒙𝟐 is
𝒅
𝒅𝒙
[𝟒𝒙𝟓 + 𝟕𝒙𝟐 ] = 𝟐𝟎𝒙𝟒 + 𝟏𝟒𝒙.
Theorem 8 (The Product Rule of Differentiation)
Let f and g be differentiable functions, then the product of f and g is also differentiable
and its derivative is given by
𝑑
[𝑓(𝑥) 𝑔(𝑥)] = 𝑓(𝑥) 𝑔′(𝑥) + 𝑔(𝑥) 𝑓′(𝑥).
𝑑𝑥
95
The derivative of a product of two functions is the derivative of the first function times
the second function, plus the derivative of the second function times the first function.
Derivation of the Product Rule
𝑑
𝑑𝑥
[𝑓(𝑥)𝑔(𝑥)] = 𝑙𝑖𝑚
(1)
[𝑓(𝑥 +∆𝑥) 𝑔(𝑥 + ∆𝑥)] − [𝑓(𝑥) 𝑔(𝑥) ]
∆𝑥→0
= 𝑙𝑖𝑚
∆𝑥
∆𝑥
= 𝑙𝑖𝑚 [𝑓(𝑥 + ∆𝑥)
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
∆𝑥
∆𝑥 →0
= 𝑙𝑖𝑚 [𝑓(𝑥 + ∆𝑥)
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
∆𝑥
∆𝑥 →0
= 𝑙𝑖𝑚 𝑓(𝑥 + ∆𝑥) ∙ 𝑙𝑖𝑚
∆𝑥 →0
(2)
𝑓(𝑥 +∆𝑥) 𝑔(𝑥 + ∆𝑥) − 𝑓(𝑥 + ∆𝑥) 𝑔(𝑥) + 𝑓(𝑥 + ∆𝑥) 𝑔(𝑥) − 𝑓(𝑥) 𝑔(𝑥)
∆𝑥→0
∆𝑥 →0
+ 𝑔(𝑥)
∆𝑥
] + 𝑙𝑖𝑚 [𝑔(𝑥)
∆𝑥
]
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥 →0
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
(3)
𝑓(𝑥 + ∆𝑥) − (𝑓(𝑥)
∆𝑥
+ 𝑙𝑖𝑚 𝑔(𝑥) ∙ 𝑙𝑖𝑚
∆𝑥 →0
(4)
]
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥 →0
∆𝑥
= 𝑓(𝑥) 𝑔′(𝑥) + 𝑔(𝑥) 𝑓′(𝑥)
In step (1), we applied the limit definition of the derivative.
In step (2), we need to introduce the term 𝑓(𝑥 + ∆𝑥) 𝑔( 𝑥). So, we added and
subtracted this term.
In step (3), 𝑓(𝑥 + ∆𝑥) is factored out from the first two terms, while 𝑔(𝑥) is factored
out from the last two terms.
In steps (4) and (5), we evaluated the limit as well as applied the limit over a sum and
product.
Example 2.3.5.
product rule.
Determine the derivative of 𝑔(𝑥) = (3𝑥 − 1)(𝑥 2 + 4𝑥) by applying the
Solution:
𝑑
𝑑
𝑑
[(3𝑥 − 1)(𝑥 2 + 4𝑥)] = (3𝑥 − 1) [𝑥 2 + 4𝑥] + (𝑥 2 + 4𝑥) [3𝑥 − 1]
𝑑𝑥
𝑑𝑥
𝑑𝑥
= (3𝑥 − 1)(2𝑥 + 4) + (𝑥 2 + 4𝑥)(3)
= 6𝑥 2 + 10𝑥 − 4 + 3𝑥 2 + 12𝑥
= 9𝑥 2 + 22𝑥 − 4
𝒅
Hence, 𝒅𝒙 [(𝟑𝒙 − 𝟏)(𝒙𝟐 + 𝟒𝒙)] = 𝟗𝒙𝟐 + 𝟐𝟐𝒙 − 𝟒.
96
(5)
Example 2.3.6.
Find the derivative of the following functions:
a. 𝑓(𝑥) = 𝑥 2 (𝑥 4 − 8𝑥)
d. 𝑠 = (2 + 5𝑡)2
b. 𝑦 = (2𝑥 + 3)(4𝑥 − 5)
e. 𝑓(𝑥) = (𝑥 2 + 2𝑥)(𝑥 2 − 4)
c. 𝑔(𝑥) = (𝑥 − 1)(𝑥 2 − 𝑥 + 2)
Solutions:
a.
𝑑
𝑑𝑥
[𝑥 2 (𝑥 4 − 8𝑥)] = 𝑥 2
𝑑
𝑑𝑥
𝑑
[𝑥 4 − 8𝑥] + (𝑥 4 − 8𝑥) 𝑑𝑥 [𝑥 2 ]
= 𝑥 2 (4𝑥 3 − 8) + (𝑥 4 − 8𝑥)(2𝑥)
= 4𝑥 5 − 8𝑥 2 + 2𝑥 5 − 16𝑥 2
= 𝟔𝒙𝟓 − 𝟐𝟒𝒙𝟐
b.
𝑑
𝑑𝑥
[(2𝑥 + 3)(4𝑥 − 5)] = (2𝑥 + 3)
𝑑
𝑑𝑥
𝑑
[4𝑥 − 5] + (4𝑥 − 5) 𝑑𝑥 [2𝑥 + 3]
= (2𝑥 + 3)(4) + (4𝑥 − 5)(2)
= 8𝑥 + 12 + 8𝑥 − 10
= 𝟏𝟔𝒙 + 𝟐
c. 𝑔′ (𝑥) = (𝑥 − 1)(2𝑥 − 1) + (𝑥 2 − 𝑥 + 2)(1)
= 2𝑥 2 − 3𝑥 + 1 + 𝑥 2 − 𝑥 + 2
= 𝟑𝒙𝟐 − 𝟒𝒙 + 𝟑
d. Since 𝑠 = (2 + 5𝑡)2 can be expressed as 𝑠 = (2 + 5𝑡)(2 + 5𝑡), then we have
𝑑𝑠
𝑑𝑡
= (2 + 5𝑡)(5) + (2 + 5𝑡)(5)
= 10 + 25𝑡 + 10 + 25𝑡
= 𝟐𝟎 + 𝟓𝟎𝒕
e. 𝑓′(𝑥) = (𝑥 2 + 2𝑥)(2𝑥) + (𝑥 2 − 4)(2𝑥 + 2)
= 2𝑥 3 + 4𝑥 2 + 2𝑥 3 + 2𝑥 2 − 8𝑥 − 8
= 𝟒𝒙𝟑 + 𝟔𝒙𝟐 − 𝟖𝒙 − 𝟖
97
Theorem 9 (The Quotient Rule of Differentiation)
Let f and g be differentiable functions, then the quotient of f and g is also differentiable at
all values of 𝑥 for which 𝑔′(𝑥) ≠ 0 and its derivative is given by
𝑑 𝑓(𝑥)
𝑔(𝑥) 𝑓′(𝑥) − 𝑓(𝑥) 𝑔′(𝑥)
[
]=
[𝑔(𝑥)]2
𝑑𝑥 𝑔(𝑥)
The derivative of the quotient of two functions is the derivative of the first function
times the second function minus the derivative of the second function times the first
function, all divided by the square of the second function.
Derivation of the Quotient Rule
𝑑
[
𝑓(𝑥)
𝑑𝑥 𝑔(𝑥)
] = 𝑙𝑖𝑚
𝑓(𝑥 + ∆𝑥) 𝑓(𝑥)
−
𝑔(𝑥 + ∆𝑥) 𝑔(𝑥)
∆𝑥→0
= 𝑙𝑖𝑚
𝑔(𝑥) 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) 𝑔(𝑥 + ∆𝑥)
∆𝑥 𝑔(𝑥) 𝑔(𝑥 + ∆𝑥)
∆𝑥→0
= 𝑙𝑖𝑚
∆𝑥
𝑔(𝑥) 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) 𝑔(𝑥) + 𝑓(𝑥) 𝑔(𝑥) − 𝑓(𝑥) 𝑔(𝑥 + ∆𝑥)
∆𝑥 𝑔(𝑥) 𝑔(𝑥 + ∆𝑥)
∆𝑥→0
=
𝑙𝑖𝑚 𝑔(𝑥)
∆𝑥→0
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
+ 𝑙𝑖𝑚 𝑓(𝑥)
∆𝑥→0
𝑔(𝑥) − 𝑔(𝑥 + ∆𝑥)
∆𝑥
lim [𝑔(𝑥) 𝑔(𝑥 + ∆𝑥)]
∆𝑥→0
=
𝑔(𝑥)𝑓′(𝑥)−𝑓(𝑥)𝑔′(𝑥)
[𝑔(𝑥)]2
Similarly, the product rule derivation, the quotient rule also uses adding and
subtracting the same expression such as 𝑓(𝑥) 𝑎𝑛𝑑 𝑔(𝑥). All other rules on limit, definition,
and other algebraic operations were used to arrive at the final expression.
Example 2.3.7.
7𝑥 + 2
𝑑
7𝑥 + 2
Given 𝑓(𝑥) = 𝑥 2 + 3𝑥, determine 𝑑𝑥 [𝑥 2+ 3𝑥] using the quotient rule.
Solution:
𝑑
𝑑
(𝑥 2 + 3𝑥)
[7𝑥 + 2] − (7𝑥 + 2)
[𝑥 2 + 3𝑥]
𝑑 7𝑥 + 2
𝑑𝑥
𝑑𝑥
[ 2
] =
(𝑥 2 + 3𝑥)2
𝑑𝑥 𝑥 + 3𝑥
=
(𝑥 2 + 3𝑥)(7)−(7𝑥 + 2)(2𝑥 + 3)
=−
(𝑥 2 + 3𝑥)2
7𝑥 2 + 4𝑥 + 6
(𝑥 2 + 3𝑥)2
𝟕𝒙+𝟐
Therefore, the derivative of 𝒇(𝒙) = 𝒙𝟐 +𝟑𝒙 is
𝒅
𝟕𝒙+𝟐
[
]=−
𝒅𝒙 𝒙𝟐 +𝟑𝒙
𝟕𝒙𝟐 +𝟒𝒙 + 𝟔
(𝒙𝟐 + 𝟑𝒙)
𝟐
.
98
Example 2.3.7.
Find the derivative of the following functions:
2𝑥 3
9
a. 𝑓(𝑥) = 𝑥 4 + 2
c. 𝑦 = 𝑥 3+𝑥 2 − 2𝑥 + 1
𝑡2+ 4
b. 𝑔(𝑡) = 𝑡 3 − 3𝑡 + 2
Solution:
𝑑
a.
b.
c.
2𝑥 3
[
]
𝑑𝑥 𝑥 4 + 2
𝑑
𝑡2 + 4
[
]
𝑑𝑡 𝑡 3 − 3𝑡 + 2
𝑑
9
[
]
𝑑𝑥 𝑥 3 +𝑥 2 −2𝑥+1
=
(𝑥 4 +2)
𝑑
𝑑
[2𝑥 3 ] − 2𝑥 3
[𝑥 4 + 2]
𝑑𝑥
𝑑𝑥
(𝑥 4 + 2)2
=
(𝑥 4 +2)(6𝑥 2 ) − 2𝑥3 (4𝑥 3 )
=
−2𝑥 6 + 12𝑥 2
=
(𝑥 4 + 2)2
(𝑥 4 + 2)2
(𝑡 3 − 3𝑡 + 2)
𝑑 2
𝑑
[𝑡 + 4] − (𝑡 2 + 4) [𝑡 3
𝑑𝑡
𝑑𝑡
(𝑡 3 − 3𝑡 + 2)2
=
(𝑡 3 − 3𝑡 + 2)(2𝑡) − (𝑡 2 + 4)(3𝑡 2 − 3)
=
(2𝑡 4 − 6𝑡 2 + 4𝑡) − (3𝑡 4 + 9𝑡 2 − 12)
=
−𝑡 4 − 15𝑡 2 + 4𝑡 + 12
=
=
− 3𝑡 + 2]
(𝑡 3 −3𝑡 + 2)2
(𝑡 3 − 3𝑡 + 2)2
(𝑡 3 −3𝑡 + 2)2
𝑑
𝑑
[9] − (9)
[𝑥 3
𝑑𝑥
𝑑𝑥
(𝑥 3 + 𝑥 2 − 2𝑥 + 1)2
(𝑥 3 + 𝑥 2 − 2𝑥 + 1)
+ 𝑥 2 − 2𝑥 + 1]
(𝑥 3 + 𝑥 2 − 2𝑥 + 1)(0) − (9)(3𝑥 2 + 2𝑥 − 2)
(𝑥 3 + 𝑥2 − 2𝑥 + 1)2
−27𝑥 2 − 18𝑥 + 18
= (𝑥 3 + 𝑥 2 − 2𝑥 + 1)2
You have seen how the derivative is used to determine the slope of the tangent to the
graph of a function. The derivative can also be used to determine the rate of change of one
variable with respect to another. Applications involving rate of change, sometimes referred
to as instantaneous rate of change, occur in different fields (e.g., population growth rate,
marginal revenue, velocity, and acceleration).
99
A common use for instantaneous rate of change is to describe the motion of an
object moving linearly. It is customary to use a horizontal or vertical line with a designated
origin to represent a line of motion in such problems. Let us try to answer example 2.3.8.
Example 2.3.8.
Find the instantaneous rate of change in the volume of a spherical
balloon which is being inflated with respect to its radius at the instance when its radius is
4
5𝑐𝑚. The volume of the spherical balloon is given by 𝑉 = 3 π𝑟 3, where V is the volume, and
r is the radius.
Solution:
𝑅𝐼𝑛𝑠𝑡 =
𝑑𝑉 4
= (3𝜋)𝑟 2 = 4𝜋𝑟 2
𝑑𝑟 3
When 𝑟 = 5𝑐𝑚, then the instantaneous rate of change in the volume of the spherical
𝑑𝑉
balloon with respect to its radius is 𝑅𝐼𝑛𝑠𝑡 = 𝑑𝑟 = 4𝜋(5)2 = 100𝜋 which means that the
volume of the balloon increases by 100π cm3 for every centimeter increase in the radius.
Derivatives of Trigonometric Functions
Theorem 10 (Derivatives of Trigonometric Functions)
The following are the derivatives of the six trigonometric functions:
𝑑
𝑑𝑥
[𝑠𝑖𝑛 𝑥] = 𝑐𝑜𝑠 𝑥
𝑑
[𝑐𝑜𝑠 𝑥] = −𝑠𝑖𝑛 𝑥
𝑑𝑥
𝑑
[𝑡𝑎𝑛 𝑥] = 𝑠𝑒𝑐 2 𝑥
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
[𝑐𝑜𝑡 𝑥] = − 𝑐𝑠𝑐 2 𝑥
[𝑠𝑒𝑐 𝑥] = 𝑠𝑒𝑐 𝑥 𝑡𝑎𝑛 𝑥
𝑑
𝑑𝑥
[𝑐𝑠𝑐 𝑥] = − 𝑐𝑠𝑐 𝑥 𝑐𝑜𝑡 𝑥
Note to Self:
This is a safe space. You
can write anything here.
100
Derivation of the Derivatives of sin 𝑥 and cos 𝑥
The following proofs give the derivatives for 𝑠𝑖𝑛 𝑥 and 𝑐𝑜𝑠 𝑥:
𝑑
𝑑𝑥
[sin 𝑥] = 𝑙𝑖𝑚
sin (𝑥 + ∆𝑥) − sin 𝑥
= 𝑙𝑖𝑚
𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 ∆𝑥 + 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥 − 𝑠𝑖𝑛 𝑥
𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 ∆𝑥 − 𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥
𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 ∆𝑥 − 𝑠𝑖𝑛 𝑥
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Algebraic Manipulation)
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Sum and Difference Identity)
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Limit definition of derivative)
∆𝑥
∆𝑥→0
∆𝑥
∆𝑥
+ 𝑙𝑖𝑚
𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚 𝑠𝑖𝑛 𝑥 ∙ 𝑙𝑖𝑚
∆𝑥→0
𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥
∆𝑥→0
𝑠𝑖𝑛 𝑥 (𝑐𝑜𝑠 ∆𝑥 − 1)
∆𝑥→0
+ 𝑙𝑖𝑚
−(1 − 𝑐𝑜𝑠 ∆𝑥)
∆𝑥
∆𝑥→0
(Algebraic Manipulation)
(Algebraic manipulation)
+ 𝑙𝑖𝑚 𝑐𝑜𝑠 𝑥 ∙ 𝑙𝑖𝑚
∆𝑥→0
𝑠𝑖𝑛 ∆𝑥
∆𝑥→0
∆𝑥
(Trigonometric Limit Theorem)
= 𝑠𝑖𝑛 𝑥 (0) + 𝑐𝑜𝑠 𝑥 (1)
= 𝒄𝒐𝒔 𝒙
𝒅
Therefore, 𝒅𝒙 [𝒔𝒊𝒏 𝒙] = 𝒄𝒐𝒔 𝒙.
𝑑
[𝑐𝑜𝑠 𝑥]
𝑑𝑥
= 𝑙𝑖𝑚
𝑐𝑜𝑠 (𝑥 + ∆𝑥) − 𝑐𝑜𝑠 𝑥
= 𝑙𝑖𝑚
𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 ∆𝑥 − 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥 − 𝑐𝑜𝑠 𝑥
𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 ∆𝑥 − 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥
𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 ∆𝑥 − 𝑐𝑜𝑠 𝑥
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Algebraic Manipulation)
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Sum and Difference Identity)
∆𝑥
∆𝑥→0
= 𝑙𝑖𝑚
(Limit definition of derivative)
∆𝑥
∆𝑥→0
∆𝑥→0
𝑐𝑜𝑠 𝑥 (𝑐𝑜𝑠 ∆𝑥 − 1)
∆𝑥→0
∆𝑥
= 𝑙𝑖𝑚 𝑐𝑜𝑠 𝑥 ∙ 𝑙𝑖𝑚
∆𝑥→0
− 𝑙𝑖𝑚
− 𝑙𝑖𝑚
∆𝑥→0
−(1 − 𝑐𝑜𝑠 ∆𝑥)
∆𝑥→0
∆𝑥
𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥
∆𝑥
𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥
∆𝑥
(Algebraic Manipulation)
(Algebraic manipulation)
− 𝑙𝑖𝑚 𝑠𝑖𝑛 𝑥 ∙ 𝑙𝑖𝑚
∆𝑥→0
∆𝑥→0
𝑠𝑖𝑛 ∆𝑥
∆𝑥
(Trigonometric Limit Theorem)
= cos 𝑥 (0) − sin 𝑥(1)
= −𝐬𝐢𝐧 𝒙
𝒅
Therefore, 𝒅𝒙 [𝐜𝐨𝐬 𝒙] = −𝐬𝐢𝐧 𝒙.
101
Derivation of the Derivative of tan 𝑥
sin 𝑥
In the case of 𝑓(𝑥) = tan 𝑥, we use the identity tan 𝑥 = cos 𝑥 then apply the quotient
rule as well as the previously derived formulas for the derivatives of sin 𝑥 and cos 𝑥.
𝑑
[
sin 𝑥
𝑑𝑥 cos 𝑥
]=
=
cos 𝑥 (cos 𝑥) − ( sin 𝑥(−sin 𝑥)
cos2 𝑥
cos2 𝑥 + sin2 𝑥
cos2 𝑥
(Using Quotient Rule)
(Algebraic Manipulation)
1
= cos2𝑥
(Using Trigonometric Identities)
2
= sec 𝑥
cos 2 𝑥 + sin2 𝑥 = 1
sec 𝑥 =
1
cos 𝑥
𝒅
Therefore, 𝒅𝒙 [𝐭𝐚𝐧 𝒙] = 𝐬𝐞𝐜 𝟐 𝒙.
The proofs for the other trigonometric functions' derivatives are done similarly,
applying differentiation rules and appropriate trigonometric identities. These are all left to
you as exercises.
Example 2.3.9.
Find the derivative of the following functions:
a. 𝑓(𝑥) = 3 sin 𝑥 + 5 cos 𝑥
d. 𝑓(𝜃) = 3 𝜃 2 cos 𝜃
b. 𝑦 = 2𝑥 2 + 5𝑥 + cot 𝑥
e. 𝑔(𝑧) = 2 sin 𝑧 + 1
sec 𝑧
c. 𝑦 = tan 𝜃 + 3 csc 𝜃
Solutions:
a. 𝑓(𝑥) = 3 sin 𝑥 + 5 cos 𝑥
𝑓 ′ (𝑥) = 3 cos 𝑥 + 5(−sin 𝑥)
𝒇′ (𝒙) = 𝟑 𝐜𝐨𝐬 𝒙 − 𝟓 𝐬𝐢𝐧 𝒙
b. 𝑦 = 2𝑥 2 + 5𝑥 + cot 𝑥
𝒅𝒚
𝒅𝒙
= 𝟒𝒙 + 𝟓 – 𝐜𝐬𝐜 𝟐 𝒙
102
c. 𝑦 = tan 𝜃 + 3 csc 𝜃
𝑑𝑦
= sec 2 𝜃 + 3(−csc 𝜃 cot 𝜃)
𝑑𝜃
= 𝐬𝐞𝐜 𝟐 𝜽 − 𝟑 𝐜𝐬𝐜 𝜽 𝐜𝐨𝐭 𝜽
d. 𝑓(𝜃) = 3𝜃 2 𝑐𝑜𝑠 𝜃, the product rule is applied.
𝑓′(𝜃) = (3𝜃 2 )
𝑑
𝑑
[cos𝜃] + (cos 𝜃)
[3𝜃 2 ]
𝑑𝜃
𝑑𝜃
= (3𝜃 2 )(− sin𝜃) + (cos 𝜃)(6𝜃)
= −𝟑𝜽𝟐 𝐬𝐢𝐧 𝜽 + 𝟔𝜽𝐜𝐨𝐬 𝜽
In factored form, 𝑓′(𝜃) = 3𝜃(2cos 𝜃 − 𝜃sin 𝜃).
sec 𝑧
e. 𝑔(𝑧) = 2 sin 𝑧 + 1
𝑑
𝑑
(2 sin 𝑧 + 1) [sec 𝑧] − (sec 𝑧)
[2 sin 𝑧 + 1]
𝑑
sec 𝑧
𝑑𝑧
𝑑𝑧
[
] =
𝑑𝑧 2 sin 𝑧 + 1
(2 sin 𝑧 + 1)2
=
=
(2 sin z + 1)(sec z tan z)−(sec z)(2 cos z)
(2 sin z + 1)2
2 sin 𝑧 sec 𝑧 tan 𝑧+sec 𝑧 tan 𝑧−2 sec 𝑧 cos 𝑧
(2 sin z + 1)2
Here, we simplify the result using the following identities: sec 𝑥 cos 𝑧 = 1 and
sin 𝑧 sec 𝑧 = tan 𝑧. Thus,
=
𝟐 𝐭𝐚𝐧𝟐 𝒛+𝐬𝐞𝐜 𝒛 𝐭𝐚𝐧 𝒛−𝟐
(𝟐 𝒔𝒊𝒏 𝒛 + 𝟏)𝟐
Note to Self:
This is a safe space. You
can write anything here.
103
Practice #10
Try this! ☺
For nos. 1-2 , find the derivative of the following functions.
1) 𝑓(𝑥) = 𝑥 3 (1 – 4𝑥)2
2) 𝑦 =
2𝑡 2 − 5𝑡 3
6𝑡 2 + 1
For nos. 3-4, solve as required.
3) A cylindrical tube expands as its radius 𝑟 and height ℎ increase. If the height of the tube
is always twice the radius, find the instantaneous rate of change of the volume of the
tube with respect to its radius when its radius is 5cm.
4) A plane consists of a rectangle surmounted by a semicircle. The width of the rectangle
coincides with the diameter of the semicircle and the height or length of the rectangle
is twice its width. Let 𝑥 inches be the width of the rectangle. Find the instantaneous
rate of change of its area with respect to 𝑥 when 𝑥 = 2 inches.
5) Differentiate 𝑓(𝑥) = 2 cos 𝑥 − 6 sec 𝑥 + 3
6) Differentiate 𝑔(𝑧) = 10 tan 𝑧 − 2 cot 𝑧
7) Differentiate 𝑓(𝑤) = tan 𝑤 sec 𝑤
8) Differentiate ℎ(𝑡) = 𝑡 3 − 𝑡 2 sin 𝑡
1
9) Find y’ of y = sin (2 𝑥 2 )
𝑑𝑣
10) Find 𝑑θ 𝑜𝑓 𝑣 = tan (2θ)
Process Questions:
Using the rules discussed in this lesson, can we possibly differentiate some functions
such as:
1. (𝑥 2 + 5)30 ? (polynomial function)
2. √𝑥 2 + 5 ?
(radical function)
Explain each.
Student’s responses:
Please write your answers on page 162.
104
My Working Space
(write your solutions to the practice problems here)
105
Lesson 2.4 Chain Rule for Differentiation
This lesson will introduce one of the more useful and essential differentiation
techniques, the chain rule for differentiation.
Suppose you are getting the derivative of a function given by 𝑓(𝑥) = (2𝑥 + 1)9 . At
first, you might try to expand the right-hand side of the equation by multiplying the binomial
by itself nine times or by applying the Binomial Expansion Theorem. Then finally, you will
get the derivative of each resulting term using the appropriate derivative rule. But this
process can be tiring and time-consuming. The chain rule for differentiation will be used to
get the derivative of such functions in a more convenient way without requiring long
methods like expanding a binomial. Before proceeding to the solution of the derivative of
𝑓(𝑥) = (2𝑥 + 1)9 using the chain rule, we define the technique first.
There are instances when a function is written in composite form; that is, 𝑓(𝑥) =
𝐹(𝐺(𝑥)), where 𝐹 is a function of 𝑢, say 𝑦 = 𝐹(𝑢), and 𝐺 is a function of 𝑥, say 𝑢 = 𝐺(𝑥). An
example of a function written in composite form is the previous function, 𝑓(𝑥) = (2𝑥 + 1)9 .
This function can be viewed as a composition 𝑓(𝑥) = 𝐹(𝐺(𝑥)), where 𝑢 = 𝐺(𝑥) = 2𝑥 + 1
and 𝐹(𝑢) = (𝑢)9 .
The following theorem will help us get the derivative of such composite functions.
Theorem 11 (The Chain Rule for Differentiation)
If there are functions defined by the equations 𝑢 = 𝐺(𝑥) and 𝑦 = 𝐹(𝑢), such that
𝐺 ′ (𝑥) 𝑎𝑛𝑑 𝐹 ′ (𝑢) both exist, then 𝑦 can be expressed as a function of 𝑥; that is,
𝑦 = ℎ(𝑥) = 𝐹(𝐺(𝑥))
and the derivative of 𝑦 with respect to 𝑥 can be computed using any of the following
formulas:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
= (𝑑𝑢) (𝑑𝑥 ), or
ℎ′ (𝑥) = 𝐹 ′ (𝑢) ∙ 𝐺 ′ (𝑥)
= 𝐹 ′ (𝐺(𝑥)) ∙ 𝐺 ′ (𝑥)
Example 2.4.1.
Determine the derivative of the function 𝑓(𝑥) = (2𝑥 + 1)9 , using the
chain rule for differentiation.
Solution:
Applying the chain rule for differentiation, we have:
𝑢 = 𝐺(𝑥) = 2𝑥 + 1, 𝑎𝑛𝑑 𝑦 = 𝐹(𝑢) = 𝑢9
106
Note that 𝑦 = 𝐹(𝐺(𝑥)) = (2𝑥 + 1)9
Applying the chain rule, we get:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
= 𝑑𝑢 ∙ 𝑑𝑥
= 𝐷𝑢 (𝑢9 ) ∙ 𝐷𝑥 (2𝑥 + 1)
= (9𝑢8 )(2)
= 18𝑢8
But
𝑑𝑦
𝑑𝑥
𝑑𝑦
must be a function of 𝑥; so, we replace 𝑢 in
𝒅𝒚
𝒅𝒙
Example 2.4.2.
𝑑𝑥
= 18𝑢8 by 2𝑥 + 1 to get
= 𝟏𝟖(𝟐𝒙 + 𝟏)𝟖 .
𝑥+1 4
Find the derivative of the function 𝑓(𝑥) = (3−𝑥) .
Solution:
Applying the chain rule for differentiation, we have:
𝐺(𝑥) =
𝑥+1
𝑎𝑛𝑑 𝐹(𝑢) = 𝑢4
3−𝑥
Then we get:
𝑓 ′ (𝑥) = 𝐹 ′ (𝑢) ∙ 𝐺 ′ (𝑥)
𝑥+1
= 4𝑢3 ∙ 𝐷𝑥 (3−𝑥)
= 4𝑢3 ∙
(3−𝑥)(1)−(𝑥+1)(−1)
(3−𝑥)2
4
= 4𝑢3 ∙ (3−𝑥)2
But note again that the required derivative must be a function of 𝑥.
Thus, 𝒇
′ (𝒙)
=
𝟏𝟔𝒖𝟑
(𝟑−𝒙)𝟐
=
𝒙+𝟏 𝟑
)
𝟑−𝒙
(𝟑−𝒙)𝟐
𝟏𝟔(
=
𝟏𝟔((𝒙+𝟏))
(𝟑−𝒙)𝟓
𝟑
.
Remark: Consider our previous solution. Most mathematicians do not tolerate writing the
variables u and x simultaneously; that is,
16u3
f ′ (x) = (3−x)2 .
To avoid this
mathematical error in our solution, we introduce a shorter process by skipping
the step of representing 𝑦 = 𝐹(𝑢) and 𝑢 = 𝐺(𝑥), as shown in the next example.
107
Example 2.4.3.
Determine the derivative of 𝑓(𝑥) = √4 − 𝑥 2 .
Solution:
From the Chain Rule for Differentiation,
𝐼𝑓 𝑦 = 𝑓(𝑥) = 𝐹(𝐺(𝑥)) 𝑡ℎ𝑒𝑛
𝑑𝑦
= 𝑓 ′ (𝑥) = 𝐹 ′ (𝐺(𝑥)) ∙ 𝐺 ′ (𝑥)
𝑑𝑥
Now,
1
𝑓(𝑥) = √4 − 𝑥 2 = (4 − 𝑥 2 )2
Getting the derivative, we have,
𝑓 ′ (𝑥) =
=
1
1
(4 − 𝑥 2 )2−1 ∙ 𝐷𝑥 (4 − 𝑥 2 )
2
1
1
(4 − 𝑥 2 )−2 ∙ (−2𝑥)
2
𝑥
=−
1
(4 − 𝑥 2 )2
𝒇′ (𝒙) = −
𝒙
√𝟒−𝒙𝟐
Remark: Observe that skipping the step of representing 𝑦 = 𝐹(𝑢) and 𝑢 = 𝐺(𝑥) as
suggested requires us to identify the inner and outer functions of the composition.
We proceeded first to get the derivative of the outer function (in this case, the
radical) while retaining the inner function as its base. Then we multiplied the
resulting derivative of the outer function to the derivative of the inner function.
Example 2.4.4.
4
Determine the derivative of 𝑓(𝑥) = √2𝑥 + 3.
Solution:
1
4
𝑓(𝑥) = √2𝑥 + 3 = (2𝑥 + 3)4
Since the radicand is a function of 𝑥, the chain rule must be applied in obtaining the
required derivative.
Thus, we have:
𝑓 ′ (𝑥) =
1
1
(2𝑥 + 3)4−1 ∙ 𝐷𝑥 (2𝑥 + 3)
4
1
3
= 4 (2𝑥 + 3)−4 ∙ (2)
108
𝒇′ (𝒙) =
𝟏
𝟑
𝟐(𝟐𝒙+𝟑)𝟒
There are cases when the chain rule may coexist with the differentiation formulas
that we had discussed. Let us consider the following examples for this case.
Example 2.4.5.
Find the derivative of 𝑓(𝑥) = (5𝑥 2 − 2𝑥 + 3)4 (2𝑥 3 − 7)3 .
Solution:
Observing the given function, its derivative can be determined by first expanding its
factors. Then proceed to get the derivative of each of the terms of the resulting expansion,
which is degree 17. This method can be tedious and time-consuming.
So again, we will make use of the chain rule, along with other derivative formulas
discussed previously, for an alternative solution to this problem. Thus, the derivative of the
given function is as follows:
𝑓(𝑥) = (5𝑥 2 − 2𝑥 + 3)4 (2𝑥 3 − 7)3
𝑓 ′ (𝑥) = (5𝑥 2 − 2𝑥 + 3)4 ∙ 𝐷𝑥 (2𝑥 3 − 7)3 + (2𝑥 3 − 7)3 ∙ 𝐷𝑥 (5𝑥 2 − 2𝑥 + 3)4
= (5𝑥 2 − 2𝑥 + 3)4 ∙ [3(2𝑥 3 − 7)2 ∙ 𝐷𝑥 (2𝑥 3 − 7)]
+(2𝑥 3 − 7)3 ∙ [4(5𝑥 2 − 2𝑥 + 3)3 ∙ 𝐷𝑥 (5𝑥 2 − 2𝑥 + 3)]
= (5𝑥 2 − 2𝑥 + 3)4 ∙ [3(2𝑥 3 − 7)2 ∙ 6𝑥 2 ] +(2𝑥 3 − 7)3 ∙ [4(5𝑥 2 − 2𝑥 + 3)3 ∙ (10𝑥 − 2)]
= 18𝑥 2 (5𝑥 2 − 2𝑥 + 3)4 (2𝑥 3 − 7)2 +8(2𝑥 3 − 7)3 (5𝑥 − 1)(5𝑥 2 − 2𝑥 + 3)3
= 2(5𝑥 2 − 2𝑥 + 3)3 (2𝑥 3 − 7)2 [9𝑥 2 (5𝑥 2 − 2𝑥 + 3) + 4(2𝑥 3 − 7)(5𝑥 − 1)]
= 2(5𝑥 2 − 2𝑥 + 3)3 (2𝑥 3 − 7)2 (45𝑥 4 − 18𝑥 3 + 27𝑥 2 + 40𝑥 4 − 8𝑥 3 − 140𝑥 + 28)
= 𝟐(𝟓𝒙𝟐 − 𝟐𝒙 + 𝟑)𝟑 (𝟐𝒙𝟑 − 𝟕)𝟐 (𝟖𝟓𝒙𝟒 − 𝟐𝟔𝒙𝟑 + 𝟐𝟕𝒙𝟐 − 𝟏𝟒𝟎𝒙 + 𝟐𝟖)
(3𝑥 2 −2)
Example 2.4.6.
3
Find the derivative of 𝐺(𝑥) = (4−2𝑥−𝑥 3)2.
Solution:
Applying the chain rule for differentiation along with the quotient rule, we have:
2
𝐺
′ (𝑥)
=
𝐺 ′ (𝑥) =
3
2
[(4−2𝑥−𝑥 3 )2 ]2
2
𝐺 ′ (𝑥) =
3
(4−2𝑥−𝑥 3 ) ∙𝐷𝑥 [(3𝑥 2 −2) ]−(3𝑥 2 −2) ∙𝐷𝑥 [(4−2𝑥−𝑥 3 ) ]
2
3
(4−2𝑥−𝑥 3 ) ∙3(3𝑥 2 −2) ∙𝐷𝑥 (3𝑥 2 −2)−(3𝑥 2 −2) ∙2(4−2𝑥−𝑥 3 )∙𝐷𝑥 (4−2𝑥−𝑥 3 )
(4−2𝑥−𝑥 3 )4
2
2
3
(4−2𝑥−𝑥 3 ) ∙3(3𝑥 2 −2) ∙(6𝑥)−(3𝑥 2 −2) ∙2(4−2𝑥−𝑥 3 )∙(−2−3𝑥 2 )
(4−2𝑥−𝑥 3 )4
109
2
𝐺 ′ (𝑥) =
2(4−2𝑥−𝑥 3 )(3𝑥 2 −2) [9𝑥(4−2𝑥−𝑥 3 )−(3𝑥 2 −2)(−2−3𝑥 2 )]
(4−2𝑥−𝑥 3 )4
2
𝐺
′ (𝑥)
𝐺
′ (𝑥)
=
2(4−2𝑥−𝑥 3 )(3𝑥 2 −2) [36𝑥−18𝑥 2 −9𝑥 4 −(−6𝑥 2 −9𝑥 4 +4+6𝑥 2 )]
(4−2𝑥−𝑥 3 )4
2
=
2(4−2𝑥−𝑥 3 )(3𝑥 2 −2) [36𝑥−18𝑥 2 −9𝑥 4 +6𝑥 2 +9𝑥 4 −4−6𝑥 2 ]
(4−2𝑥−𝑥 3 )4
2
𝐺 ′ (𝑥) =
2(4−2𝑥−𝑥 3 )(3𝑥 2 −2) (−18𝑥 2 +36𝑥−4)
(4−2𝑥−𝑥 3 )4
𝟐
𝑮′ (𝒙) =
Example 2.4.7.
𝟒(𝟑𝒙𝟐 −𝟐) (−𝟗𝒙𝟐 +𝟏𝟖𝒙−𝟐)
(𝟒−𝟐𝒙−𝒙𝟑 )
𝟑
Find the derivative of 𝐻(𝑥) = (3 − 𝑥)11 √5 + 2𝑥.
Solution:
1
The given function can be written as 𝐻(𝑥) = (3 − 𝑥)11 (5 + 2𝑥)2 . Applying the
product rule, together with the chain rule, we have:
1
1
𝐻 ′ (𝑥) = (3 − 𝑥)11 𝐷𝑥 [(5 + 2𝑥)2 ] + (5 + 2𝑥)2 𝐷𝑥 [(3 − 𝑥)11 ]
1
1
1
= (3 − 𝑥)11 (2) (5 + 2𝑥)− 2 (2) + (5 + 2𝑥)2 (11)(3 − 𝑥)10 (−1)
1
1
= (3 − 𝑥)11 (5 + 2𝑥)− 2 − 11(5 + 2𝑥)2 (3 − 𝑥)10
1
= (3 − 𝑥)10 (5 + 2𝑥)− 2 [(3 − 𝑥) − 11(5 + 2𝑥)]
=
(3 − 𝑥)10 (3 − 𝑥 − 55 − 22𝑥)
1
(5 + 2𝑥)2
=−
(𝟑 − 𝒙)𝟏𝟎 (𝟐𝟑𝒙 + 𝟓𝟐)
𝟏
(𝟓 + 𝟐𝒙)𝟐
The chain rule for differentiation applies to algebraic functions and trigonometric
functions. The following theorem presents the chain rule formulas in getting the derivative
of trigonometric functions:
110
Theorem 12 (Chain Rule for Trigonometric Functions)
If 𝑢 is a function of 𝑥, then:
1. 𝐷𝑥 (sin 𝑢) = cos 𝑢 ∙ 𝐷𝑥 𝑢
2. 𝐷𝑥 (cos 𝑢) = − sin 𝑢 ∙ 𝐷𝑥 𝑢
3. 𝐷𝑥 (tan 𝑢) = sec 2 𝑢 ∙ 𝐷𝑥 𝑢
4. 𝐷𝑥 (cot 𝑢) = −csc 2 𝑢 ∙ 𝐷𝑥 𝑢
5. 𝐷𝑥 (sec 𝑢) = sec 𝑢 tan 𝑢 ∙ 𝐷𝑥 𝑢
6. 𝐷𝑥 (csc 𝑢) = − csc 𝑢 cot 𝑢 ∙ 𝐷𝑥 𝑢
The above formulas are almost similar to the formulas we encountered in the
previous section, except that the formulas in the theorem above have a multiplier of 𝐷𝑥 𝑢.
This multiplier implies that instead of the variable 𝑥 serving as the argument of the
trigonometric function, it is now the variable u that depends on other forms of the function
of 𝑥.
Example 2.4.8.
Find the derivative of 𝑓(𝑥) = sin(𝑥 3 )
Solution:
Since the argument is not simply x but a function of 𝑢 = 𝑥 3 , we apply the formula
𝐷𝑥 (sin 𝑢) = cos 𝑢 ∙ 𝐷𝑥 𝑢. Then we have:
𝑓 ′ (𝑥) = cos(𝑥 3 ) ∙ 𝐷𝑥 (𝑥 3 )
= cos(𝑥 3 ) ∙ 3𝑥 2
= 𝟑𝒙𝟐 𝐜𝐨𝐬(𝒙𝟑 )
Example 2.4.9.
2
Find the derivative of 𝐺(𝑥) = tan (𝑥)
Solution:
Using the formula 𝐷𝑥 (tan 𝑢) = sec 2 𝑢 ∙ 𝐷𝑥 𝑢, we have:
2
2
𝐺 ′ (𝑥) = sec 2 ( ) ∙ 𝐷𝑥 ( )
𝑥
𝑥
2
2
= sec 2 ( ) ∙ (− 2 )
𝑥
𝑥
=−
𝟐
𝟐𝐬𝐞𝐜 𝟐 (𝒙)
𝒙𝟐
111
Example 2.4.10.
Determine the derivative of 𝑦 = csc √𝑥
Solution:
Applying the formula 𝐷𝑥 (csc 𝑢) = − csc 𝑢 cot 𝑢 ∙ 𝐷𝑥 𝑢, we have:
𝑦′ =
𝑑𝑦
= − csc √𝑥 cot √𝑥 ∙ 𝐷𝑥 (√𝑥)
𝑑𝑥
= − csc √𝑥 cot √𝑥 ∙
=−
1
2√𝑥
𝐜𝐬𝐜 √𝒙 𝐜𝐨𝐭 √𝒙
𝟐√ 𝒙
Practice #11
Try this! ☺
Determine the derivatives of the following functions by using the chain rule. Show a
complete, neat & systematic solution. Express final answers in simplest form.
3
1. 𝐹(𝑥) = √4𝑥 2 − 2𝑥 + 1
2. 𝑓(𝑥) = (4 − 𝑥 3 )2 (2𝑥 2 + 1)3
1
3. 𝐺(𝑥) = cos ( 2 )
𝑥
4. 𝑔(𝑥) = cot(4𝑥 2 + 7)
5. 𝐻(𝑥) = sec 5 𝑥
Process Questions:
1. How do you differentiate using the chain rule?
2. What is the difference between chain rule and power rule?
Student’s responses:
Please write your answers on page 163.
Worksheet #5: (See page 156)
112
My Working Space
(write your solutions to the practice problems here)
113
114
Lesson 2.5
Optimization and Higher Order
Derivatives
Nowadays, business industries are trying to optimize the services they provide to the
people by maximizing their productivity and minimizing waste and other liabilities. Indeed,
by the advancement of technologies, this goal is made possible.
It is fascinating how mathematics has greatly influenced today’s advancement in
technology. Using technology, our work becomes easy, and productivity is doubled
compared to doing a specific task by hand. In this lesson, we will discuss one of the
applications of derivatives – optimization. By using the first derivative test, we can
determine the optimality of something in the context of the different fields of study such as
engineering, biology, economics, etc. How does this topic in Calculus applicable in the reallife setting? Let us see as we go further.
Increasing and Decreasing Functions
You might have an idea already about the terms increasing and decreasing. If an
employer tells the employee that his/her salary is increasing, the employee will think that
his/her salary is growing steadily over the term of the employment. On the other hand, if you
take out a loan from a bank, your debt will gradually decrease over time once you start
paying back the loan. A graph can represent the former and the latter situations for better
visualization. Now, we will formally define what increasing, and decreasing functions are.
Definition 4 (Increasing and Decreasing Functions)
i) A function 𝑓 is increasing on an interval (𝑎, 𝑏) if for every 𝑥1 , 𝑥2 ∈ (𝑎, 𝑏) with 𝑥1 < 𝑥2 , then
𝑓(𝑥1 ) < 𝑓(𝑥2 ).
ii) A function 𝑓 is decreasing on an interval (𝑎, 𝑏) if for every 𝑥1 , 𝑥2 ∈ (𝑎, 𝑏) with 𝑥1 < 𝑥2 , then
𝑓(𝑥1 ) > 𝑓(𝑥2 ).
Illustration 21
Note. From Calculus Applications of the Derivative, 2022, Math24
(https://math24.net/increasing-decreasing-functions.html). Copyright Math24.
115
In the graphs shown in illustration 21, we can immediately observe what part of the
graph the function is increasing or decreasing. But how about if we are only given a
mathematical formula for the function, can we describe the function's behavior
immediately?
Inequality 𝑥1 < 𝑥2 may somehow suggest a technique for us to determine whether
the function is increasing or decreasing, but it will be difficult since the given inequality
indicates that there are infinitely many values of 𝑥1 and 𝑥2 that we need to exhaust on a given
interval and the values of 𝑥1 and 𝑥2 that we are going to compare change from time to time.
Note that some functions may not satisfy the given definition of increasing and
decreasing functions over a broadly defined interval. But we can restrict the length of the
interval for us to determine whether a function is increasing or decreasing. Consider the
graph shown in illustration 22.
Illustration 22
Shown in illustration 22 is the graph of the function 𝑦 = 𝑥 3 + 4𝑥 2 . As we can see, we
cannot say that the function is increasing or decreasing in general in the interval (−5,2); but
if we are going to restrict the length of the interval, say (0,2), we can say that for 𝑥1 , 𝑥2 ∈
(0,2), such that 𝑥1 < 𝑥2 , we have 𝑓(𝑥1 ) < 𝑓(𝑥2 ), thus it is increasing.
116
Illustration 23
The graph of the function 𝑦 = 𝑥 3 + 4𝑥 2 is increasing on the interval (0, 2) as shown
in illustration 23. The graphical method may also be unreliable in the given definition of
increasing and decreasing functions. The definition requires us to test the function for all
pairs on the open interval, making it a strenuous task. So, to address this concern, derivatives
are used to check whether the function is increasing or decreasing.
Theorem 13 (Increasing and Decreasing Functions Involving Derivatives)
Suppose 𝑓 is differentiable on the open interval (𝑎, 𝑏)
i) If 𝑓′(𝑥) > 0 for all 𝑥 ∈ (𝑎, 𝑏), then 𝑓 is increasing on the interval (𝑎, 𝑏).
ii) If 𝑓′(𝑥) < 0 for all 𝑥 ∈ (𝑎, 𝑏), then 𝑓 is decreasing on the interval (𝑎, 𝑏).
Take note that the theorem suggests us to get a value of 𝑥 that belongs to the interval
(𝑎, 𝑏) and substitute it to the first derivative of the function, after which we will compare the
resulting value to 0.
117
Example 2.5.1.
Given the function 𝑓(𝑥) = 𝑥 2 + 2𝑥, determine whether the function is
increasing or decreasing on the given intervals.
a) (−3, −1)
b) (−1,5)
c) (−3,5)
Solutions:
a) (−3, −1)
First, we need to get the first derivative of 𝑓(𝑥) = 𝑥 2 + 2𝑥, which is 𝑓′(𝑥) = 2𝑥 + 2.
Now, we will pick any number, say −2, that belongs to the interval (−3, −1). Substitute −2
to the first derivative of the function.
𝑓′(𝑥) = 2𝑥 + 2
𝑓′(−2) = 2(−2) + 2
𝑓′(−2) = −2
So, 𝑓′(−2) = −2 < 0.
Since −2 represents all the values on the interval (−3, −1), and 𝑓′(−2) < 0, it will
follow that the function is decreasing on the interval (−3, −1) as shown in illustration 24.
Illustration 24
118
b) (−1,5)
Get the first derivative of 𝑓(𝑥) = 𝑥 2 + 2𝑥, which is 𝑓′(𝑥) = 2𝑥 + 2. Now, we will pick
any number, say 0, that belongs to the interval (−1,5). Substitute 0 to the first derivative of
the function.
𝑓′(𝑥) = 2𝑥 + 2
𝑓′(0) = 2(0) + 2
𝑓′(0) = 2
So, 𝑓′(0) = 2 > 0.
Since 0 represents all the values on the interval (−1,5), and 𝑓′(0) > 0, it will follow
that the function is increasing on the interval (−1,5) as shown in illustration 25.
Illustration 25
119
c) (−3,5)
Study the graph shown in illustration 26.
Illustration 26
As shown in illustration 25, the graph illustrates that the function 𝑓(𝑥) = 𝑥 2 + 2𝑥 is
decreasing on the interval (−3, −1) and increasing on the interval (−1,5). At a given value
𝑥 = −1, the behavior of the graph of the function changes, increasing to decreasing. So, what
can we say, in general, about the function's behavior? This situation will conclude that the
function is neither increasing nor decreasing on the specified interval (−3,5). It is said that
𝑥 = −1 serves as the critical value for the function 𝑓(𝑥) = 𝑥 2 + 2𝑥.
Critical Number of a Function
Definition 5 (Critical Number of a Function)
A number 𝒄 in the domain of function 𝑓 is called a critical number of 𝑓 if 𝒇′(𝒄) = 𝟎 or
𝒇′(𝒄) is undefined.
Example 2.5.2.
Find the critical numbers of the following functions.
a) 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5
𝑥
b) 𝑓(𝑥) = 𝑥 2 +4
d) 𝑓(𝑥) = cos 𝑥
3
e) 𝑓(𝑥) = √𝑥
2𝑥
c) 𝑓(𝑥) = 𝑥 2 −1
120
Solutions:
a) 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5
We will determine the first derivative of the function and find its critical number/s.
𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5
𝑓′(𝑥) = 6𝑥 2 − 6𝑥 − 12
From this, we can see that 𝑓′(𝑥) is defined for all values of 𝑥. Now, we will set 𝑓 ′ (𝑥) =
0 to find the critical numbers.
6𝑥 2 − 6𝑥 − 12 = 0
6(𝑥 2 − 𝑥 − 2) = 0
6(𝑥 − 2)(𝑥 + 1) = 0
𝒙 = 𝟐 𝑜𝑟 𝒙 = −𝟏
Therefore, the critical numbers are −𝟏 and 𝟐. Shown in illustration 27 is the graph
of the function y = 2x 3 − 3x 2 − 12x + 5 with critical numbers −1 and 2.
Illustration 27
121
𝑥
b) 𝑓(𝑥) = 𝑥 2 +4
Determine the first derivative of the function.
𝑓(𝑥) =
𝑓′(𝑥) =
𝑥2
𝑥
+4
(𝑥 2 + 4)(1) − (𝑥)(2𝑥)
(𝑥 2 + 4)2
𝑓′(𝑥) =
4 − 𝑥2
(𝑥 2 + 4)2
Even if the given function is rational, we can see that it is defined for values of 𝑥. Now,
let us proceed to finding the critical numbers of the function by letting 𝑓 ′ (𝑥) = 0.
𝑓′(𝑥) =
0=
4 − 𝑥2
(𝑥 2 + 4)2
4 − 𝑥2
(𝑥 2 + 4)2
0 = 4 − 𝑥2
𝒙 = −𝟐 𝑜𝑟 𝒙 = 𝟐
Therefore, the critical numbers are −𝟐 and 𝟐.
2𝑥
c) 𝑓(𝑥) = 𝑥 2 −1
Determine the first derivative of the function.
𝑓(𝑥) =
𝑓′(𝑥) =
2𝑥
𝑥2 − 1
(𝑥 2 − 1)(2) − (2𝑥)(2𝑥)
(𝑥 2 − 1)2
𝑓′(𝑥) =
−2𝑥 2 − 2
(𝑥 2 − 1)2
122
The function will be undefined if 𝑥 = −1 or 𝑥 = 1. But take note that these values are
not critical numbers because they are the exceptions in the domain of the function. Now, let
𝑓 ′ (𝑥) = 0.
−2𝑥 2 − 2
𝑓′(𝑥) = 2
(𝑥 − 1)2
0=
−2𝑥 2 − 2
(𝑥 2 − 1)2
0 = −2𝑥 2 − 2
2𝑥 2 = −2
𝑥 2 = −1
Clearly, we can say that there is no real solution to the equation above. Thus, it will
2𝑥
imply that 𝑓(𝑥) = 𝑥 2 −1 has no critical numbers as shown in illustration 28. It can be
observed that the given function is decreasing on the interval (−∞, −1) ∪ (−1, 1) ∪ (1, ∞).
Illustration 28
123
d) 𝑓(𝑥) = cos 𝑥
Determine the first derivative of the function.
𝑓(𝑥) = cos 𝑥
𝑓′(𝑥) = −sin 𝑥
The function is defined for all values of 𝑥 in the interval (−∞, ∞). Now, let 𝑓 ′ (𝑥) = 0.
𝑓′(𝑥) = −sin 𝑥
0 = −sin 𝑥
We know that a sine function will be 0 when the angle measure is a multiple of 𝜋.
From this, we can say that the critical numbers are values of 𝑥 such that 𝒙 = 𝒌𝝅, where 𝑘 ∈
𝑍.
3
e) 𝑓(𝑥) = √𝑥
Determine the first derivative of the function.
3
𝑓(𝑥) = √𝑥
2
𝑥 −3
𝑓′(𝑥) =
3
𝑓′(𝑥) =
1
3
3√𝑥 2
Looking at the derivative of the function, there is no such value of 𝑥 that will make
𝑓 ′ (𝑥) = 0. If we let 𝑥 = 0, the first derivative of the function will be undefined, but take note
3
that 0 is part of the domain of 𝑓(𝑥) = √𝑥. So, we can say that 0 is a critical number of the
function 𝑓.
Note to Self:
This is a safe space. You
can write anything here.
124
Absolute (Global) and Relative (Local) Maximum and Minimum of a Function
Definition 6 (Absolute Maximum and Minimum of a Function)
For a function 𝑓 defined on a set 𝑆 of real numbers and a number 𝑐 ∈ 𝑆,
i) 𝑓(𝑐) is the absolute maximum of 𝑓 on 𝑆 if 𝑓(𝑐) ≥ 𝑓(𝑥) for all 𝑥 ∈ 𝑆; and
ii) 𝑓(𝑐) is the absolute minimum of 𝑓 on 𝑆 if 𝑓(𝑐) ≤ 𝑓(𝑥) for all 𝑥 ∈ 𝑆.
An absolute maximum or minimum is referred to as an absolute extremum.
Definition 7 (Relative Maximum and Minimum of a Function)
i) 𝑓(𝑐) is a relative (local) maximum of 𝑓 if 𝑓(𝑐) ≥ 𝑓(𝑥) for all 𝑥 ∈ (𝑎, 𝑏) that contains 𝑐.
ii) 𝑓(𝑐) is a relative (local) minimum of 𝑓 if 𝑓(𝑐) ≤ 𝑓(𝑥) for all 𝑥 ∈ (𝑎, 𝑏) that contains 𝑐.
In either case, we call 𝑓(𝑐) as a relative extremum of 𝑓.
Note. From Calculus topics made easy, 2022, Cool Creative Calculus
(https://coolcreativecalculus.wordpress.com/topic-1/). Copyright Cool Creative Calculus.
125
Example 2.5.3.
Given 𝑓(𝑥) = 𝑥 2 − 4,
a) Locate any absolute extrema of the function on the interval (−∞, ∞)
b) Locate any absolute extrema of the function on the interval (−2,2)
c) Locate any absolute extrema of the function on the interval [−2,2]
Solutions:
a) Locate any absolute extrema of the function on the interval (−∞, ∞)
The given function is quadratic. So, we would expect that its graph would be a
parabola that opens upward as shown in illustration 29.
Illustration 29
Looking at the graph of 𝑓(𝑥) = 𝑥 2 − 4, we can see that its vertex is at (0, −4). From
this information, we can say that the absolute minimum (and relative minimum) of the
function is −4. How about its absolute maximum (or relative maximum), is there any? Now,
please observe that the endpoints of the graph of the function keep on going up; by this, we
can say that the function has no absolute maximum (nor relative maximum). This situation
will lead us to conclude that there is no maximum or minimum value that exists for a function
at times.
126
b) Locate any absolute extrema of the function on the interval (−2,2)
Illustration 30
Looking at the graph shown in illustration 30, we can say that the absolute minimum
(as well as the relative minimum) is still −4, and there is no absolute maximum (nor relative
maximum). You might be thinking that 0 is the absolute maximum value. But take note that
we have the open interval (−2,2), and in such cases, both −2 and 2 are not included as the
endpoints of the function's graph.
c) Locate any absolute extrema of the function on the interval [−2,2]
Illustration 31
It is observable from the graph shown in illustration 31 that the absolute minimum
(as well as the relative minimum) is −4. Now, let us check whether there will be an absolute
127
maximum (and relative maximum) given the interval [−2,2]. In this case, the endpoints −2
and 2 are in the interval [−2,2]. Here, 𝑓 assumes its absolute maximum at two points, (−2,0)
and (2,0). Hence, the absolute maximum (and the relative maximum) is 0. This example will
lead us to the following theorem.
Extreme Value Theorem (EVT)
A continuous function 𝑓 defined on a closed and bounded interval [𝑎, 𝑏] attains both an
absolute maximum and an absolute minimum on that interval.
Theorem 14
Suppose that 𝑓 is continuous on the closed interval [𝑎, 𝑏]. Then, the absolute extrema of
𝑓 must occur at an endpoint or at a critical number.
Theorem 14 gives us a simple procedure for finding the absolute extrema of a
continuous function on a closed and bounded interval:
a. Find all critical numbers in the interval and compute function values at these
points.
b. Compute function values at the endpoints.
c. The largest function value is the absolute maximum, and the smallest function
value is the absolute minimum.
Example 2.5.4.
2
Find the absolute extrema of 𝑓(𝑥) = 𝑥 on the interval [1,4].
Solution:
Observe that on the closed interval [1,4], function 𝑓 is continuous. Citing the EVT, we
can say that 𝑓 has absolute maximum and absolute minimum on the interval [1,4]. Looking
at the graph below, we can say that the function reaches its maximum at 𝑥 = 1 with the point
(1,2) and its minimum at 𝑥 = 4 with the point (4,0.5).
Illustration 32
128
Example 2.5.5.
interval [−2,4].
Find the absolute extrema of 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5 on the
Solution:
You might want to graph the given function first for you to have an idea about the
location of the absolute extrema and the relative extrema.
Illustration 33
From the graph shown in illustration 33, the maximum appears to be at the endpoint
𝑥 = 4, and the minimum seems to be at the relative minimum at 𝑥 = 2. Now, let us find the
critical numbers by determining its first derivative.
𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5
𝑓′(𝑥) = 6𝑥 2 − 6𝑥 − 12
𝑓′(𝑥) = 6(𝑥 2 − 𝑥 − 2)
𝑓′(𝑥) = 6(𝑥 − 2)(𝑥 + 1)
Critical Numbers: 𝑥 = −1 and 𝑥 = 2
We can see that both critical numbers belong to the interval [−2,4]. Now, we are going
to compare the values of 𝑓 by solving and comparing the values at the endpoints and the
values at the critical numbers to find the absolute extrema as well as the relative extrema.
𝑓(−2) = 1
𝑓(4) = 37
𝑓(−1) = 12
𝑓(2) = −15
129
In illustration 34, it appears to be that at 𝑓(4) = 37, we have our absolute maximum
and at 𝑓(2) = −15, we have our absolute minimum. It can be said that the absolute extrema
are also relative extrema on the interval [−2,4].
absolute maximum (also
a relative maximum)
relative maximum
relative minimum
absolute minimum (also
a relative minimum)
Illustration 34
First Derivative Test
Theorem 15
Suppose that 𝑓 is continuous on the interval [𝑎, 𝑏] and 𝑐 ∈ (𝑎, 𝑏) is a critical number.
i) If 𝑓 ′ (𝑥) > 0 for all 𝑥 ∈ (𝑎, 𝑐) and 𝑓 ′ (𝑥) < 0 for all 𝑥 ∈ (𝑐, 𝑏) (i.e. 𝑓 changes from
increasing to decreasing at 𝑐), then 𝑓(𝑐) is a relative maximum.
ii) If 𝑓 ′ (𝑥) < 0 for all 𝑥 ∈ (𝑎, 𝑐) and 𝑓 ′ (𝑥) > 0 for all 𝑥 ∈ (𝑐, 𝑏) (i.e. 𝑓 changes from
decreasing to increasing at 𝑐), then 𝑓(𝑐) is a relative minimum.
iii) If 𝑓 ′ (𝑥) has the same sign on (𝑎, 𝑐) and (𝑐, 𝑏), then 𝑓(𝑐) is not a relative extremum.
Example 2.5.6.
derivative test.
Find the relative extrema of 𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 24𝑥 − 10 using first
Solution:
Determine the first derivative of the function.
𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 24𝑥 − 10
130
𝑓′(𝑥) = 6𝑥 2 + 18𝑥 − 24
𝑓′(𝑥) = 6(𝑥 − 1)(𝑥 + 4)
We can say that there are two critical numbers of the function, namely −4 and 1. Now,
for us to determine which is the relative maximum or minimum, we are going to pick test
numbers from the interval (−∞, −4), (−4,1), and (4, ∞) and plug it in in the first derivative
of the function.
+
𝟎
𝟎
−
−𝟒
𝟎
+
𝟏
Observing the number line above the “plus” and “minus” signs describes the slope of
the function at a given critical number. Let us start first on the interval (−∞, −4). If we will
pick a test number, say −5, from this interval and plug it in 𝑓′(𝑥) = 6(𝑥 − 1)(𝑥 + 4), the
result would be 36, a positive number. So, in this case 𝑓 ′ (𝑥) > 0 and the function is increasing
(slope is positive) towards −4. Let us pick another test number from the interval (−4,1), say
0. If we plug in 0 in the function's first derivative, the result would be a negative number. So,
it means that the function is decreasing after −4. From the theorem, it is stated that if the
function is increasing at the start and after passing the critical number, it starts to decrease.
It is said that the corresponding critical number is a relative maximum. In this case, −4 is a
maximum. Citing the previous theorem, it can be concluded that 1 is a relative minimum
since the function is decreasing from its left and increasing from its right.
We can further observe this by looking at the graph of the function shown in
illustration 35.
Slope is positive from
the right of 1
Slope is 0 at 𝑥 = −4
Slope is negative from
the right of −4 and from
the left of 1
Slope is 0 at 𝑥 = 1
Slope is positive from the left
of −4
Illustration 35
131
Application of Derivative - Optimization
Example 2.5.7.
You have 40 linear feet of fencing to enclose a rectangular space for a
garden. Find the largest area that can be enclosed with this much fencing and the dimensions
of the corresponding garden.
Solution:
There are a lot of possible dimensions that we can configure from a 40 linear feet
fencing. The question now is how are we to decide which configuration is optimal? Let us
illustrate a picture and label it appropriately.
𝒚
𝑨 = 𝒙𝒚
𝒙
Figure 1
From figure 1, we want to find the dimensions with the largest possible area. Now, we
will maximize 𝐴 = 𝑥𝑦, the formula for finding the area of the garden.
We can observe that the given function contains two variables, 𝑥 and 𝑦. With what we
have discussed so far, we cannot proceed yet because the function involves two variables.
So, we will use the information given in the problem that the perimeter of this garden will be
40 linear feet. Now,
2𝑥 + 2𝑦 = 40
2𝑦 = 40 − 2𝑥
𝑦 = 20 − 𝑥
Now we can use this new information to maximize the garden's area.
𝐴 = 𝑥𝑦
𝐴 = 𝑥(20 − 𝑥)
As you can see, we now have our new function, which involves a single variable only.
But before we maximize 𝐴(𝑥), we need to determine the interval on which 𝑥 must lie. Since
𝑥 is a garden dimension, we must have 𝑥 ≥ 0. Since the perimeter is 40 feet, we must have
𝑥 ≤ 20, which will imply that we want to find the maximum value of 𝐴(𝑥) on the closed
interval [0,20].
132
Let us graph 𝐴(𝑥) = 𝑥(20 − 𝑥) for us to have an idea of the maximum value of the
function.
Illustration 36
The graph in illustration 36 shows that the maximum appears to occur around 𝑥 =
10. Now, let us verify this one by finding the critical number of the function.
𝐴(𝑥) = 𝑥(20 − 𝑥)
𝐴′ (𝑥) = (1)(20 − 𝑥) + (𝑥)(−1)
𝐴′ (𝑥) = 20 − 2𝑥
𝐴′ (𝑥) = 2(10 − 𝑥)
We can say that the function has only one critical number, 𝑥 = 10. Given the interval
[0,20] and the critical number 𝑥 = 10, we will compare the function values.
𝐴(𝑥) = 𝑥(20 − 𝑥)
𝐴(0) = 0
𝐴(20) = 0
𝐴(10) = 100
From the given function values above, we can say that the absolute maximum is 100
with 𝑥 = 10. It implies that the maximum area enclosed with 40 feet of fencing is 100 square
feet. To determine the dimensions of the garden, we will use 𝑥 = 10 and solve for 𝑦.
𝑦 = 20 − 𝑥
𝑦 = 10
Therefore, the garden should have a dimension of 10ft on each side to have an optimal
area which is 100 𝑓𝑡 2 .
133
Example 2.5.8.
A square sheet of cardboard 18" on a side is made into an open box by
cutting squares of equal size out of each corner and folding up the sides along the dotted
lines. Find the dimensions of the box that will yield a maximum volume.
Solution:
We can visualize the given problem more clearly if we are going to illustrate it. Study
the figure below.
𝟏𝟖
𝒙
𝒙
𝟏𝟖 − 𝟐𝒙
𝒙
𝒙
Figure 2
From figure 2, if the sides are folded up, we can make a rectangular prism with a
volume of
𝑉 = 𝑙𝑤ℎ
After folding up the sides, we can see that ℎ = 𝑥. This information will give us an idea
that length and width will be 18 − 2𝑥. Now, we can construct our new function defined as
𝑉(𝑥) = (18 − 2𝑥)(18 − 2𝑥)(𝑥)
𝑉(𝑥) = (18 − 2𝑥)2 (𝑥)
𝑉(𝑥) = 4𝑥(𝑥 − 9)2
In the given function, since 𝑥 is a measurement, we cannot allow that 𝑥 will be
negative, so we have 𝑥 ≥ 0. As for the corners of the cardboard, we have 𝑥 ≤ 9. Note that if
we cut squares with side 9" on each corner of the cardboard, we would end up cutting the
entire sheet of cardboard. We will now find the absolute maximum of the function on the
closed interval [0,9]. Let us illustrate first the graph of the function on the interval [0,9] so
that we will have an idea of the maximum value.
134
Illustration 37
It seems like the maximum value is 432 with 𝑥 = 3 as shown in illustration 37. Now,
let us verify this one by finding the critical number of the function.
𝑉(𝑥) = 4𝑥(𝑥 − 9)2
𝑉′(𝑥) = 4(𝑥 − 9)2 + (2)(𝑥 − 9)(1)(4𝑥)
𝑉′(𝑥) = 4(𝑥 − 9)[(𝑥 − 9) + 2𝑥]
𝑉′(𝑥) = 4(𝑥 − 9)(3𝑥 − 9)
From the first derivative of the function, we can see that it has two critical numbers,
3 and 9. These critical numbers belong to the interval [0,9]. Now, we will compare the
function values based on the values on the endpoints and the critical numbers to find the
absolute maximum of the function.
𝑉(𝑥) = 4𝑥(𝑥 − 9)2
𝑉(0) = 0
𝑉(9) = 0
𝑉(3) = 432
We can conclude from the given function values that the maximum possible volume
is 432 cubic inches. We can achieve this volume by cutting 3" out of each cardboard corner.
Further, we can say that for the open box to have an optimal volume; it should have the
following dimensions:
Length = 12 inches
Width = 12 inches
Height = 3 inches
135
Example 2.5.9.
You want to construct a box whose base length is 3 times the base
width. The material for building the top and bottom of the box costs Php 500 per square foot,
and the material to be used for the sides costs Php 300 per square foot. If the box must have
a volume of 80 cubic feet, determine the dimensions that will minimize the cost to build the
box.
Solution:
First, we need to illustrate this box to have a clear picture of its dimensions as shown
in figure 3.
H
L
W
Figure 3
The dimensions of this box are as follows:
𝐿𝑒𝑛𝑔𝑡ℎ = 3𝑥
𝑊𝑖𝑑ℎ𝑡 = 𝑥
𝐻𝑒𝑖𝑔ℎ𝑡 = 𝑦
Now, we need to have our working equation for the volume of the box.
𝑉 = 𝑙𝑤ℎ
𝑉 = (3𝑥)(𝑥)(𝑦) = 80
3𝑥 2 𝑦 = 80
Since we need to have a single variable for our first derivative, we need to do
something about our working equation above.
3𝑥 2 𝑦 = 80
𝑦=
80
3𝑥 2
Now, we will construct our working equation for the cost (which is to be minimized)
in making the box.
136
𝐶 = (2)(500)(3𝑥 2 ) + (2)(300)(3𝑥𝑦) + (2)(300)(𝑥𝑦)
𝐶 = 3000𝑥 2 + 1800𝑥𝑦 + 600𝑥𝑦
𝐶 = 3000𝑥 2 + 2400𝑥𝑦
80
Substituting 𝑦 = 3𝑥 2 , we have
80
𝐶 = 3000𝑥 2 + 2400𝑥 ( 2 )
3𝑥
𝐶 = 3000𝑥 2 +
64000
𝑥
𝐶(𝑥) = 3000𝑥 2 +
64000
𝑥
We are now ready to determine the derivative of the function 𝐶 in terms of 𝑥.
𝐶(𝑥) = 3000𝑥 2 +
64000
𝑥
𝐶(𝑥) = 3000𝑥 2 + 64000𝑥 −1
𝐶 ′ (𝑥) = 6000𝑥 − 64000𝑥 −2
𝐶 ′ (𝑥) = 6000𝑥 −
64000
𝑥2
Now, we will let 𝐶 ′ (𝑥) = 0 to determine the critical numbers of the function and
eventually find the absolute minimum.
6000𝑥 −
64000
=0
𝑥2
6000𝑥 =
64000
𝑥2
6000𝑥 3 = 64000
𝑥3 =
3
32
3
𝑥=√
32
3
137
3
32
Since we now have our critical number, let us test whether √ 3 is a minimum or
3
32
maximum value using the first derivative test. Take note that √ 3 ≈ 2.20. So, we might want
3
32
3
32
to pick test numbers from these intervals: (0, √ 3 ) and (√ 3 , ∞). Note that from the given
intervals, we do not allow values below 0 because the function will be undefined, and we
3
32
cannot observe the behavior of the function. Picking 1 from the left of √ 3 and 3 from the
3
32
right of √ 3 will tell us that our critical number is the relative minimum of the function. We
can verify this one by looking at the graph in illustration 38.
Illustration 38
Now, let us compute the minimum cost of constructing the box.
𝐶(𝑥) = 3000𝑥 2 +
64000
𝑥
138
2
3
𝐶 (√
32
32
64000
) = 3000 ( √ ) +
3
3
3 32
√
3
3
3
𝐶 (√
32
) ≈ 43,610.89
3
Therefore, the minimum cost to construct a box with a volume of 80 cubic feet is
Php 43,610.89.
Higher-Order Derivatives
A higher-order derivative is a notion such that after solving for the first derivative of
a function, which is also a function, we can still proceed with solving the derivative of the
derivative of the original function. We can continue this process of solving the derivatives up
to 𝑛𝑡ℎ times. Study table 8 below.
Order
Prime Notation
1
𝑦′ = 𝑓′(𝑥)
Leibniz Notation
𝑑𝑓
𝑑𝑥
𝑦′′ = 𝑓′′(𝑥)
𝑑2𝑓
𝑑𝑥 2
𝑦′′′ = 𝑓′′′(𝑥)
𝑑3𝑓
𝑑𝑥 3
2
3
4
5
n
(𝑥)
𝑑4𝑓
𝑑𝑥 4
(𝑥)
𝑑5𝑓
𝑑𝑥 5
𝑦 (𝑛) = 𝑓 (𝑛) (𝑥)
𝑑𝑛 𝑓
𝑑𝑥 𝑛
𝑦
(4)
𝑦
(5)
=𝑓
(4)
=𝑓
(5)
Table 8
139
Example 2.5.10.
possible.
If 𝑓(𝑥) = 2𝑥 4 − 4𝑥 3 − 5𝑥 2 − 3, compute as many derivatives as
Solution:
𝑓(𝑥) = 2𝑥 4 − 4𝑥 3 − 5𝑥 2 − 3
𝑓 ′ (𝑥) =
𝑑𝑓
= 8𝑥 3 − 12𝑥 2 − 10𝑥
𝑑𝑥
𝑓 ′′ (𝑥) =
𝑑2𝑓
= 24𝑥 2 − 24𝑥 − 10
𝑑𝑥 2
𝑑3𝑓
𝑓 (𝑥) = 3 = 48𝑥 − 24
𝑑𝑥
′′′
𝑓
(4)
𝑑4𝑓
(𝑥) = 4 = 48
𝑑𝑥
𝑓 (5) (𝑥) =
𝑑5 𝑓
=0
𝑑𝑥 5
We can continue this process and gets the same answer which is 0. So instead of
exhausting and listing all the derivatives up to the nth order, we will just write the following:
𝑓 (𝑛) (𝑥) =
𝑑𝑛 𝑓
= 0, 𝑓𝑜𝑟 𝑛 ≥ 5.
𝑑𝑥 𝑛
Second Derivative Test
Theorem 16
Suppose that 𝑓 is continuous on the interval (𝑎, 𝑏) and 𝑓 ′ (𝑐) = 0, for some number 𝑐 ∈ (𝑎, 𝑏).
i) If 𝑓 ′′ (𝑐) < 0, then 𝑓(𝑐) is a relative maximum.
ii) If 𝑓 ′′ (𝑐) > 0, then 𝑓(𝑐) is a relative minimum.
Example 2.5.11.
𝑥 4 − 8𝑥 2 + 10.
Use the second derivative test to find the relative extrema of 𝑓(𝑥) =
Solution:
𝑓(𝑥) = 𝑥 4 − 8𝑥 2 + 10
𝑓′(𝑥) = 4𝑥 3 − 16𝑥
140
From the first derivative, we see that the critical numbers are −2, 0, and 2. Now we
shall continue solving for the second derivative.
𝑓′(𝑥) = 4𝑥 3 − 16𝑥
𝑓′′(𝑥) = 12𝑥 2 − 16
𝑓 ′′ (−2) = 32 > 0
𝑓 ′′ (0) = −16 < 0
𝑓 ′′ (2) = 32 > 0
Looking at the solutions above, we can say that 𝒇(𝟎) is a relative maximum while
𝒇(−𝟐) and 𝒇(𝟐) are relative minima. This is evident in the function’s graph shown in
illustration 39.
Illustration 39
Example 2.5.12.
You have 40 linear feet of fencing to enclose a rectangular space for a
garden. Find the largest area that can be enclosed with this much fencing and the dimensions
of the corresponding garden.
Solution:
We have solved this problem already in the previous sections. Based on the
conditions given in the problem, we have our working equation with its first derivative and
critical number.
𝐴(𝑥) = 𝑥(20 − 𝑥)
𝐴′ (𝑥) = (1)(20 − 𝑥) + (𝑥)(−1)
𝐴′ (𝑥) = 20 − 2𝑥
141
𝐴′ (𝑥) = 2(10 − 𝑥)
From this, we can say that the critical number of the function is 10. But how do we
know that this value will be a relative maximum or relative minimum? Yes, we will use the
second derivative test and check whether it is a maximum or a minimum. Now,
𝐴′ (𝑥) = 2(10 − 𝑥)
𝐴′′ (𝑥) = −2
𝐴′′ (10) = −2 < 0
Based on the second derivative test, we can see that 𝐴′′ (𝑐) = 𝐴′′ (10) < 0. It means
that the critical number 10 is a relative maximum of the function, and we have verified that
using the second derivative test. Therefore, the garden should have a dimension of 10ft on
each side to have an optimal area which is 100 𝑓𝑡 2 .
Note:
If 𝑓 ′′ (𝑐) = 0 or 𝑓 ′′ (𝑐) is undefined, the Second Derivative Test will have no meaning. That is,
𝑓(𝑐) may be a relative maximum, relative minimum, or neither. If this happens, we must rely
on the information from the First Derivative Test to determine whether 𝑓(𝑐) is a relative
extremum.
Practice #12
Try this! ☺
For numbers 1 & 2, determine the following:
i.
critical point/s
ii.
interval/s where the function is increasing
iii.
interval/s where the function is decreasing
iv.
relative maximum point/s
v.
relative minimum points/s
vi.
absolute extrema
1. 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
𝑥
2. 𝑔(𝑥) = 𝑥 2 −1
For numbers 3 to 5, answer as required.
3. A box with no top is to be built by taking 16"-by-10" sheet of cardboard and cutting
𝑥 inches square out of each corner and folding up the sides. Find the value of 𝑥 that
maximizes the volume of the box.
35𝑥−𝑥 2
4. A company’s revenue for selling 𝑥 (thousand) items is given by 𝑅(𝑥) = 𝑥 2+35 . Find
the value of 𝑥 that maximizes the revenue and find the maximum revenue.
5. Given 𝑓(𝑥) = 𝑥 4 + 3𝑥 2 − 2, find 𝑓 4 (𝑥).
142
Process Questions:
1. What does it mean when we say the function is increasing or decreasing?
2. What is the significance of having critical values in a function? Explain.
3. How does testing the first derivative help us find the absolute extrema of the
function? Explain.
4. What is the purpose of determining the first derivative of a function? How
about the second derivative?
5. Do you think a function may have no derivative? If so, cite an example and
explain.
Student’s responses:
Please write your answers on page 163.
Worksheet #6: (See page 157)
Note to Self:
This is a safe space. You can write
anything here.
143
My Working Space
(write your solutions to the practice problems here)
144
REFLECTION
Reflective Essay
Guide Questions:
1. What part of the lesson did you find easy?
2. What part of the lesson did you find challenging that you think you need to focus on?
3. How will you relate the concept you have learned to your life right now?
4. Why is it necessary for STEM learners to master the concepts of limits, continuity, and
derivatives?
5. In general, how can you relate limits, continuity, and derivatives with what is
happening today?
Please write your reflection on page 164.
Answer the L column of the KWL chart
Write three (3) specific ideas you have learned from the topics discussed.
K
(What I know)
W
(What I want to know)
L
(What I learned)
145
146
ACTION
MINI TASK 1
Solving Problems Involving Limits and Continuity Through Video Presentation
You will be given problem sets involving Limits and Continuity. You will answer
these problem sets and create a pre-recorded video on how you solved each problem in the
background. You will also explain the solution process by emphasizing essential topics
discussed in class. The video presentation may include concept sharing, solving techniques,
applications, etc. You may use applications like MS PowerPoint, online whiteboard,
physical whiteboard, manila paper, etc., depending on the availability of resources you
have at home.
File Type: MPEG-4 (.mp4)
File Name: STEM 308_MT1_Section_Group Number
Processing Questions:
RUBRIC FOR MINI TASK 1
How did you find the mini- task?
CRITERIA
EXEMPLARY
EFFICIENT
PROFICIENT
LIMITED
SCORE
________________________________________________________________________________________________________________
________________________________________________________________________________________________________________
The video covers all the
The video covers all the
The video covers only
The video is missing essential
problems given in the
problems given in the
some of the problems given
contents based on the given
problem set, explicitly shows
problem set, shows and
in the problem set and
problem sets; showed the
How did the lessons
help
real-world
thethe
topic?
and explains
the you
processsee
of the
explains
the process ofuse ofshows
process of
process of solving the
solving each problem,
solving each problem,
solving each problem; the
problem but is incorrect or
evidently planned the
plans the outline of the
outline of the video
unclear; outline of the video
________________________________________________________________________________________________________________
outline of the video
video presentation, and
presentation is evident but
presentation is not evident,
CONTENT AND
________________________________________________________________________________________________________________
presentation by smoothly
emphasizes major
could be improved, and
and lacks emphasis in
ORGANIZATION
transitioning from one
components of the topic.
some major components of
explaining the major
_
problem to another. Major
the topic are not given
components of the topic.
components of the topic are
emphasis.
given emphasis.
15
PRESENTATION SKILLS
CREATIVITY
13
12
10
Excellent transitions to
connect a topic to the other
and sustain the viewer’s
attention
Segments of the video are
connected well;
transitions are effectively
observed and
communicate with the
viewers.
Segments of the tutorial
are not connected
efficiently; there is limited
to no use of transitions and
verbal cues to link ideas.
The tutorial presenter is not
linking ideas and is just barely
reading scripts.
15
12
10
8
Attractive and creative
design of slides/board.
Effective way of using
images, text, and effects to
deliver their message. It
leaves a lasting impact on
the viewer.
Attractive and creative
design of slides/board.
The use of images, text,
and effects are
appropriately used to
deliver their message.
The effort is shown in
creating attractive and
appealing slide/board
designs. Able to use
images, text, and effects
together.
Shows little effort in
portraying creativity in the
slides/boards. The use of text,
images, and effects have little
to no impact on the delivery
of the message.
10
8
TOTAL
6
4
/40
147
BASIC CALCULUS PERFORMANCE TASK
The BASCAL Company is one of the popular packaging industries in the city. You are working in
that company as a paper designer where your job is to come up with different designs and sizes of
boxes. One client contracted your company to come up with a box design for a certain product. The
condition of your manager is to maximize the material to be used per box so that no material
will be wasted as well as to identify the maximum volume of the box which it can hold. As a
paper designer, you are tasked to come up with the dimensions of the box, its maximum volume,
and the proposed price per box which you will submit to your manager. Your written report will
be evaluated according to the organization of your output, use of mathematical concepts, accuracy
of computation, promptness of submission, and completeness of necessary information.
The design of your box must include the following:
a. The dimensions of the material to be used in making the box
b. The actual area of the box used
c. The dimensions of the box at its maximum volume
d. The maximum volume of the box
e. The proposed price of the box
PERFORMANCE TASK RUBRIC
CRITERIA
EXEMPLARY
EFFICIENT
PROFICIENT
LIMITED
Sufficient mathematical
concepts support the
report. The use of ideas is
evident in each
statement.
The use of the concepts in
certain parts are confusing
and erroneous.
Little or no attempt has been
made to utilize concepts
USE OF MATHEMATICAL
CONCEPTS (CONTENT)
The report is supported
extensively by appropriate
mathematical concepts and
in detail. The use of the ideas
is awe-inspiring.
ACCURACY OF
COMPUTATIONS
COMPLETENESS OF
INFORMATION AND
ORGANIZATION
15
13
12
10
All computations are
correct with detailed
solutions, and the graphical
presentation is wellconstructed and easy to
understand.
All computations are
made with all correct
solutions with graphical
representation.
There are minor errors in
the computations.
Most of the computations are
erroneous.
15
12
10
8
The report contains all the
information needed; labeling
is evident and very
organized.
The report contains
complete information
with all the labels but
organization is not
evident.
The report lacks some
information though the
organization is evident.
The written output does not
contain the necessary
information or data of
graphical representation and
references.
10
8
6
4
Submit the written report on
or before the deadline.
Submit the output 1-2
days late from the
deadline.
Submit the output 3 or 4
days late from the day of
the deadline.
Submit the output 5 or more
days late from the day of the
deadline.
5
4
3
2
PROMPTNESS
TOTAL
SCORE
/45
148
ANSWERS TO PRE-TEST
1)
2)
3)
4)
5)
b
b
d
c
b
6) a
7) a
8) c
9) b
10) c
11) a
12) b
13) c
14) d
15) b
16) a
17) a
18) c
19) a
20) b
21) a
22) c
23) d
24) a
25) b
ANSWERS TO PRACTICE PROBLEMS
Practice #1:
1) 3
2a) 1
2b) −1
2c) 𝐷𝑁𝐸
2d) 𝐷𝑁𝐸
2e) 5
1
3) 0
4) 0
5) 𝐷𝑁𝐸
1
3) − 8
Practice #2:
1) 12
2) − 6
Practice #3:
1) − 1
3
2) − 8
1
4) 4
Practice #4:
Test A
a)
b)
c)
d)
e)
−2
−2
−2
1
1
f) 1
g) −2
h) 1
i) 𝐷𝑁𝐸
j) 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
k) 1
l) −2
2) 5
3) +∞
Test B
1) −∞
149
Practice #5:
1)
1
5) 𝐷𝑁𝐸
2
2) 1
3) 4
4)
9) 1
1
6) 3
7) 0
1+√3
10) 1
8) 𝐷𝑁𝐸
2
Practice #6:
1) 𝑓 is not continuous at 𝑎 = −1. Since it is not defined at 𝑎 and it does not satisfy (ii),
therefore the type of discontinuity involved is essential.
2) 𝑔 is a continuous function at 𝑎 = 4.
Practice #7:
1) continuous
2) continuous
3) discontinuous
4) discontinuous
5) continuous
Practice #8:
3
1) a. 𝑚 = −2, 𝑦 = −2𝑥 − 1
2) a. 𝑦 ′ = 6𝑥
3
b. 𝑚 = − 2 , 𝑦 = − 2 𝑥 + 14
15
b. 𝑓 ′ (𝑥) = − 𝑥 4
𝑑𝑦
11
c. 𝑑𝑥 = − (3𝑥−2)2
Practice #9:
a) i. The function is differentiable at 𝑎 = 0.
ii. The function is continuous at 𝑎 = 0.
iii. The function is not differentiable at [−1,1].
iv. The function is not continuous at [−1,1].
b) i. The function is not differentiable at 𝑎 = 1.
ii. The function is continuous at 𝑎 = 1.
iii. The function is differentiable at (1, 2).
iv. The function is continuous at (1, 2).
150
Practice #10:
1) 𝑓′(𝑥) = 𝑥 2 (4𝑥 − 1)(20𝑥 − 3)
2) 𝑦′ =
−30𝑡 4 − 15𝑡 2 + 4𝑡
7) 𝑓′(𝑤) = sec 3 𝑤 + sec 𝑤 tan2 𝑤
(6𝑡 2 + 1)2
3) 𝑉 = 2π𝑟3 and RInst =150𝜋
4) 𝐴 =
πx2
8
6) 𝑔′(𝑧) = 10 sec 2 𝑧 + 2 csc 2 𝑧
8) ℎ′(𝑡) = 3𝑡 2 − 2𝑡 sin 𝑡 − 𝑡 2 cos 𝑡
π
+ 2𝑥 2 and RInst = 2 + 8 ≈ 9.57
5) 𝑓 ’(𝑥) = −2 sin 𝑥 − 6 sec 𝑥 tan 𝑥
𝑥2
9) 𝑦′ = 𝑥 cos ( 2 )
𝑑𝑣
10) 𝑑𝜃 = 2 sec 2(2𝜃)
Practice #11:
1) 𝐹 ′ (𝑥) =
8𝑥−2
5) 𝐻 ′ (𝑥) = 5 sec 5 𝑥 tan 𝑥
2
3(4𝑥 2 −2𝑥+1)3
2) 𝑓 ′ (𝑥) = 6𝑥(𝑥 3 − 4)(2𝑥 2 + 1)2 (4𝑥 3 + 𝑥 − 8)
1
3) 𝐺
′ (𝑥)
=
2 sin( 2 )
𝑥
𝑥3
4) 𝑔′ (𝑥) = −8𝑥 csc 2 (4𝑥 2 + 7)
Practice #12:
𝑥
−𝑥 2 −1
1) 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2; 𝑓 ′ (𝑥) = 3𝑥 2 − 3
2) 𝑔(𝑥) = 𝑥 2 −1; 𝑔′ (𝑥) = (𝑥 2 −1)2
a.
b.
c.
d.
e.
f.
a. none
b. none
c. (−∞, −1), (−1,1), & (1, +∞)
d. none
e. none
f. none
𝑥 = −1 and 𝑥 = 1
(−∞, −1) 𝑜𝑟 (1, ∞)
(−1,1)
(−1,4) relative maximum
(1,0) relative minimum
No absolute extrema
3) 𝑥 =
8−√19
3
≈ 1.21
4) 𝑥 = 5 and 𝑅(𝑥) = 2.5 (maximum revenue)
5) 24
151
EVALUATION
Worksheet #1: Limits Part 1
Evaluate the following limits. (2pts each)
1. lim (𝑥 2 + 2𝑥 − 5)3
𝑥→−8
3. lim
𝑥→
8−4𝑥
𝑥→2 2𝑥 2 −𝑥−6
5. lim
𝑥 2 −2𝑥
𝑥→5 𝑥−5
√2𝑥−5−1
𝑥 2 −9
𝑥→3
7. lim
9. lim
𝑥→2
2. lim3
3−√𝑥 2 +5
𝑥 2 −2𝑥
−𝑥 4 +5𝑥+6
𝑥−3
2
3
4. lim √
𝑥→−3
6. lim
2𝑥 2 −3𝑥+5
𝑥+6
𝑥 2 −4𝑥
𝑥→0 𝑥 3 −3𝑥 2 −2𝑥
8. lim
5𝑥−4
𝑥→7 𝑥 2 −3𝑥−28
10. lim (
3𝑥 2 +2𝑥
𝑥→2 3𝑥 2 −2𝑥−3
3
)
152
Worksheet #2: Limits Part 2
Evaluate the following limits.
One-sided limit
⎹ 𝑥 + 2 ⎸ + 1, 𝑥 < −1
1. Given 𝑓(𝑥) = { −𝑥 + 1, − 1 ≤ 𝑥 ≤ 1
𝑥 2 − 2𝑥 + 2,
𝑥>1
a.
(3 pts)
b. lim+ 𝑓(𝑥)
lim 𝑓(𝑥)
𝑥→ 1−
𝑥→ 1
c. lim 𝑓(𝑥)
𝑥→ 1
Infinite Limits
2. 𝑓(𝑥) =
𝑥 2 − 3𝑥 + 2
a.
𝑥+1
(3 pts)
lim 𝑓(𝑥)
𝑥→− 1−
b. lim + 𝑓(𝑥)
𝑥→ −1
c. lim 𝑓(𝑥)
𝑥→ −1
Limits at Infinity
5𝑥 2 − 3
3. lim
(2 pts)
𝑥→∞ 8𝑥 2 −2𝑥 + 1
Infinite Limits at Infinity
1 + 𝑥2
4. lim
𝑥→∞ 1 + 𝑥
(2 pts)
Limits of Trigonometric Functions
𝑡𝑎𝑛2 𝑥 cos 3𝑥
5. lim𝜋
𝑥→
𝑥
2
sec −cot 2𝑥
3
sin 2𝑥
(2 pts)
1−cos 8𝑡
(2 pts)
6. lim sin 3𝑥
𝑥→0
7. lim
(2 pts)
𝑡→0 3 sin 3𝑡
𝑠𝑖𝑛2 𝑟
(2 pts)
tan 4𝑦
(2 pts)
8. lim 4𝑟 2+3𝑟
𝑟→0
9. lim
𝑦→0
4𝑦
153
Worksheet #3: Continuity
A. Determine if each of the given function is continuous at the given value of 𝑎. If the function
is discontinuous, state which of the conditions is/are not satisfied and determine what type
of discontinuity is involved. (2pts each)
1. 𝑓(𝑥) =
3𝑥 2 +2
𝑥+2
2. 𝑔(𝑥) =
;𝑎 = 2
2
3. 𝑘(𝑥) = {√9 −2 𝑥 , −3 ≤ 𝑥 < 3 ; 𝑎 = 3
𝑥
, 𝑥≥3
𝑥 2 −1
𝑥−1
;𝑎 = 1
3𝑥
4. ℎ(𝑥) = √𝑥 3
+𝑥 2
;𝑎 = 0
1
5. 𝐽(𝑥) = cos (𝑥) ; 𝑎 = 0
B. Determine if each of the given functions is continuous on each given intervals. (5pts each)
4
1. 𝑓(𝑥) = 𝑥+1
a. (−∞, −1)
c. (−2, −1]
b. (−2, −1)
d. [−1,1]
e. (−2, ∞)
𝑥 2 − 1 , 𝑥 < −5
2. 𝑔(𝑥) = {√25 − 𝑥 2 , −5 ≤ 𝑥 ≤ 5
5−𝑥 , 𝑥 >5
a. (−∞, −5)
c. [−5,5]
b. (−6, −4]
d. [−5,5)
e. (4, ∞)
154
Worksheet #4: Derivatives Part 1
Answer what is asked.
1. Find the slope of the graph of 𝑦 = 2𝑥 − 3 at the point (2,1). (2pts)
2. Find the slope of the tangent line to the graph of 𝑓(𝑥) = √𝑥 at the point (1,1). (2pts)
3. Find the equation of the tangent line to the curve defined by 𝑓(𝑥) = 𝑥 − 2𝑥 2 at the
point (1, −1). (3pts)
2𝑥
4. Determine the slope of the tangent line to the graph of 𝑓(𝑥) = 3𝑥−1 at the point
(1,1). Then give the equation of the tangent line in slope-intercept form. (𝟑𝐩𝐭𝐬)
𝑥+5
5. The function 𝑓(𝑥) = 𝑥 2+2𝑥+1 is not differentiable at what value of 𝑎? Explain your
answer. (1pt)
6. Which of the following is a function continuous at the specified value of 𝑎 but not
differentiable at 𝑎? Explain your answer. (1pt)
a. 𝑓(𝑥) = 𝑥 2 − 2𝑥 ; 𝑎 = 1
b. 𝑓(𝑥) = |𝑥| ; 𝑎 = 0
1
c. 𝑓(𝑥) = 2−𝑥 ; 𝑎 = 3
3
d. 𝑓(𝑥) = 𝑥+1 ; 𝑎 = 0
For numbers 7 and 8, consider each given function, number 𝑎, and the interval. Then
answer the following questions and show your solutions:
a. Is the function differentiable at the given number?
b. Is the function continuous at the given numbers?
c. Is the function differentiable at the given interval?
d. Is the function continuous at the given interval?
2
if 𝑥 > 0
7. 3. 𝑓(𝑥) = { 3 − 𝑥
; 𝑎 = 0 ; [ 0, 2 )
−2𝑥 + 3 if 𝑥 ≤ 0
2𝑥
8. 4. 𝑓(𝑥) = 3−𝑥 ; 𝑎 = 2 ; ( 0, 3 )
(4pts)
(4pts)
155
Worksheet #5: Derivatives Part 2
Solve as required. (2 points each)
𝑑𝑦
1. 𝑦 = 𝑥 2 + 3, find 𝑑𝑥 .
2. 𝑓(𝑥) = (𝑥 + 1)(𝑥 2 + 3), find 𝑓′(𝑥).
3. 𝑦 =
𝑥 2 + 3𝑥 − 4
𝑑𝑦
2𝑥 + 1
𝑑𝑥
, find
.
4. Differentiate: 𝑓(𝑥) = 4sec𝑥 − 2csc𝑥
5. Find the derivative: 𝑦 = 3sec𝑥 (tan𝑥)
𝑑𝑦
6. Use the Chain Rule to find
𝑦 = (𝑢 − 2)3 𝑎𝑛𝑑 𝑢 =
𝑑𝑥
in terms of 𝑥.
1
2𝑥 + 1
𝑑𝑦
7. Solve for 𝑑𝑥 and simplify the result.
(𝑎) 𝑦 = (3𝑥 − 2)5 (2𝑥 2 + 5)6
5𝑥−1 3
(𝑏) 𝑦 = (
)
2𝑥+3
(𝑐) 𝑦 = (cos 3𝑥)2
(𝑑) 𝑦 = tan(sin 2𝑥)
156
Worksheet #6: Optimization & Higher Order Derivatives
A. Determine the critical number/s of each function if there is/are any. (2pts each)
1. 𝑓(𝑥) = 2𝑥 + 3
4. 𝑓(𝑥) = 𝑥 4 − 4𝑥 3 + 12
2. 𝑓(𝑥) = 6𝑥 2 − 𝑥 − 2
5. 𝑓(𝑥) = 𝑥 5 − 5𝑥 3 + 10𝑥 + 4
3. 𝑓(𝑥) = 2𝑥 3 + 6𝑥 + 7
B. For each function, determine the following: a) critical point/s; b) interval/s where the
function is increasing; c) interval/s where the function is decreasing; d) relative maximum
point/s; e) relative minimum points/s; f) absolute extrema, if there are any. (7pts each)
1
1. 𝑓(𝑥) = − 3 (𝑥 3 + 2𝑥 2 + 𝑥 − 1)
2. 𝑓(𝑥) = 2𝑥 3 + 3𝑥 2 − 12𝑥 − 10
C. Problem Solving (6pts)
1. A rectangular field is to be covered by 800-meter of hog wire in such a way that one side
has a river for its boundary. What is the largest area that can be enclosed and what are the
dimensions of the field?
Length=?
Area=?
Width=?
RIVER
157
PROCESS QUESTIONS
Process Questions (from page 25)
1. How to determine the limit of a function through table of values? graphing?
2. How to determine that the limit of a function does not exist through tables of
values? graphing?
3. Why is it incorrect to say that say that the lim 𝑓(𝑥) “equals DNE”, nor do we write
𝑥→𝑎
“lim 𝑓(𝑥) = 𝐷𝑁𝐸?
𝑥→𝑎
4. Is it correct to say that the lim 𝑓(𝑥) = 𝑓(𝑎) since 𝑎 is in the domain of 𝑓? Why? Why
not?
𝑥→𝑎
Student’s responses:
Process Questions (from page 30)
1. What could be the possible cases that the limit of a function does not exist?
2. In which real-life situation can you apply the practical (may not be mathematical)
concept of limit?
3. How does limit describes the behavior of a function?
4. What concepts in algebra are needed in evaluating limits of a function?
Student’s responses:
158
Process Questions (from page 32)
1. What are the strategies that need to be mindful of in solving the actual limit of the
0
function when its initial limit is indeterminate of type " 0 ".
2. How important are the concepts of factoring or rationalizing when the limit of the
0
function is indeterminate of type " 0 "?
Student’s responses:
Process Questions (from page 49)
1. When we conclude that the limit of a certain function is −∞ or +∞, does the limit
really exist? Explain.
0 ∞
2. Is it meaningful or possible to operate infinities such as: 0, ∞, or ∞ − ∞? Explain
your answer for each of the three situations mentioned.
Student’s responses:
159
Process Questions (from page 58)
1. Is there a limit involving trigonometric functions that will not exist? Cite an
example.
2. Is there any case where the limit of trigonometric functions can be determined by
observing some values of the function or its graph? Cite an example.
Student’s responses:
Process Questions (from page 64)
1. By just looking at the graph of any function, can you really tell whether continuity
or discontinuity exists? Explain.
2. Do you think all quadratic functions are continuous? Why?
Student’s responses:
160
Process Questions (from page 69)
1. What do you need to be mindful of when determining the continuity of a rational
function on a given interval? Why?
2. What do you need to be mindful of when determining the continuity of a piecewise
function on a given interval? Why?
Student’s responses:
Process Questions (from page 81)
1. How to describe the relationship between the slope of the tangent line to a curve
and the limit definition of the derivative?
2. What is the difference between derivative and differentiation?
Student’s responses:
161
Process Questions (from page 87)
1. How to describe the relationship between differentiability and continuity of a
function?
2. How to determine if the given function is differentiable at the given point?
3. How to determine if the given function is differentiable at the given interval?
Student’s responses:
Process Questions (from page 104)
Using the rules discussed in this lesson, can we possibly differentiate some functions
such as:
1. (𝑥 2 + 5)30 ? (polynomial function)
2. √𝑥 2 + 5 ?
(radical function)
Explain each.
Student’s responses:
162
.
Process Questions (from page 112)
1. How do you differentiate using the chain rule?
2. What is the difference between chain rule and power rule?
Student’s responses:
Process Questions (from page 143)
1. What does it mean when we say the function is increasing or decreasing?
2. What is the significance of having critical values in a function? Explain.
3. How does testing the first derivative help us find the absolute extrema of the
function? Explain.
4. What is the purpose of determining the first derivative of a function? How
about the second derivative?
5. Do you think a function may have no derivative? If so, cite an example and
explain.
Student’s responses:
163
Reflective Essay (from page 145)
Guide Questions:
1. What part of the lesson you find easy?
2. What part of the lesson did you find challenging that you think you need to focus on?
3. How will you relate the concept you have learned to your life right now?
4. Why is it necessary for STEM learners to master the concepts of limits, continuity, and
derivatives?
5. In general, how can you relate limits, continuity, and derivatives with what is
happening today?
164
Congratulations!
You are about to complete this learning packet. Now, answer the postassessment to check how well you learned.
----POST TEST---1. Which of the following is the value of
a. 0
b. 1
c. 2
2. Which of the following is the
a.
√2
3
1
b. 2
c.
b. 4
6. Given that
a. (−∞, −2)
d.
√42
6
.
c. 2
d. 0
c. −∞
d. +∞
.
b. 1
5. Evaluate
a. 0
√3
2
, evaluate
4. Evaluate
a. −1
d. 3
?
3. Given
a. 8
?
.
b. 1
c.
√2
4
d. −
√2
4
, the x-values where the function is continuous are ________.
b. (2, +∞)
c. (−∞, 2) ∪ (2, +∞)
d. [2, +∞)
165
7. Which of the following is the value of
a. 0
b. 1
c. 2
8. Which of the following is the
a.
−√252
b.
6
9. Given that
?
d. 3
?
3
− √252
c.
6
3
√252
6
d.
√252
6
, the x values where the function is continuous are ________.
a. (−∞, 3)
b.(3, +∞)
10. Evaluate
c. [3, +∞)
d. (−∞, 3) ∪ (3, +∞)
c. −∞
d. +∞
.
a. −1
b. 1
11. Which of the following is the derivative of 𝑓(𝑥) = √100𝜋 ?
a. 0
b. −5
12. If
then what is
b. − 𝑥 9
13. If
d. 5𝜋
c. 𝑥 8
d. 𝑥 9
c. 12(3𝑥 − 5)3
d. 𝑥
c. 8𝑥 + 1
d.
?
16
a. 8𝑥 7
c. 𝜋
16
then what is 𝑦′?
a. 4(3𝑥 − 5)3
14. What is
b. −5
if
a. 4𝑥 2 − 3𝑥
?
b.
1
2√4𝑥 2 −3𝑥+1
8𝑥−3
2√4𝑥 2 −3𝑥+1
166
15. What is 𝐷𝑥 [(8𝑥 + 3)2 (4 − 5𝑥)]?
a. −960𝑥 2 + 272𝑥 + 237
c. −80𝑥 + 17
b. −960𝑥 2 + 32𝑥 + 147
d. 4 − 5𝑥
16. What is
a.
?
−47
(8𝑥+3)2
b. −
−47
c.
1
5
2(4−5𝑥)2 (8𝑥+3)2
47
−5𝑥
d. (8𝑥+3)2
3
1
2(8𝑥+3)2 (4−5𝑥)2
17. Which of the following is the derivative of
a. cos 3𝑥 + sin 4𝑥
b. cos 𝑥
?
c. 3 cos 3𝑥 + 4 sin 4𝑥
d. sin 2𝑥
3
18. Which of the following is the derivative of 𝑓(𝑥) = √tan 3𝑥 ?
a. 𝑥(sec 𝑥)2
b.
sec2 3𝑥
1
(tan 3𝑥)3
c. 𝑥(sec 𝑥)2 + tan 𝑥
b. 1
4−5𝑥
20. If 𝑓(𝑥) = 8𝑥+3, then what is
a.
47
1
3
±
−47
√151
3
1
d. 4
?
55
1
b. 3
−55
c. 144
21. Find the critical number/s of
a.
2
at 𝑥 = 1.
c. −1
b. 121
121
sec 2 3𝑥
(tan 3𝑥)3
19. Find the slope of the tangent line to the curve
a. 0
d.
d. 144
.
1
c. 2
22. Determine the absolute maximum of the function
d. 1
over the interval,
, if it exists.
a. (0.5, 15)
b. (2.5,17.4)
c. (2, 15)
d. (1,12)
167
23. Which of the following is the local maximum of the function
the interval,
?
a. (0,1)
b. (2, −3)
24. Given the function
a. 0
c. (1, −1)
over
d. (−2, −19)
, solve for the 4th derivative.
b. 192
c. 24 + 192𝑥
d. 96𝑥 2 + 24𝑥 − 12
25. A farmer has 2400𝑓𝑡 of fencing and wants to fence off a rectangular field that borders a
straight river. He needs no fence along the river. What are the dimensions of the field that
has the largest area?
a. 2,200𝑓𝑡 𝑏𝑦 100𝑓𝑡
c. 1,000𝑓𝑡 𝑏𝑦 700𝑓𝑡
b. 400𝑓𝑡 𝑏𝑦 1,000𝑓𝑡
d. 1,200𝑓𝑡 𝑏𝑦 600𝑓𝑡
168
SELF-ASSESSMENT
Before saying that you are done with STEM 308, kindly complete the table below.
Lesson
Worksheet No.
Definition and
Illustration of Limits;
Limit Theorems;
Indeterminate Forms
1
One-sided Limits,
Infinite Limits, Limits at
Infinity, and Infinite
Limits at Infinity; Limits
of Trigonometric
Functions
2
Continuity at a Point;
Continuity on an
Interval
3
Actual Date of
Completion
Were you
able to
complete the
task on
time?
Reason(s):
Mini Task 1
The Tangent Line and
the Derivative;
Differentiability and
Continuity of Functions
4
Theorems of
Differentiation and
Instantaneous Rate of
Change; Chain Rule for
Differentiation
5
Optimization and
Higher-order
Derivatives
6
Performance
Task
Congratulations! You have completed the learning
packet and finished the Fourth Quarter.
169
REFERENCES
Arceo, C. P. P., Lemence, R. S., Ortega Jr., O. M., & Vallejo, L. J. D. (2016). Teaching Guide for
Senior High School Basic Calculus (J. E. C. Lope & M. P. Roque, Eds.; Issue June).
Commission on Higher Education, 2016.
Azad, Kalid. (2022, February 17). BetterExplained. Retrieved from An Intuitive
Introduction to Limits: https://betterexplained.com/articles/an-intuitiveintroduction-to-limits/
(2021, January 27). MathIsFun. Retrieved from Limits (Formal Definition):
https://www.mathsisfun.com/calculus/limits-formal.html
(2022). Definition of Continuity [Graph]. Math Warehouse.
https://www.mathwarehouse.com/calculus/continuity/continuity-definitions.php
(2022). Calculus Applications of the Derivative [Graph]. Math24.
https://math24.net/increasing-decreasing-functions.html
(2022). Calculus topics made easy [Graph]. Cool Creative Calculus.
https://coolcreativecalculus.wordpress.com/topic-1/
Egarguin, N. J., Fontanil, L., & Lawas, V. (2017). Basic CALCULUS for Senior High School. C&E
Publishing, Inc.
170