Yong-Jung Kim
25 Lectures for
Undergraduate Calculus I
February 24, 2025
Department of Mathematical Sciences, KAIST
2
Copyright 2024. Dept. of Mathematical Sciences, KAIST. All rights reserved.
To the ones who question
Foreword
The need for revising the first-year calculus curriculum has been raised for a long time.
This is not due to changes in calculus itself but rather because the current curriculum
does not reflect the high school education experienced by incoming students and because
recent technological advancements have shifted the aspects of calculus that need to be
emphasized.
Since calculus is one of the first major mathematical subjects encountered in college,
the primary focus has been on increasing students’ interest and motivation in their studies.
To achieve this, a new structure was adopted that emphasizes learning mathematical facts
through the process of achieving key objectives rather than simply listing them. The first
half of Calculus I is centered around understanding Kepler’s laws mathematically. In Part
I, the concept of limits is introduced, leading to the study of differentiation and integration. Newton invented differentiation to explain planetary motion. In Part II, students learn
about vectors, including velocity, acceleration, and gravity in three-dimensional space.
Through this, Kepler’s laws are proven using the two-body problem. The second half of
Calculus I focuses on the development and understanding of approximation techniques.
Part III covers the core mathematical techniques of integration and differentiation, and
Part IV approaches sequences and series from the perspective of approximation methods.
In particular, the goal is to provide a mathematical understanding of essential approximation techniques for engineering students.
The long-term practice of achieving scientific objectives may feel more challenging
since it is an experience not commonly encountered in secondary education. Unlike a simple enumeration of fragmented facts, the process of building a conceptual tower—where
understanding previous concepts is necessary to grasp subsequent ones—will be a valuable experience and greatly benefit students in their scientific research careers.
This lecture note is revised annually, reflecting feedback from professors who previously taught the course, students, teaching assistants, and the department. This is the advantage of having an in-house textbook. Students are encouraged to share their feedback
through their teaching assistants.
Daejeon, August 2023
KAIST Department of Mathematical Sciences Calculus
Curriculum Reform Committee
vii
Preface
Lecture notes are created for lectures. They are written with students in mind, not for
a vague general audience. The key difference from academic textbooks is that they are
designed to help students follow the course. To this end, the goal was to create a text that
feels like a conversation between the instructor and students. Efforts have been made to
present appropriate questions that encourage students to think independently. Sometimes,
rather than providing detailed explanations, posing suitable hints and motivating questions
can be more effective in helping students find answers on their own.
Most books used in actual lectures include elements of both academic textbooks and
lecture notes, but they sometimes fall into an ambiguous category. ”25 Lectures for Undergraduate Calculus I” stays true to the characteristics of a lecture note, adopting a format that fosters interaction between students and instructors. The main components are
well-structured questions and problems designed to help students achieve understanding
through independent thinking. Additionally, considering that many universities structure
a semester as approximately 25 lectures of 75 minutes each, this book is organized into
25 lectures to fit within a 14-week schedule, with 1-2 lectures per week.
Questions are the driving force of learning and the starting point of creative thinking.
It is hoped that through contemplating and answering the questions and problems in this
lecture note, students will deepen their understanding of calculus. This lecture note was
originally written in Korean, and the English version was translated using ChatGPT and
subsequently reviewed and edited.
Daejeon, August 2024
Yong-Jung Kim
ix
To KAIST Students Taking This Course
A semester typically consists of about 25 lectures of 75 minutes each, and this lecture note
is structured accordingly. Reading the lecture material beforehand and preparing questions will be beneficial for your studies. Even if time is limited, at least review the lecture
objectives before attending. The first two weeks of Calculus I are crucial. Unlike courses
that rely heavily on memorization, falling behind in the first two weeks makes it difficult
to catch up later. Attempting to solve the problems in each lecture independently before looking at the solutions will aid comprehension. Each lecture includes several questions—take the time to think about them. Mathematics is best learned at a steady pace;
taking time to reflect will be more effective. At the end of each lecture, there are practice
problems that help reinforce and review the material. If you feel additional problems are
needed, you are encouraged to work on exercises from standard calculus textbooks.
Questions are the driving force of learning and the starting point of creative thinking.
It is hoped that through contemplating and answering the questions and problems in this
lecture note, students will deepen their understanding of calculus.
Daejeon, August 2024
Professor Yong-Jung Kim, KAIST Department of Mathematical Sciences
xi
Differences Between College and High School
Learning
The following is a summary of a column by Professor Steven Zucker of Johns Hopkins
University, published in the March 2003 issue of AMS Notices. This column discusses key
aspects that help students transitioning into college understand the differences between
high school and college learning and adapt successfully. Although this article was written
22 years ago with American students in mind, consider its relevance as you read.
1. A New Level of Responsibility
• In college, students are responsible for their own learning.
• Professors and advisors provide guidance, but the primary responsibility lies with
the students themselves.
2. A New Peer Group
• Most students are no longer ahead of their peers; they will compete with students
of similar abilities.
• Excelling in college math courses (earning A or B grades) requires consistent effort,
not just talent.
• Failures (students who fail usually do so due to a lack of effort, poor time management, or reliance on high school study habits) prove this point.
3. Changes in Learning Expectations
• College learning aims to enable flexible application of knowledge in new situations.
• Conceptual thinking is crucial in mathematics, providing a strong foundation for
other fields.
4. Changing Roles of Professors and Students
• Professors guide learning rather than simply teaching all content; students must
engage with textbooks and actively learn.
• Students should spend at least two hours each week reviewing lectures, reading
textbooks, and solving exercises. Classes move faster and with less repetition.
xiii
xiv
Differences Between College and High School Learning
• Exams cover several weeks or an entire semester’s content, requiring cumulative
knowledge rather than short-term memorization.
5. Utilizing Learning Resources
• Using textbooks effectively is essential in college.
• While textbooks may have been less emphasized in high school, they are crucial for
deeply understanding concepts and examples in college.
6. Conclusion
• College demands greater autonomy and responsibility. Understanding and preparing for these differences is essential for success.
Contents
Part I Differentiation: Mathematical Description of Motion
1
Limit and Continuity #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Common Language Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Quality Control and ε-δ Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Limit and Continuity #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 Rigorous definitions using ε-δ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Limits as x → ∞ and f (x) → ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.1 Rate of Increase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 Intermediate and Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4
Chain Rule and Variable Centered Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Variable Centered Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5
Integration & Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . 37
5.1 Antiderivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.2 Area Function as an Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.3 Riemann Sum and Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6
Inverse Functions and Their Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
6.1 Bijection (One-to-One and Onto Function) . . . . . . . . . . . . . . . . . . . . . . . . . . 47
6.2 Inverse Functions and Variable-Oriented Notation . . . . . . . . . . . . . . . . . . . . 50
6.3 Integration by Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
7
Logarithm, Exponential, & Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . 57
7.1 Natural Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
7.2 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
7.3 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3
4
7
xv
xvi
Contents
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
Part I
Differentiation: Mathematical Description of
Motion
During Newton’s era (1643–1727), the most captivating scientific topic was the motion of
celestial bodies. Galileo (1564–1642) had already dismantled the geocentric model and
paved the way for the heliocentric theory. Kepler (1571–1630) described the orbits and
motions of celestial bodies. The study of celestial motion was one of the most pressing scientific issues of the time. However, these descriptions were based solely on observational data and did not provide an underlying explanation. More importantly, there
was no mathematical framework capable of explaining such motion. Newton, who developed a mathematical tool to describe dynamic phenomena beyond static ones like the
Pythagorean theorem, created calculus as a result. Using calculus, Newton explained the
motion of celestial bodies.
The space age has now arrived in Korea. The subject best suited to rigorously and
deeply teach the motion of celestial bodies is not general physics but calculus. The key
concepts of calculus, such as differentiation, integration, the chain rule, polar coordinates,
and variable transformations, are essential to understanding Newton’s celestial theories.
In Part I, we develop the concept of differentiation, a tool for describing dynamic motion. Studying calculus while imagining how Newton developed it to explain planetary
motion makes understanding its content and value much more engaging. In Part II, we
apply these concepts to mathematically express and determine the orbits of planets and
satellites, ultimately explaining celestial motion, including Kepler’s laws. Through this
process, we become familiar with differentiation and gain experience in applying it to
specific situations.
Differentiation is not limited to celestial motion; it is widely used in many fields. We
will explore the fundamental concepts of calculus from various perspectives, not just in
relation to planetary motion. Specifically, we will address concepts such as convergence,
continuity, and differentiability using the ε-δ method. This method, developed after Newton’s time, allows us to approach differentiation with greater rigor than Newton himself
did.
Lecture 1
Limit and Continuity #1
1 Calculus extensively uses the concept of limits. Not only are differentiation and integration defined through limits, but more fundamental concepts like convergence and continuity are also understood via limits. In real-world applications, we often seek approximate
values rather than exact ones. A key question is: ”If we improve the accuracy, does the
approximation converge to the true value, or is there an inherent limitation preventing it
from reaching the exact value?” Since limits and convergence appear repeatedly in many
areas beyond calculus, it is crucial to understand them clearly.
In this lecture, we will explore the concepts of limits, convergence, and continuity using
everyday language.
Use of Symbols
Mathematics is often described as the study of numbers, but in reality, it heavily relies on
symbols. Functions are commonly represented by f , g, variables by x, y, z, and constants
by a, b, c. Since the Latin alphabet alone is insufficient, Greek letters are frequently used
as well. The commonly used Greek lowercase letters include:
1. α (alpha), β (beta), γ (gamma), ω (omega), σ (sigma), θ (theta), ρ (rho), φ (phi),
2. ε (epsilon), δ (delta), λ (lambda), τ (tau).
The commonly used Greek uppercase letters include:
1. Ω (Omega), Σ (Sigma), ∆ (Delta).
Distinguishing between different symbols helps clarify meaning, which is especially
useful when advancing to deeper mathematical concepts. Becoming familiar with these
notations while studying the textbook is essential.
1 As in most mathematics courses, the first two weeks of a calculus course are the most critical. A clear
understanding of the early concepts makes the later material easier to follow. If one fails to grasp the
beginning, the subsequent content also becomes challenging.
3
4
1 Limit and Continuity #1
1.1 Common Language Definitions
Let us explore the properties of limits and continuity using everyday language. Suppose a
function f is defined on an interval [a, b] ⊂ R. This is denoted as:
f : [a, b] → R
The interval [a, b], which includes both a and b and all numbers between them, is called a
closed interval. Using set notation, it is written as [a, b] = {x ∈ R : a ≤ x ≤ b}. Excluding
the endpoints gives an open interval, written as (a, b).
If a variable x ∈ [a, b] approaches a specific value c ∈ (a, b) and the function value f (x)
approaches a specific value L, we call L the limit of f (x) as x approaches c. This is written
as:
lim f (x) = L
x→c
We also say that f (x) converges to L as x approaches c. For a function to have a limit, x
must approach c from both directions and yield the same value L.
In one-dimensional space R, there are two directions: left and right. The right-hand
limit, when x approaches c from the right, is denoted as:
lim f (x) = L
x→c+
Similarly, the left-hand limit is written as limx→c− f (x) = L. The limit exists if and only
if both the left and right limits are equal:
lim f (x) = L
x→c
⇔
lim f (x) = lim f (x) = L
x→c−
x→c+
Here, the symbol ⇔ means ”if and only if.”
The concept of continuity at a point c depends on the function value f (c). A function
f is continuous at c if the following four values exist and are equal:
lim f (x) = lim f (x) = lim f (x) = f (c)
x→c+
x→c−
x→c
The following is the first problem of this lecture note. Generally, mathematical textbooks provide examples and theorems, but we mostly handle these in the form of problems. This approach encourages students to think about the meaning and principles from
the perspective of solving problems before explanations or proofs are provided. Furthermore, framing concepts as questions can make their meaning clearer. Problems should
not be solved once and forgotten; most of them should be continuously utilized. Among
them, some are called theorems, rules, or laws, which are particularly important.
Attempting to answer a problem before reading the solution or proof can lead to a faster
understanding and is beneficial for developing mathematical skills.
Problem 1.1. Determine the left-hand and right-hand limits of the following functions
at the given point c, and verify their continuity at that point according to the definition
discussed earlier.
1.1 Common Language Definitions
5
2
−1
(1) sin x at c = 0.
(2) cos x at c = 0.
(3) xx−1
at c = 1.
f (x) at c = 0, when f (x) is the Heaviside function,
(
0, if x < 0
f (x) =
1, if x ≥ 0.
1
(4) x−1
at c = 1.
(5)
Solution 1.1 Knowing the graph of a function allows for an easier computation of onesided limits.
(1,2) lim sin x = 0 and lim cos x = 1. In both cases, the limit exists at c = 0 and equals the
x→0
x→0
function value, thus the functions are continuous.
2 −1
2 −1
(3) limx→1 xx−1
= 2. Since xx−1
has an undefined denominator at x = 1, it is not defined
at that point. However, the limit exists. To claim continuity, the function must be defined
2 −1
= x + 1 at x = 1.
as xx−1
1
(4) The limit lim
does not exist since the function diverges as x approaches 1, meanx→1 x − 1
ing no finite limit L exists. Therefore, it cannot be made continuous at x = 1 regardless of
how the function is defined there.
(5) The limit lim f (x) does not exist since f (x) approaches 1 from the right and 0 from
x→0
the left at x = 0. Because the left and right limits differ, the limit does not exist, making
the function discontinuous at x = 0. Furthermore, redefining the function value at x = 0
cannot make it continuous. ⊔
⊓
Mathematical theorems and laws consist of conditions and conclusions, which is common in most scientific disciplines. However, sometimes, by examining a conclusion, one
can infer the minimal conditions required for it to hold, even without explicitly stating
them.2 Let us practice reading mathematical contexts by examining the following rules.
Problem 1.2 (Rules of Limits). The following are some fundamental rules concerning
limits. Identify the conditions and conclusions required for these expressions to hold, and
explain their significance.
(1) lim ( f (x) + g(x)) = lim f (x) + lim g(x).
x→c
x→c
x→c
(2) lim ( f (x) − g(x)) = lim f (x) − lim g(x).
x→c
x→c
x→c
(3) lim (k f (x)) = k(lim f (x)),
x→c
x→c
where k is a constant.
(4) lim ( f (x)g(x)) = (lim f (x))(lim g(x)).
x→c
x→c
x→c
Solution 1.2 Consider the conditions and conclusions necessary for these expressions
to hold. The condition for (1) is that the limits lim f (x) and lim g(x) must exist. If these
x→c
x→c
conditions hold, the conclusion is that the function f + g also has a limit at x = c, and that
limit is the sum lim f (x) + lim g(x). The same reasoning applies to the other cases. ⊔
⊓
x→c
x→c
Question 1.1. Consider proving the four rules above. What is the difference between a
proof and an explanation?
2 Sometimes minimal conditions are sufficient, but not always. This makes mathematics even more in-
triguing.
6
1 Limit and Continuity #1
Explaining and understanding mathematical definitions and facts in one’s own words is
crucial for learning mathematics. However, explanations alone sometimes lack the rigor
required for mathematical proofs. A proof must adhere to the formal structure and level
of rigor expected by mathematicians. The ε-δ arguments in the next section illustrate this
rigor, which is not always required in calculus but is essential in certain proofs.
Problem 1.3 (Limit of Quotient). The quotient rule for limits is stated as follows:
lim f (x)
f (x) x→c
=
.
x→c g(x)
lim g(x)
lim
x→c
Identify the necessary conditions and explain the conclusion.
Solution 1.3 Unlike Problem 1.2, this case has a specific issue: division by zero. The
conditions are that lim f (x) and lim g(x) exist, but additionally, lim g(x) ̸= 0 must hold
x→c
x→c
x→c
to ensure the right-hand expression is well-defined. This also implies that g(x) ̸= 0 as x
approaches c. ⊔
⊓
One of the most commonly used and important functions is the power function, defined
as f (x) = xα for a given real exponent α. When dealing with limits of such functions,
special considerations arise.
Problem 1.4 (Limit of Power Functions). Identify the necessary conditions for the following power rule to hold:
α
lim ( f (x))α = lim f (x) .
x→c
x→c
Solution 1.4 When a power function is given, the necessary conditions depend on the
exponent α.
(1) If α is a positive integer, applying the product rule from Problem 1.2 α times suffices,
so the only requirement is that the limit lim f (x) exists.
x→c
(2) If α is a negative integer, the function takes the form of a fraction. Therefore, the
denominator must not be zero, meaning an additional requirement is that the limit exists
and is nonzero.
(3) If α is a positive real number, we need the condition lim f (x) ≥ 0, since raising a
x→c
negative number to a real exponent is not defined.
(4) If α is a negative real number, we require the condition lim f (x) > 0. ⊔
⊓
x→c
Determining whether a function is continuous is important. The above limit laws are
equally applicable to verifying the continuity of the following functions.
Problem 1.5 (Rules of Continuity). Let k ∈ R be a real number, and let f and g be
functions that are continuous at c. Show that the following functions are continuous at c.
Note that some cases require additional conditions. Identify those cases and specify the
necessary conditions.
(1) f + g
(2) f − g
(3) k f
(4) f g
(5) f /g
(6) f k
1.2 Quality Control and ε-δ Arguments
7
Solution 1.5 (1-4) The first four cases hold without additional conditions, as in Problem
1.2.
(5) In the case of the fraction f /g, the denominator must not be zero, so we require the
additional condition g(c) ̸= 0.
(6) For the power function f k , additional conditions depend on k. If k is a positive integer,
no extra condition is needed. If k is negative, we require f (c) ̸= 0, and if k is a real number,
f must be nonnegative. ⊔
⊓
Problem 1.6 (Continuity of a Composition). ””Let g : R → R be continuous at c ∈ R and
f : R → R be continuous at g(c) ∈ R. Then, the composition function ( f ◦g)(x) := f (g(x))
is continuous at c.”” Does this statement seem correct? Explain in your own words.
Solution 1.6 Since g is continuous at c, as x approaches c, g(x) approaches g(c). Since f
is continuous at g(c), f (g(x)) approaches f (g(c)). Thus, the composite function f (g(x))
is continuous at c. ⊔
⊓
The second sentence in the solution treats g(x) as if it were a single variable, stating that
g(x) approaches g(c). This is an appropriate perspective for dealing with composite functions. However, something still seems missing—like the fine line between an explanation
and a proof.
1.2 Quality Control and ε-δ Arguments
This section is written to aid the understanding of ε-δ arguments in the next chapter. The
ε-δ argument is a useful and natural concept. For example, when controlling the quantity
or quality of products in a factory, one must properly regulate the materials and production
environment. In this context, we can say, ””To control the output quality within an error
of ε, we must control the input quality within an error of δ .””3
Problem 1.7 (Quality Control). When a factory produces products, if there are no impurities in the raw material, the defect rate √
is 0. Suppose that when the impurity level is x µg
per gram of material, the defect rate is x%. (1) The factory manager receives an order
from the CEO to reduce the defect rate to at most 4% (ε = 4%). What impurity level must
be maintained to achieve this? (2) If the order changes to reducing the defect rate to at
most 0.4%, what should the impurity level be?
√
Solution 1.7 (1) The condition x < 4 implies x < 16. Thus, the impurity level must be
maintained below δ = 16µg √
per gram.
(2) Similarly, the condition x < 0.4 gives x < 0.16. Thus, the impurity level must be
maintained below δ = 0.16µg per gram. In this case, reducing the defect rate by a factor
of 10 required reducing the impurity level by a factor of 100. ⊔
⊓
In the problem above, when an order is given to reduce the defect rate below a certain
threshold, the factory manager must determine the necessary impurity level. Such scenarios are common in real life. In mathematics, the desired margin of error in the output
3 When there is a single input factor, a univariate function is used; when there are multiple input factors,
a multivariable function is used. In Calculus I, we deal with univariate functions.
8
1 Limit and Continuity #1
is traditionally denoted by ε > 0, while the corresponding control range for the input is
denoted by δ > 0. In many cases, eliminating impurities entirely is impossible. The best
one can do is minimize the control range as much as possible. This situation frequently
occurs in various practical contexts.
Problem 1.8 (Quality Control with Continuity). A traditional vinegar brewery aims for
an optimal acidity level of pH 3. To achieve this, the ingredients must be properly fermented, and under the factory’s conditions, an acidity of pH 3 is reached at a fermentation rate of 36%. Let y denote the
√ acidity (pH) and x the fermentation rate, with their
relationship given by y = f (x) = x/2. The factory owner instructs the manager to keep
the acidity within a range of pH 3 ± 0.1. What range of fermentation rates should be
maintained?
√
Solution 1.8 First, the upper bound of the fermentation rate is given
by y = x/2 < 3.1,
√
leading to x < (6.2)2 = 38.44. The lower bound is given by y = x/2 > 2.9, leading to
x > (5.8)2 = 33.64. Therefore, the fermentation rate must be maintained within a range
of 36% ± 2.36%. ⊔
⊓
In the above example, the difference between the optimal fermentation rate of 36% and
the upper bound of 38.44 is 2.44, while the difference with the lower bound of 33.64
is 2.36. Choosing the smaller value, let us set δ = 2.36. Additionally, let the permissible
error range for acidity set by the factory owner be ε = 0.1. We can then write the following
equation:
| f (x) − 3| < ε if |x − 36| < δ .
(1.1)
When the upper and lower bounds differ, selecting the smaller value might seem like a
loss of information, but it is sometimes necessary or more convenient. Many mathematical
definitions are based on such relationships, collectively known as the ε-δ argument.
Question 1.2. Can we always find a δ > 0 corresponding to any given small permissible
error ε > 0?
If we assume an arbitrary ε (0 < ε < 3) instead of a specific number, we can determine
δ as a function of ε using the above calculations. This approach eliminates the need for
repeated numerical substitution. Let’s verify:
√ since the output y must remain within ±ε of
3, we have 3 − ε < y < 3 + ε. Using y = x/2, we can rewrite this as
√
3 − ε < x/2 < 3 + ε ⇒ 36 − (24ε − 4ε 2 ) < x < 36 + (24ε + 4ε 2 )
Thus, δ can be set to (24ε − 4ε 2 ) or a smaller value. Typically, given a permissible error
range ε, the adjustment range δ is determined accordingly.4
Question 1.3. The reason
√ why we can always find a δ > 0 for any ε > 0 in the previous
problem is that f (x) = x/2 is continuous at c = 36. Can you explain why?
In the following problem, we consider a case where the defect rate behaves differently
depending on whether there are no impurities at all or even a small amount of impurities.
4 In the ε-δ argument, ε and δ are of interest when they are sufficiently small but positive. In this case,
choosing δ = 24ε − 4ε 2 leads to negative δ when ε > 6. Therefore, adding a condition such as ε < 6
would be appropriate, though it is not strictly necessary. The goal is not to find the optimal δ .
1.2 Quality Control and ε-δ Arguments
9
Problem 1.9 (Quality Control with Discontinuity). Suppose the probability of defective
products
√ occurring when xµg of impurities are mixed per gram of input material is given
by 1 + x%, and when there are no impurities, the probability is 0 (see left figure). Solve
Problem 1.7 (1) and (2) again under this assumption.
√
Solution 1.9 (1) The condition 1 + x < 4 implies x < 9. Thus, the impurity level must
be controlled to 9µg/g(= δ ). √
√
(2) Similarly,
the condition 1 + x < 0.4 results in x < −0.6, which is not possible.
√
Since x cannot be negative, there is no valid control range. That is, no δ > 0 exists. ⊔
⊓
In the above problem, a tolerance limit of ε = 4 allowed a valid δ , whereas a stricter
limit of ε = 0.4 did not. This is because the defect probability function f (x) is discontinuous at x = 0. If it were continuous, such issues would not arise.5
Problem 1.10. If the function f (x) in Problem 1.8 is also discontinuous at c, the factory
owner could assign an impossible task to the manager (see figure). How small must ε > 0
be set to ensure that no δ > 0 satisfies (1.1)? (Why is it impossible to create such a
situation if f (x) is continuous?)
Solution 1.10 If f (x) is discontinuous at c, then either the left-hand or right-hand limit
differs from the function value f (c). If ε is set smaller than the difference between these
limits and f (c), then no matter how small δ > 0 is chosen, the error cannot be kept within
ε. ⊔
⊓
Exercises
1. Find the following limits.
x2 − c2
(1) lim
(2) lim xπ
x→c x − c
x→0
(3) lim π x
x→0
2. Suppose the output quantity is given by f (x) = x2 for an input quantity x. If the desired
output is 100 with an error tolerance of ε = 1, what is the maximum range of adjustment around x = 10? (Find δ > 0 such that | f (x) − f (10)| < ε whenever |x − 10| < δ .)
5 Some concepts cannot be fully understood just by explanation. Sometimes, one needs to reflect and re-
alize, ”Yes, that was it!” This is one of those concepts. Recognizing that the difference between continuity
and discontinuity aligns with the meaning of the ε − δ definition is such a realization. It takes time.
10
1 Limit and Continuity #1
3. If lim f (x) = f (c), explain why a control range δ > 0 exists for any arbitrarily small
x→c
error tolerance ε > 0.
4. Prove that a function f (x) being continuous at x = c is equivalent to the existence of
a control range |x − c| < δ ensuring that f (x) remains within an error bound ε > 0 of
f (c).
5. Given a function f (x) and an error tolerance ε > 0, determine if there exists a δ > 0
such that the output at x = c can be adjusted by a real number L. If not, prove why.
(1) f (x) = x + 2, c = 1, L = 3, and ε = 0.1.
x2 − 4
, c = 2, L = 4, and ε = 0.1
(2) f (x) =
x−2
2
x −4
(3) f (x) =
, c = 1, L = 3, and ε = 0.1
x−2
(4) f (x) = ⌊x⌋, c = 1, L = 1, and ε = 0.05
6. Given the functions f and g defined as follows:
(
(
x,
x < 0,
−x,
x < 0,
f (x) =
g(x) =
x + 1, x ≥ 0.
x − 1, x ≥ 0.
(a)Discuss the continuity of f and g.
(b)Discuss the continuity of f + g.
(c)Explain how this problem differs from Problem 1.5, and consider similar examples
for cases (2)-(6).
7. Find an example where the composition of two discontinuous functions f ◦ g is continuous.
Lecture 2
Limit and Continuity #2
To move beyond explanations and provide rigorous proofs, we need well-defined definitions that everyone agrees upon. In this lecture, we define the concepts of limits, convergence, and continuity using the ε-δ argument. This method determines whether we can
find an input adjustment range δ to control the output error range ε within a given bound.
Although this approach may initially feel unfamiliar, with practice, one can become comfortable with it.
2.1 Rigorous definitions using ε-δ
We now aim to rigorously define limits, convergence, and continuity using the ε-δ argument. Given an allowed error ε > 0, definitions are based on whether a corresponding adjustment range δ > 0 can be found. Understanding that these definitions align with those
in natural language and becoming familiar with expressing them in one’s own words is
crucial.
Continuity is a fundamental concept that appears in many contexts. A classical definition states that ”a small change in x near a point c results in a small change in f (x)
near f (c).” However, this is still an explanatory definition and is insufficient for rigorous
proofs. The following precise definition is given using the ε-δ argument.
Definition 2.1 (Limit and Continuity). Let f : R → R, c ∈ R, and L ∈ R be given. If for
any given ε > 0, there exists a δ > 0 such that
| f (x) − L| < ε
whenever
0 < |x − c| < δ ,
(2.1)
then we say that the limit of f (x) as x approaches c is L, written as lim f (x) = L. Moreover,
x→c
if this limit equals L = f (c), we say that f is continuous at c. If f is continuous at every
point in an interval (a, b), then we say that f is continuous on (a, b).
Question 2.1. What would change if we slightly modified (2.1) to require that for any
given ε > 0, there exists a δ > 0 such that
| f (x) − L| < ε
whenever
|x − c| < δ ?
(2.2)
11
12
2 Limit and Continuity #2
The difference lies in replacing 0 < |x − c| < δ with |x − c| < δ , meaning that the
condition must also hold for |x − c| = 0. This implies that | f (c) − L| < ε must be satisfied
for any ε > 0, which occurs only if L = f (c). Thus, this definition applies exclusively to
continuous cases. If one only needs continuity and not limits separately, this definition
could be used instead.
Sometimes, using shorthand notation makes writing on boards or in answers more concise:
lim f (x) = L
x→c
⇔
∀ε > 0, ∃δ > 0 s.t. | f (x) − L| < ε w.e. 0 < |x − c| < δ .
Here, the symbol ‘⇔’ means both sides are equivalent. It can be read as ”if and only if”
or ”the left-hand side is defined by the right-hand side.” The notation ‘∀ε > 0, ∃δ > 0’
means ”for all ε > 0, there exists δ > 0,” while ‘s.t.’ stands for ”such that” and ‘w.e.’ for
”whenever.”
Once the definition is understood, the next step is to perform simple calculations using
it. The problems and answers may seem obvious, but the goal is to experience firsthand
how the definition captures the intuitive meaning of limits and continuity. Becoming proficient requires practice.
Problem 2.1. For the function f (x) = 2x + 1, prove the following:
(1) lim f (x) = 5. (2) lim f (x) ̸= 2. (3) f is continuous at any point c ∈ R.
x→2
x→1
Solution 2.1 Let us analyze the function f (x) = 2x + 1 using the ε-δ argument.
(1) Suppose an arbitrary ε > 0 is given. We find δ satisfying (2.1) so that whenever
|x − 2| < δ , we have
| f (x) − 5| = |2x + 1 − 5| = |2(x − 2)| = 2|x − 2| < ε.
For this to hold, choosing δ = ε/2 suffices, though any smaller value also works.
(2) To show that limx→1 f (x) ̸= 2, we must demonstrate that there exists some ε > 0
for which no δ > 0 can satisfy the condition. Since f (1) = 3, the difference from 2 is 1.
Choosing ε = 0.5, we consider values of x near 1:
| f (x) − 2| = |2x + 1 − 2| = |2(x − 1) + 1| = 2|x − 1| + 1 > 0.5.
Thus, regardless of how small δ is, there exist x within |x −1| < δ such that | f (x)−2| > ε.
This proves that limx→1 f (x) ̸= 2.
(3) For any c ∈ R, we show that f (x) = 2x + 1 is continuous at c. Given ε > 0, set
δ = ε/2. Then, for |x − c| < δ , we have
| f (x) − f (c)| = |2x + 1 − 2c − 1| = 2|x − c| < 2δ = ε.
Since we can always find such a δ > 0 for any given ε > 0, f is continuous at c by
definition. ⊔
⊓
Question 2.2. Does the continuity defined in Definition 2.1 align with our intuitive understanding of continuity?
2.1 Rigorous definitions using ε-δ
13
The ε-δ argument is a dynamic concept, much like a game. If an opponent provides
an error bound ε > 0, one must always find an adjustment range δ > 0 that satisfies
(2.1) or (2.2). The ability to do so signifies that a function is convergent or continuous.
Thus, proving continuity means finding such a δ > 0 that satisfies these conditions. In
other words, if a factory manager is given an arbitrary defect tolerance ε > 0, they must
be able to find an appropriate input adjustment range δ > 0 to meet this requirement.
Proving a statement in this context is about finding δ , not about verbal persuasion. If for
some ε > 0, no such δ > 0 can be found, then the function is not continuous or does not
converge. Proofs thus shift from the realm of explanation to the realm of discovery.
For large ε > 0, finding a suitable δ > 0 is generally easier. However, in cases of continuity or convergence, we must be able to find δ > 0 for arbitrarily small ε > 0. Hence,
we assume an arbitrary ε > 0 is given and find the corresponding δ > 0. The ability to
do this implies that L is the limit of f (x) or that f is continuous at c. Conversely, if L is
not a limit or if f is not continuous, then there exists some ε for which no δ satisfies the
definition.
The definition of limits discussed earlier assumes that the function is defined on the
entire real number set R. However, even if f is only defined on an open interval (a, b),
limits can still be well-defined at any c in the interval. This is because if δ is small enough,
all points satisfying |x − c| < δ remain in (a, b), where f is well-defined. However, if the
domain is a closed interval [a, b] and c is one of its endpoints, finding such a δ > 0 is not
possible. Thus, for functions defined on a closed interval where c is an endpoint, limits
must be defined using one-sided limits.
Problem 2.2. Given a function f : R → R and a point c ∈ R, a limit L may or may not
exist, depending on the behavior of the function. However, if it exists, it must be unique.
Prove this uniqueness property.
Solution 2.2
⊔
⊓
We can now rigorously define left-hand and right-hand limits, previously described in
informal terms. Defining them independently is a useful exercise.
Problem 2.3. Define left-hand and right-hand limits using the ε-δ argument.
Solution 2.3 Definitions may vary slightly, but the core idea remains the same. (1) Definition of the left-hand limit: Given a function f : (a, b) → R, a point c ∈ (a, b), and a real
number L ∈ R, if for any ε > 0,
| f (x) − L| < ε
whenever
−δ < x−c < 0
is satisfied for some δ > 0, then L is called the left-hand limit of f (x) at c, written as
lim f (x) = L. (2) For the right-hand limit, replace −δ < x − c < 0 with 0 < x − c < δ .
x→c−
⊔
⊓
Problem 2.4 (Left Continuity and Right Continuity). Define left continuity and right
continuity using the ε-δ argument.
Solution 2.4 Left and right continuity can be defined in terms of one-sided limits as
follows:
14
2 Limit and Continuity #2
Definition A: Given a function f : (a, b) → R and a point c ∈ (a, b), the function f (x)
is said to be left continuous at c if lim f (x) = f (c). That is, the left-hand limit exists and
x→c−
equals the function value f (c). If f (x) is left continuous at all c ∈ (a, b), then it is said to
be left continuous on (a, b). Right continuity is defined similarly.
Alternatively, without using limits, left and right continuity can be defined using the
ε-δ argument:
Definition B: Given a function f : (a, b) → R and a point c ∈ (a, b), f (x) is said to be
left continuous at c if for any ε > 0,
| f (x) − f (c)| < ε
whenever
−δ < x−c ≤ 0
is satisfied for some δ > 0. Similarly, f (x) is said to be right continuous at c if for any
ε > 0,
| f (x) − f (c)| < ε whenever 0 ≤ x − c < δ
is satisfied for some δ > 0.
The difference between these definitions and the limit-based definition is that L is replaced with f (c), and the condition 0 < x − c < δ is replaced with 0 ≤ x − c < δ to include
x = c. ⊔
⊓
2.2 Examples
The ε-δ argument can be used to prove the limit and continuity laws from Problems 1.2
and 1.5. However, some technical skills are required.
Problem 2.5 (Sum Law). (1) Prove the following identity:
lim ( f (x) + g(x)) = lim f (x) + lim g(x).
x→c
x→c
x→c
(2) Show that if functions f and g are continuous at c ∈ R, then f + g is also continuous
at c ∈ R.
Solution 2.5 (1) and (2) are closely related. Suppose an arbitrary ε > 0 is given. The goal
is to find a corresponding δ > 0 for the given ε > 0. Let lim f (x) = L and lim g(x) = M.
x→c
x→c
By definition, there exist δ1 > 0 and δ2 > 0 such that:
| f (x) − L| < 0.5ε
whenever
0 < |x − c| < δ1 ,
|g(x) − M| < 0.5ε
whenever
0 < |x − c| < δ2 .
Note that we use 0.5ε instead of ε in finding δ . Since δ1 and δ2 may differ, we define
δ = min(δ1 , δ2 ). Then, for 0 < |x − c| < δ , we have
| f (x) + g(x) − (L + M)| ≤ | f (x) − L| + |g(x) − M| < 0.5ε + 0.5ε = ε.
By the definition of limits, this implies that L + M is the limit of f + g at c, hence:
2.2 Examples
15
lim ( f (x) + g(x)) = L + M = lim f (x) + lim g(x).
x→c
x→c
x→c
For continuity, if f and g are continuous at c, then L = f (c) and M = g(c), so:
lim ( f (x) + g(x)) = f (c) + g(c).
x→c
Thus, f + g is continuous at c. ⊔
⊓
The above proof of convergence implies that if two functions f and g are continuous,
then their sum f + g is also continuous. Now, we will prove the continuity of composite
functions.
Problem 2.6 (Continuity of a composition). Let f : R → R be continuous at c ∈ R,
and let g : R → R be continuous at f (c) ∈ R. Then, prove that the composite function
(g ◦ f )(x) := g( f (x)) is continuous at c.
Solution 2.6 Suppose an arbitrary ε > 0 is given. The goal is to find a δ > 0 that satisfies
(2.2) for the composite function. Since g is continuous at f (c), there exists a δ1 > 0 such
that:
|g(y) − g( f (c))| < ε whenever |y − f (c)| < δ1 .
Since this holds for all y, we can substitute y = f (x), yielding:
|g( f (x)) − g( f (c))| < ε
whenever
| f (x) − f (c)| < δ1 .
The key step is to treat δ1 as if it were ε in the definition of continuity for f . Since f is
continuous at c, there exists a δ2 > 0 such that:
| f (x) − f (c)| < δ1
whenever |x − c| < δ2 .
Setting δ = δ2 completes the proof. (This proof, though constructed after much thought,
turns out to be quite simple. However, it surpasses the informal attempt made in Lecture
1 to justify the continuity of composite functions. This highlights the elegance of the ε-δ
argument.) ⊔
⊓
x, if x < 1
Problem 2.7. For the function f (x) =
prove the following:
2x, if x ≥ 1
(1) lim f (x) = 2
(2) lim f (x) = 1.
x→1−
x→1+
Problem 2.8 (One-sided limits). Prove the following statement:
lim f (x) = L
x→c
if and only if
lim f (x) = L = lim f (x).
x→c−
x→c+
Problem 2.9 (Sandwich Theorem). Suppose functions f , g, h : (a, b) → R satisfy g(x) ≤
f (x) ≤ h(x) for all x ∈ (a, b), and for some real number c ∈ (a, b), we have: lim g(x) =
x→c
lim h(x) = L. Then, prove that lim f (x) = L.
x→c
x→c
In each of the above problems, given an arbitrary ε > 0, the goal is to find a corresponding δ > 0. In Problem 2.7, this requires examining the function’s structure, while in
Problems 2.8 and 2.9, the given conditions must be used to determine δ > 0.
16
2 Limit and Continuity #2
2.3 Limits as x → ∞ and f (x) → ∞
Sometimes, as a variable approaches infinity (x → ∞ or x → −∞), a function may either
converge to a limit or diverge to infinity. We can define such cases rigorously. Let us
establish formal definitions for these limits.
Problem 2.10. Using the ε-δ argument, formulate precise definitions for the following
limits for a function f : R → R.
(1) lim f (x) = L. (2) lim f (x) = L. (3) lim f (x) = ∞. (4) lim f (x) = −∞.
x→∞
(5) lim f (x) = ∞.
x→c+
x→−∞
(6) lim f (x) = ∞.
x→c−
x→c
x→c
(7) lim f (x) = −∞.
x→c+
(8) lim f (x) = −∞.
x→c−
Solution 2.10 A useful exercise is to attempt formulating these definitions independently
and then compare them with the following standard definitions. Verify that they align with
the intended interpretations.
(1) lim f (x) = L ⇔ ∀ε > 0, ∃N ∈ R s.t. | f (x) − L| < ε w.e. x > N.
x→∞
(2) lim f (x) = L
⇔
∀ε > 0, ∃N ∈ R s.t. | f (x) − L| < ε w.e. x < N.
(3) lim f (x) = ∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) > N w.e. 0 < |x − c| < δ .
(4) lim f (x) = −∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) < N w.e. 0 < |x − c| < δ .
(5) lim f (x) = ∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) > N w.e. 0 < x − c < δ .
(6) lim f (x) = ∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) > N w.e. 0 < c − x < δ .
(7) lim f (x) = −∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) < N w.e. 0 < x − c < δ .
(8) lim f (x) = −∞
⇔
∀N ∈ R, ∃δ > 0 s.t. f (x) < N w.e. 0 < c − x < δ . ⊔
⊓
x→−∞
x→c
x→c
x→c+
x→c−
x→c+
x→c−
Exercises
1. Prove the following limits using the formal definition:
(1) lim x−1 = 0
(2) lim x−1 = ∞
(3) lim |x − 1|−1 = ∞
x→∞
x→0+
x→1
2. Determine the following limits if they exist:
x+1
x+1
x3 − 1
(1) lim 2
(2) lim 2
(3) lim 2
x→∞
x→1 x − 1
x→−1 x + 4x + 3
x +1
3. Prove continuity at the given points using the definition:
(1) f (x) = 1x at c ̸= 0 (2) f (x) = |x + 3| at c = −3 (3) f (x) = sin x at c = 0
4. Verify whether the following functions are continuous
p at x = 0:
(1) f (x) = x sin(1/x) with f (0) = 0
(2) f (x) = |x|
(3) f (x) = x2 sin(1/x) with f (0) = 0
5. (1) Plot the graph of the function:
2.3 Limits as x → ∞ and f (x) → ∞
17
(
0
if x ≤ 0,
f (x) =
−1
cos(x ) if x < 0.
(2) Show that this function is continuous at all points except c = 0. (Use the fact that
cos x is continuous and that the composition of continuous functions is also continuous.)
6. Find all values

 of a that ensure continuity for all x:
x+1
 x + 1 if x ̸= −1,
3
(2) f (x) = a + 3
(1) f (x) = x + 1
ax2 + a
a
if x = −1.

 ax − 1 if x ≤ a,
(3) f (x) = x − a
1
if x > a.
if x ≤ 0,
if x > 0.
7. Suppose f satisfies lim f (x) = 2, lim f (x) = 1, f (0) = 12 . Compute the following
x→0−
x→0+
limits:
(1) lim f (x − x2 ) (2) lim f (−x3 + 3x − 2) (3) lim f (1 + x − 2x3 )
x→0−
x→1
x→1−
8. Functions f and g are defined as follows:
(Function definitions omitted for brevity)
(1) Find conditions on a and b so that g ◦ f is continuous over all real numbers and plot
its graph.
(2) Prove that g ◦ f is continuous at x = −1 using the ε-δ argument.
Lecture 3
Differentiation
In this lecture, we will explore the fundamental concept of differentiation, one of the
most essential topics in calculus. A clear understanding of differentiation is crucial, as it
serves as a tool for describing motion mathematically. The type of motion we consider
involves continuous changes in position—we do not deal with instantaneous jumps. Having already learned the important mathematical concept of continuity, we now introduce
differentiation, which allows us to quantify changes in motion, such as velocity and acceleration.
3.1 Rate of Increase
Consider a function f (x) defined as follows:
f : R → R,
x∈R
Here, the variable x represents real numbers, and the function f also takes real values.
This function can represent the position of an object moving along a one-dimensional line
or describe other quantities in different contexts. The choice of variable notation depends
on the situation. For instance, if x represents time, we often use t ∈ R instead. If f (t)
represents the position of a runner after t seconds, it describes motion. However, t could
also represent other quantities, such as the production output of a factory over t hours.
The input variable does not necessarily have to denote time; it can signify various other
parameters. While we may use different symbols, the fundamental concept remains the
same, and choosing appropriate notation helps avoid confusion.
When the variable x increases from a to b, the function value increases from f (a) to
f (b). We define the changes in x and f as △x = b − a and △ f = f (b) − f (a). The ratio
of these changes, given by
f (b) − f (a) △ f
≡
b−a
△x
Mean Growth Rate
19
20
3 Differentiation
is called the mean growth rate mean growth rate 평균 증가율. For example, if the mean
growth rate is 10, it means that for every unit increase in x from a, the function value
increases by ten times that amount, i.e., 10△x. Similarly, if x decreases, f (x) decreases at
the same rate. A positive mean growth rate implies that as x increases, f (x) also increases,
and as x decreases, f (x) decreases. Conversely, a negative mean growth rate indicates
that f (x) decreases as x increases and vice versa. Rewriting the formula with a = c and
b = c + h, we obtain
f (c + h) − f (c)
Mean Growth Rate
h
where h can be either positive or negative.
Differentiation has multiple equivalent definitions. The definition chosen often depends
on its usefulness, and proving the equivalence of different definitions is a fundamental
exercise.
Definition 3.1 (Derivative). A function f : (a, b) → R, a point c ∈ (a, b), and a real number v ∈ R are given. If, for any ε > 0, there exists a corresponding δ > 0 such that
| f (c + h) − f (c) − hv| ≤ |h|ε
whenever
|h| < δ
(3.1)
holds, then f is said to be differentiable at c, and the real number v is called the derivative
derivative of f at c. This relationship is written as f ′ (c) = v.
ootnoteHow does this definition differ from the previous definition of limits? How
would it look if written in the same format?
Let us now explore the relationship between this definition of differentiation and the
mean growth rate.
Problem 3.1. Show that the differentiability of f at c is equivalent to the existence of the
limit of the mean growth rate, and that this limit equals the derivative f ′ (c) defined above.
That is,
f (c + h) − f (c)
f ′ (c) = lim
.
(3.2)
h→0
h
Solution 3.1 The expressions (3.1) and (3.2) are almost identical, with the only difference
being that (3.2) has h in the denominator, requiring us to exclude h = 0. We first prove
the forward direction (⇒). To establish the limit in (3.2), assume that an arbitrary ε > 0
is given. From (3.1), we obtain a corresponding δ > 0. For nonzero h, dividing (3.1) by
|h| gives
3.1 Rate of Increase
21
f (c + h) − f (c)
− v < ε whenever 0 < |h| < δ .
h
This precisely matches the definition of the limit in (3.2).
(3.3)
For the reverse direction (⇐), assume (3.2) holds. Given any ε > 0, take the corresponding δ > 0 from the definition of the limit. Multiplying both sides of (3.3) by |h|
yields (3.1). The special case of h = 0 is trivial since both sides of (3.1) become zero. ⊔
⊓
Thus, differentiation is simply the limit of the mean growth rate. Therefore, the derivative at a point c represents the instantaneous growth rate instantaneous growth rate
순간 증가율. Geometrically, it corresponds to the slope of the tangent line to the graph of
f at (c, f (c)).
Problem 3.2 (Tangent Line Formula). If a function f is differentiable at c, derive the
equation of the tangent line to the graph of f at the point (c, f (c)).
Solution 3.2 Using the fact that the equation of a line with slope a passing through the
point (x0 , y0 ) is given by y − y0 = a(x − x0 ), we derive the equation of the tangent line.
The tangent line has a slope of f ′ (c) and passes through the point (c, f (c)), thus it can be
written as:
y − f (c) = f ′ (c)(x − c) or y = f ′ (c)(x − c) + f (c)
Tangent Line Formula
⊔
⊓
Instead of the expression in equation (3.2), the following form is also commonly used:
f ′ (c) = lim
b→c
f (b) − f (c)
b−c
Derivative (Instantaneous Rate of Change)
Remark 3.1. The derivative of a function is also called the rate of change, but this terminology can be misleading. For example, what does it mean if the rate of change is −1? If
one interprets it as a decrease, then verbally it is called a ”rate of change,” but mentally it
is thought of as an ”increase rate” in reverse.
Not all functions have derivatives. For the instantaneous rate of change to exist, the
left-hand and right-hand limits of the average rate of change must both exist and be equal.
If this limit exists, we say that the function f is differentiable at c. Thus, it is essential to
distinguish between functions that are differentiable and those that are not.
Problem 3.3 (Examples).
(1) Show that the function f (x) = |x| is not differentiable at x = 0.
(2) Show that the function f (x) = x2 is differentiable at x = 1.
(3) Show that a function f is not differentiable at points of discontinuity.
Solution 3.3 (1) We show that the right-hand and left-hand limits of the average rate of
change are different. These limits are given by:
lim
h→0+
f (0 + h) − f (0)
|h|
h
= lim
= lim = lim 1 = 1.
+
+
h
h→0 h
h→0 h
h→0+
22
3 Differentiation
lim
h→0−
|h|
−h
f (0 + h) − f (0)
= lim
= lim
= lim −1 = −1.
h
h→0+
h→0− h
h→0− h
Since these values are different, the limit does not exist, and thus f (x) is not differentiable
at x = 0.
(2) The limit is given by:
(1 + h)2 − 1
h2 + 2h
f (1 + h) − f (1)
= lim
= lim
= lim (h + 2) = 2.
h→0
h→0
h→0
h→0
h
h
h
lim
Since the limit exists and is equal to 2, the derivative at x = 1 is 2.
(3) Suppose the function f is discontinuous at c. This means that there exists some
ε > 0 such that no δ > 0 can be chosen to satisfy continuity. That is, for every natural
number n > 0, there exists 0 < |hn | < n1 such that
| f (c + hn ) − f (c)| > ε.
1 Thus, for any M > 0, if n > M , then
ε
| f (c + hn ) − f (c)|
ε
≥
≥ εn ≥ M.
|hn |
|hn |
This implies that hn approaches 0, but the average rate of change can become arbitrarily
large, meaning that the limit does not exist. (Try drawing a graph to explain this idea
further.) ⊔
⊓
Problem 3.4 (Differentiability Implies Continuity). Show that if a function f (x) is differentiable at x = c, then it must be continuous at x = c.
Solution 3.4 There are several ways to prove this. Since Problem 3.3(3) is equivalent to
this statement, its proof suffices. However, try thinking of other methods as well. ⊔
⊓
3.2 Differentiation Rules
Differentiation is a frequently performed calculation, and it is essential to understand and
memorize some key cases to differentiate accurately. This section covers eight differentiation rules.
Question 3.1 (Is a Number a Function?). A function f : R → R maps a real number x
to a function value f (x). If a constant function always returns the same real value 3 for
all x ∈ R, the most straightforward way to represent it is simply by writing 3 instead of
using a function symbol like f . Although a number is just a number, it can also be used
as function notation. Using a single notation with dual meanings is convenient, and such
notation will be frequently employed in various contexts.
1 Think carefully about why this is true and make sure to understand it.
3.2 Differentiation Rules
23
Problem 3.5 (Rules of Differentiation). Prove the following differentiation rules and
state the conditions necessary for each rule to hold.
1. The derivative of a constant function is 0.
2. f (x) = x ⇒ f ′ (x) = 1.
3. f (x) = xn for n ∈ N ⇒ f ′ (x) = nxn−1 .
4. f (x) = 1x ⇒ f ′ (x) = −x−2 .
5. Sum Rule: ( f + g)′ (x) = f ′ (x) + g′ (x).
6. Product Rule: ( f g)′ (x) = ( f ′ g + f g′ )(x).
f ′
f ′ g − f g′
7. Quotient Rule:
(x) =
(x).
g
g2
8. Power Rule: f (x) = xα for α ∈ R ⇒ f ′ (x) = αxα−1 .
Solution 3.5 The eight differentiation rules above form the foundation of differentiation.
First, let us examine the conditions for their application. The first three rules require no
additional conditions. Rule (4) requires x ̸= 0 since division by zero is undefined. Rules
(5), (6), and (7) require that both functions f and g are differentiable. Additionally, Rule
(7) requires that g(x) ̸= 0. The Power Rule (8) requires careful attention. If α is not a
positive integer, the validity of xα and xα−1 depends on x. If α − 1 is negative, then x ̸= 0
is required, and if α is not an integer, then x ≥ 0 must hold.
Now, let us prove these rules. The first four rules are special cases of the Power Rule
(8). (1) If f is a constant function, then f (x) = f (x + h). Thus,
f (x + h) − f (x)
0
= lim = lim 0 = 0.
h→0
h→0 h
h→0
h
f ′ (x) = lim
(2) If f (x) = x, then f (x + h) − f (x) = h. Hence,
f ′ (x) = lim
h→0
f (x + h) − f (x)
= lim 1 = 1.
h→0
h
n−2 h2 + · · · + hn .
(3) If f (x) = xn , then f (x + h) − f (x) = (x + h)n − xn = nxn−1 h + n(n+1)
2 x
Therefore,
n−2 h2 + · · · + hn
nxn−1 h + n(n+1)
2 x
f ′ (x) = lim
= nxn−1 .
h→0
h
(4) If f (x) = 1x and x ̸= 0, then
1 1
1
1 x − (x + h)
1
−
= lim
= − 2.
h→0 h x + h
h→0 h x(x + h)
x
x
f ′ (x) = lim
(5) The differentiation of a sum is straightforward. If f and g are differentiable, then
f (x + h) + g(x + h) − f (x) − g(x)
h
f (x + h) − f (x)
g(x + h) − g(x)
= lim
+ lim
= f ′ (x) + g′ (x).
h→0
h→0
h
h
( f + g)′ (x) = lim
h→0
24
3 Differentiation
(6) The differentiation of a product is slightly more complex but highly significant. The
key technique used in this proof is ”subtracting and then adding back the subtracted term.”
If f and g are differentiable, then
f (x + h)g(x + h) − f (x)g(x)
h→0
h
f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x)
= lim
h→0
h
f (x + h)g(x + h) − f (x + h)g(x)
f (x + h)g(x) − f (x)g(x)
= lim
+ lim
h→0
h→0
h
h
g(x + h) − g(x)
f (x + h) − f (x)
= lim f (x + h)
+ lim
g(x)
h→0
h→0
h
h
= f (x)g′ (x) + f ′ (x)g(x).
lim
Thus, the limit exists and ( f g)′ (x) = f (x)g′ (x) + f ′ (x)g(x).
(7) The differentiation of a quotient can be derived similarly to the product rule but
requires extra caution. The functions f and g must be differentiable, and g(x) ̸= 0 for the
rule to hold. Since gf = f · 1g , we can first differentiate g1 and then apply the product rule
(calculation omitted).
(8) The power rule is one of the most commonly used differentiation rules. If α is not
an integer, logarithms simplify the proof. Using logarithmic differentiation, we start with
y = xα and take the logarithm:
ln y = ln xα = α ln x.
Differentiating both sides with respect to x using the derivative of the logarithm gives
1
y′
= α ⇒ y′ = αx−1 y = αxα−1 .
y
x
⊔
⊓
Problem 3.6. Find the derivative of the following functions.
(2) f (t) = t 2 + t 3 , t ∈ R
(1) f (x) = 3xα , x > 0.
Solution 3.6 (1) Considering the function as the product of the constant function 3 and
the power function xα , we apply the derivative of a product:
(3xα )′ = 0xα + 3αxα−1 .
Since the derivative of a constant is 0, we ignore the constant coefficient and differentiate
only the function part, then multiply by the constant coefficient. ⊔
⊓
Trigonometric functions are frequently used and will appear often in the future. Remember them well.
Problem 3.7. Memorize the following:
(1) sin′ x = cos x
(2) cos′ x = − sin x
′
2
(4) cot x = − csc x
(5) sec′ x = sec x tan x
(3) tan′ x = sec2 x
(6) csc′ x = − csc x cot x
Solution 3.7 (1) Using the sum formula for the sine function, we obtain:
3.3 Intermediate and Mean Value Theorem
25
cos h − 1
sin h
sin(x + h) − sin x sin x cos h + cos x sin h − sin x
=
= sin x
+ cos x
h
h
h
h
Taking the limit as h → 0, and using the facts that cos′ 0 = 0 and sin′ 0 = 1, we obtain
sin′ x = cos x. ⊔
⊓
Problem 3.8. Show that if a function f : [a, b] → R is continuous and attains a maximum
or minimum at some c ∈ (a, b), then f ′ (c) = 0.
Solution 3.8 Let’s compute the right-hand and left-hand limits in the case of a maximum.
Since f (c + h) − f (c) ≤ 0, we get
lim
h→0+
f (c + h) − f (c)
≤ 0,
h
lim
h→0−
f (c + h) − f (c)
≥0
h
Thus, for the derivative to exist, both limits must be equal, implying they must be 0. A
similar argument holds for a minimum. ⊔
⊓
3.3 Intermediate and Mean Value Theorem
Theorem 3.1 (Intermediate Value Theorem). If a function f : [a, b] → R is continuous
and a real number m lies between f (a) and f (b), then there exists some c in the open
interval (a, b) such that f (c) = m.
The Intermediate Value Theorem is accepted as an axiom without proof. In fact, this
theorem is equivalent to the completeness axiom of real numbers.
Theorem 3.2 (Rolle’s Theorem). If a function f : [a, b] → R is continuous, differentiable
on (a, b), and satisfies f (a) = f (b), then there exists some c ∈ (a, b) such that f ′ (c) = 0.
Proof. Suppose there exists some x such that f (x) > f (a). Then f (x) attains a maximum
in the open interval (a, b). At the maximum, the derivative is 0, satisfying the theorem. If
there exists some x such that f (x) < f (a), then f (x) attains a minimum in (a, b). At the
minimum, the derivative is 0, satisfying the theorem. If neither case occurs, then f (x) is a
constant function, whose derivative is 0 everywhere, satisfying the theorem. ⊔
⊓
Theorem 3.3 (Cauchy’s Mean Value Theorem). If functions f , g : [a, b] → R are continuous, differentiable on (a, b), and g′ (x) ̸= 0 for all x ∈ (a, b), then there exists some
c ∈ (a, b) such that
f ′ (c)
f (b) − f (a)
=
(3.4)
g′ (c)
g(b) − g(a)
Proof. (Using Rolle’s Theorem 3.2 for proof.) Since g′ (x) ̸= 0 everywhere, we conclude
that g(b) − g(a) ̸= 0 by Rolle’s theorem. Define the function
h(x) = f (x) − f (a) −
f (b) − f (a)
(g(x) − g(a))
g(b) − g(a)
26
3 Differentiation
This function is well-defined and differentiable. Also, h(a) = 0 and h(b) = 0. Thus, there
exists some c where h′ (c) = 0, which implies
h′ (c) = f ′ (c) −
f (b) − f (a) ′
g (c) = 0
g(b) − g(a)
Rearranging, we obtain (3.4). ⊔
⊓
The key to the proof above is finding the function h(x). More important than identifying
the function itself is deriving the relationship expressed in equation (3.4).
The final step involves the Mean Value Theorem.
As illustrated in the figure, this theorem states that
there exists some point c between a and b where the
derivative, or instantaneous rate of change, f ′ (c),
is equal to the average rate of change. In other
words, the Mean Value Theorem is not actually
about the mean value but rather about the mean
rate of change. However, it has traditionally been
referred to as the Mean Value Theorem rather than
the Mean Growth Rate Theorem.
Theorem 3.4 (Mean Value Theorem). If a function f : [a, b] → R is continuous and
differentiable on (a, b), then there exists some c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
b−a
(3.5)
Proof. The Mean Value Theorem (MVT) can be easily proven using Cauchy’s MVT.
Setting g(x) = x, we get g′ (x) = 1 ̸= 0, so applying Theorem 3.3 yields (3.5). ⊔
⊓
The Mean Value Theorem is a fundamental theorem that will be frequently used in the
future.
Exercises
1. Find the tangent lines to the given functions at the specified√points.
(1) y = 4 + x2 at (1, 5) (2) y = x−2 at √
(2, 0.25) (3) y = x3 at (4, 8)
x
3
2
(4) f (t) = t − t at t = 2 (5) f (x) = t 2 + 1 at x = 2 (6) f (x) = x−2
at x = 1
1
x−2
3
0.5
(7) f (s) = s − s at s = 1 (8) f (t) = t−1 at t = 2 (9) f (x) = x+1 at x = 2
2. Determine whether the following functions are differentiable at x = 0, given that
f (0) = 0.
p
(1) f (x) = x sin(1/x) (2) f (x) = x2 sin(1/x) (3) f (x) = |x|
3. The circumference of a circle of radius x is 2πx, its area is πx2 , the surface area of a
sphere is 4πx2 , and its volume is 43 πx3 . Compute the instantaneous rate of change at
x = r for each.
3.3 Intermediate and Mean Value Theorem
27
4. Find the domain of definition and the points where the following functions are not
differentiable.
p
p
2 +1
(4) √ 1
(1) |x|2 + 1
(2) |x| + 1
(3) xx−1
x2 −1
5. Use the Mean Value Theorem to prove the following.
(a)Show that for all real numbers x, y, we have | sin x − sin y| ≤ |x − y|.
(b)Show that for all natural numbers n, we have 1n (2n − 1) ≤ 2n−1 .
(c)Let f : R → R be differentiable at all real numbers and satisfy f ′ (x) ̸= 0. Show that
f has at most one root.
6. Given a sequence of real numbers x1 , x2 , . . . , xn chosen from the interval [0, 2], define
the function f as follows:
f (x) =
1 n
∑ |x − xi |,
n i=1
x ∈ R.
(a)Compute f (0) + f (2).
(b)Show that f is continuous on the entire real line.
(c)Prove that there exists some c ∈ [0, 2] such that f (c) = 1.
7. Determine whether the following statements are true or false, and provide a justification or counterexample.
(a)If f (x) + g(x) is differentiable at x = a, then both f and g are differentiable at x = a.
(b)If f (x) + g(x) and g(x) are differentiable at x = a, then f is differentiable at x = a.
(c)If f (x)g(x) and g(x) are differentiable at x = a, then f is differentiable at x = a.
8. Suppose a function f is defined on the entire real line and satisfies the following condition for all x, y ∈ R and λ ∈ [0, 1]:
f (λ x + (1 − λ )y) ≤ λ f (x) + (1 − λ ) f (y).
(a)Show that for all real numbers x < y < z, we have
f (y) − f (x)
f (z) − f (y)
≤
y−x
z−y
(b)If f (x) is differentiable at some point x0 , show that for all real numbers x, we have
f (x) ≥ f (x0 ) + f ′ (x0 )(x − x0 ).
9. For distinct θ1 , θ2 ∈ [0, π/4], use the Mean Value Theorem 3.4 to show that
ln(cos θ1 ) − ln(cos θ2 )
≤1
θ1 − θ2
Lecture 4
Chain Rule and Variable Centered Notation
Chain Rule is one of the most fundamental differentiation rules. It is powerful and
widely applicable. The true power of the chain rule becomes evident in Calculus II, when
dealing with multiple variables and vector-valued functions. In Calculus I, we consider
the simplest case of one-dimensional functions, as illustrated in the figure:
g : (a1 , b1 ) → R, f : (a2 , b2 ) → R, c ∈ (a1 , b1 ), g(c) ∈ (a2 , b2 ).
The chain rule states that ”if g is differentiable at c and f is differentiable at g(c), then
the composite function f ◦ g is differentiable at c, and its derivative is the product of the
individual derivatives,” i.e.,
( f ◦ g)′ (c) = f ′ (g(c))g′ (c).
(4.1)
Question 4.1. Can you explain the mathematical relationship given in (4.1) in everyday
language?
Mathematics is often considered the language of science. Equation (4.1) conveys a
mathematical concept. Understanding it in your own words is the first step towards comprehension. When this equation starts to feel intuitive, we call that intuition. For example,
if g′ (c) = 10, it means that when x increases slightly near c, the function g increases tenfold. Similarly, if f ′ (g(c)) = 10, then when x increases slightly near g(c), f increases
tenfold. Hence, for the composite function f (g(x)), a small increase in x around c leads
to an overall increase of 10 × 10 = 100 times, which is exactly what the chain rule (4.1)
states. Even without proof, (4.1) seems intuitively correct.
4.1 Chain Rule
The derivative of a function f is defined as the limit of the average rate of change:
29
30
4 Chain Rule and Variable Centered Notation
f ′ (c) = lim
z→c
f (z) − f (c)
.
z−c
f (c)
The ratio f (z)−
represents how much f (x) increases relative to △x when x increases
z−c
slightly near c. That is, f ′ (c) times △x gives the corresponding increase.
Now, consider the definition of a composite function. Given two functions,
g : (a1 , b1 ) → R,
and
f : (a2 , b2 ) → R,
their composite function is defined as
( f ◦ g)(x) = f (g(x)),
x ∈ Ω.
(4.2)
Problem 4.1. In general, the composite function f ◦ g given in (4.2) is not defined for all
x ∈ (a1 , b1 ). Why is this the case? What is the largest possible domain Ω for this function?
Solution 4.1 If g(x) ̸∈ (a2 , b2 ), then f (g(x)) is not defined. Hence, the largest possible
domain for the composite function is Ω := {x ∈ (a1 , b1 ) : g(x) ∈ (a2 , b2 )}. ⊔
⊓
For convenience, assume that the given functions f and g satisfy the following conditions:
g : (a1 , b1 ) → (a2 , b2 ), f : (a2 , b2 ) → R.
(4.3)
This ensures that we do not have to worry about the domain, and the composite function
is simply f ◦ g : (a1 , b1 ) → R.
Problem 4.2. Consider the two functions f and g in (4.3). Suppose that g is differentiable
at c ∈ (a1 , b1 ) and f is differentiable at g(c) ∈ (a2 , b2 ). Using their derivatives, explain
how much the composite function f ◦ g increases when x increases slightly near c.
Solution 4.2 Let’s analyze this step by step. ⊔
⊓
Theorem 4.1 (Chain Rule). Let two functions g : (a1 , b1 ) → (a2 , b2 ) and f : (a2 , b2 ) → R
be given. Suppose that the function g is differentiable at c ∈ (a1 , b1 ) and the function f is
differentiable at g(c) ∈ (a2 , b2 ). Then, the composite function f ◦ g is differentiable at c,
and its derivative is given by:
( f ◦ g)′ (c) = f ′ (g(c))g′ (c)
(4.4)
Proof. Let us prove
this usingDefinition 3.1 and equation (3.1). First, let ε > 0 be given.
ε
. Since g is differentiable at c, there exists δ1 > 0 such that:
Define ε1 := min 1, 2| f ′ (g(c))|
|g(c + h) − g(c) − hg′ (c)| < |h|ε1
whenever |h| < δ1 .
Letting H = g(c + h) − g(c), we obtain:
|H − hg′ (c)| < |h|ε1 ⇒ |H| < |h|(|g′ (c)| + ε1 ).
Thus, for |h| < δ1 , we have |H| < |h|(|g′ (c)| + 1).
4.1 Chain Rule
31
Since f is differentiable at g(c), there exists δ2 > 0 such that for |H| < δ2 ,
| f (g(c) + H) − f (g(c)) − H f ′ (g(c))| < |H|ε2 ,
ε2 :=
ε
.
2(|g′ (c)| + 1)
δ2
Now, let δ = min(δ1 , |g′ (c)|+1
) and assume |h| < δ . Then, |H| < |h|(|g′ (c)| + ε1 ) < δ2 , and
thus:
| f (g(c + h)) − f (g(c)) − hg′ (c) f ′ (g(c))|
= | f (g(c) + H) − f (g(c)) − H f ′ (g(c)) − (hg′ (c) − H) f ′ (g(c))|
≤ | f (g(c) + H) − f (g(c)) − H f ′ (g(c))| + |(hg′ (c) − H) f ′ (g(c))|
ε
ε
≤ |H|ε2 + |h|ε1 | f ′ (g(c))| ≤ |h| + |h| = |h|ε.
2
2
Thus, equation (3.1) holds. ⊔
⊓
The key part of the proof lies in the last four lines of calculations. The first equality
and the first inequality are the crucial points, while the preceding part of the proof is
preparatory. What if we use equation (3.2) instead of (3.1)? We can write it as:
f (g(c + h)) − f (g(c)) g(c + h) − g(c)
f (g(c + h)) − f (g(c))
=
.
h
g(c + h) − g(c)
h
Using limh→0 g(c + h) = g(c), we obtain the chain rule as follows:
( f ◦ g)′ (c) = lim
h→0
f (g(c + h)) − f (g(c))
g(c + h) − g(c)
lim
= f′
g′ .
h→0
g(c + h) − g(c)
h
g(c)
c
This equation provides an intuitive understanding of the chain rule. The same relationship
can be observed in the calculations of the proof. However, this is not a complete proof,
as we would need to separately consider cases where the denominator is zero, making the
proof more complicated. In contrast, equation (3.1) does not involve fractions and allows
for a unified proof.
Problem 4.3. Compute the derivative of the composite function ( f ◦ g)(x) using the chain
rule and compare it with direct differentiation after explicitly forming the composite function.
(1) f (x) = x2 , g(x) = 2x + 1
(2) f (y) = y2 , g(x) = 2x + 1
2
(3) f (g) = g , g(x) = 2x + 1
(4) f (x) = x4 , g(x) = x3
10
2
(6) f (x) = x2 , g(y) = cos y
(5) f (x) = x , g(x) = x + 1
Solution 4.3 (1,2,3) are essentially the same problem. The variable x in f (x) and in the
composite function ( f ◦ g)(x) are different. Instead, ( f ◦ g)(x) has g corresponding to x in
f (x). In (5), applying the chain rule is feasible, but explicitly forming and differentiating
the composite function is impractical. In (6), using the chain rule is necessary. ⊔
⊓
Question 4.2 (Butterfly effect and chain-reaction). A small change causing a large impact is often called the butterfly effect. Suppose a phenomenon is represented by the composition of n functions as follows:
32
4 Chain Rule and Variable Centered Notation
H(x) = ( f1 ◦ f2 ◦ · · · ◦ fn )(x) = f1 ( f2 (· · · ( fn (x)) · · · )).
Under what conditions does the butterfly effect occur? Can it be explained using the chain
rule?
The butterfly effect occurs when H ′ (c) is very large. In such cases, even a slight change
in x near c results in a significant change in H(x). When does this happen? By repeatedly
applying the chain rule:
H ′ (c) = f1′ ( f2 (· · · ( fn (c)) · · · )) × f2′ (· · · ( fn (c)) · · · ) × · · · × fn′ (c).
′ ( f (c)), . . . , f ′ ( f (· · · , ( f (c)) · · · ))
Thus, if there exists a c where each derivative fn′ (c), fn−1
n
n
1 2
is large, then the butterfly effect is maximized at that point. For example, if c is mapped
by fn to a point where fn−1 ’s derivative is large, and this continues in a chain reaction,
H ′ (c) can become extremely large, leading to a butterfly effect.
Understanding the chain rule and its proof is not enough; one must also know how to
select f and g appropriately in a given situation.
Problem 4.4. (1) Find h′ (x) when h(x) = (3x2 + 1)2 . (2) Find h′ (x) when h(x) = (3x2 +
1)6 . (3) Find x′ (t) when x(t) = cos(t 2 + 1).
Solution 4.4 (1) First, let’s perform this task without using the chain rule. Expanding, we
get h(x) = 9x4 + 6x2 + 1. Thus, h′ (x) = 36x3 + 12x. Now, let’s use the chain rule. Define
the outer function as f (g) = g2 and the inner function as g(x) = 3x2 + 1. Then, f ′ (g) = 2g
and g′ (x) = 6x. Therefore, by the chain rule,
h′ (x) = ( f ◦ g)′ (x) = f ′ (g)g′ (x) = 2(3x2 + 1)6x = 36x3 + 12x.
Both results match. However, in this case, using the chain rule appears slightly more
complex than direct expansion. But this is only true for simple cases; in most cases, the
chain rule is more convenient.
(2) Expanding (3x2 + 1)6 would be too much work, so we only use the chain rule. Let
f (g) = g6 and g(x) = 3x2 + 1, then f ′ (g) = 6g5 and g′ (x) = 6x. Thus, we have h′ (x) =
6(3x2 + 1)5 6x.
(3) Of course, the cosine function cannot be expanded like a polynomial. We must use
the chain rule. Let f (g) = cos(g) and g(t) = t 2 + 1. Then, f ′ (g) = − sin(g) and g′ (t) = 2t.
Thus, x′ (t) = f ′ (g(t))g′ (t) = − sin(t 2 + 1)2t. ⊔
⊓
Problem 4.5. (1) Find h′ (x) when h(x) = sin(x2 + x). (2) Find h′ (t) when h(t) = tan(5 −
sin(2t)).
Solution 4.5 (1) In this case, let f (g) = sin(g) and g(x) = x2 + x. Then,
h′ (x) = ( f ◦ g)′ (x) = f ′ (g(x))g′ (x) = cos(x2 + x)(2x + 1).
(2) In this case, let f (g) = tan(g) and g(t) = 5 − sin(2t). Then, g′ (t) = − cos(2t)2 and
f ′ (g) = sec2 (g). Thus,
4.2 Variable Centered Notation
33
h′ (t) = − sec2 (5 − sin(2t)) cos(2t)2.
⊔
⊓
4.2 Variable Centered Notation
In this section, we introduce variable-centered notation, a highly flexible notation based
on Leibniz notation. Traditional notation has been function-centered, and moving forward, we will use both approaches. Until now, functions have been represented with symbols such as f , g, variables with symbols such as x, y, and constants with symbols such as
a, b, c. The value of a function f at x = c is written as f (c), and for a general variable x, it
is written as f (x). However, in variable-centered notation, there is no distinction between
functions and variables—everything is a variable. More precisely, there is no separate
notation for representing functions:
f , g, h, x, y, · · ·
All are variables.
For example, these variables can represent specific physical quantities such as position,
velocity, momentum, kinetic energy, or acceleration of an object. Alternatively, they may
denote economic indicators such as per capita income, gross domestic product, economic
growth rate, savings rate, or consumption rate.
Sometimes, certain variables influence others. For instance, consider a case where a
variable x determines another variable y. We can understand their relationship as a function and express it as y = f (x). But do we really need to introduce a new symbol f ? Would
it not be sufficient to simply write y = y(x) or just mentally acknowledge that y is determined by x? We will adopt this approach and refer to it as variable-centered notation. In
this context, Leibniz notation becomes especially useful.
Question 4.3. Can any two arbitrary variables x and y always be considered functions of
each other?
If the value of x uniquely determines the value of y, then we can interpret y as a function
of x. (This does not imply a causal relationship where x causes y.) Sometimes, while y may
not be expressible as a function of x globally, it may be possible locally for small regions
of x. In such cases, we say that y is locally a function of x. However, there exist cases
where no functional relationship is possible even locally.
If three variables f , g, x are locally functionally related, and if these functional relationships are differentiable, then their derivatives can be expressed as follows:
df
,
dx
df
,
dg
dg
,
dx
dg
,
df
dx
,
df
dx
dg
(4.5)
There are a total of six different derivatives. This method of notation is called Leibniz
notation (). The original meaning of ddxf as intended by Leibniz is the ratio of the infinitesimal increase d f in f to the infinitesimal increase dx in x. This is essentially the same
34
4 Chain Rule and Variable Centered Notation
concept as the definition of the derivative as the limit of the average rate of change from
previous lectures. We use it to denote the derivative of f with respect to x.
Differentiating f as a function of x and differentiating f as a function of g yield different
df
df
results, i.e.,
̸=
. However, if f is considered as a function of g, and g as a function
dx
dg
of x, then their composite function should be the same as considering f as a function of
x. By applying the chain rule from equation (4.4), we obtain
d f dg
df
=
.
dx
dg dx
(4.6)
This expresses the chain rule using Leibniz notation.1 With this notation, the chain rule
appears as if it follows fraction cancellation. While dg in these differentiation expressions
is not an actual number that can be canceled, Leibniz notation justifies treating dg as if it
can be canceled when applying the chain rule.
Problem 4.6. (1) The relationship in equation (4.6) is not always satisfied. What are the
necessary conditions?
(2) Using the derivatives from equation (4.5), derive additional relations similar to equation (4.6). How many such relations can be formed?
Solution 4.6 (1) The chain rule must be applicable. That is, the derivative ddgf must be
df
defined at g(c), and the derivative dg
dx must be defined at c. Then, dx is defined at c and
satisfies the given equation.
(2) Including equation (4.6), a total of six such relations can be derived. ⊔
⊓
If the relationship between two variables is understood as a function, then for three
variables, 3 × 2 = 6 functions are required, and for ten variables, 10 × 9 = 90 functions
are needed. Even without introducing new symbols for these functions, their meaning can
be inferred from context. Sometimes, to clarify meaning, we use the notation f (g), which
indicates that f is considered as a function of g, while f (x) means that f is considered as
a function of x. This notation actually makes the meaning clearer. From the perspective of
function-centered notation, such usage may seem excessive and is sometimes referred to
as notation abuse. However, variable-centered notation simplifies expressions even in
complex situations, which is a significant advantage.
Velocity and Acceleration
Let us describe the motion of an object in a one-dimensional space. Consider three variables x, v, a ∈ R, which represent the position, velocity, and acceleration of a specific
object in one-dimensional space, respectively. Can we consider these three variables as
1 Equation (4.6) appears in every calculus textbook. This equation follows a variable-centered notation.
While the f on the left-hand side is the same variable as the f on the right-hand side, they are functionally
different. In function-centered notation, we would introduce a new function symbol, such as h, or express
f ◦g)
it as d(dx
= ddgf dg
dx .
4.2 Variable Centered Notation
35
functions of each other? Locally, this might be possible, but such a relationship is meaningless. The fourth variable, time t ∈ R, is extremely important and special. The unit of
time is arbitrarily given, and the reference point t = 0 is also an arbitrary moment. We
consider the three variables x, v, a as functions of time t. Then, at time t, the position, velocity, and acceleration of a specific object are given. If we only consider functions of the
variable t, there is no difference between function-centered notation and variable-centered
notation.
The velocity v of the object is not determined by the position x itself but by its derivarepresents
tive. That is, x(t + h) − x(t) represents the difference in position, x(t+h)−x(t)
h
the average velocity over time interval h, and the instantaneous velocity at time t, v(t), is
given by the limit:
x(t + h) − x(t)
v(t) = lim
h→0
h
Specifically, when the differentiation variable is time, the derivative is sometimes represented by ẋ instead of x′ . Thus, the following relationships hold:
v = ẋ,
a = v̇ = ẍ
Remark 4.1 (Preview of Part II). Space is three-dimensional. To express the position of a
planet, three coordinate functions x, y, z are required. If we simply denote these functions
as x(t), y(t), z(t), we can express position as r(t) = (x(t), y(t), z(t)). The velocity and acceleration are the first and second derivatives of this position vector function, respectively.
That is, v(t) = ṙ(t) = (ẋ(t), ẏ(t), ż(t)), and a(t) = v̇(t) = r̈(t) = (ẍ(t), ÿ(t), z̈(t)). If the
mass of an object is denoted as m and the force applied to the mass is denoted as F, then
Newton’s Second Law of Motion can be expressed as:
F = ma = mv̇ = mr̈.
(Newton’s Second Law of Motion)
Remark 4.2 (Random Variable). Let X be a random variable representing the price of a
particular stock, and let t be the time variable. Viewing X as a function of t, X(t) (sometimes denoted as Xt ) represents the stock price at time t. The fact that X is a random
variable means that its values are assigned randomly. In what sense is it random? It means
that the change dXt over an infinitesimal time interval follows a certain probability distribution, rather than Xt itself.
Exercises
1. Use the chain rule to differentiate
the composite
√ function f ◦ g.
√
(1) f (u) = u2 , g(x) = x + 1
(2) f (x) = x2 − 1, g(x) = sin x
(3) f (u) = u12 , g(t) = t 3 − t
(4) f (u) = sec u, g(t) = cost
√
(5) f (x) = x, g(u) = cos u
(6) f (s) = 2s2 , g(u) = 5u − 1
2. Differentiate the following
√ using the chain rule.3
(1) (2x + 1)3
(2) t 3 − 2t + 1
(3) cos x
1
(4) 3(cos x)−3
(5) sin(3πx) + cos(2x2 )
(6) cost+sint
36
4 Chain Rule and Variable Centered Notation
3. Differentiate the following q
using the chain rule.
p
√
(1) 1 + tan4 (t 2 )
(2) 2t + 1 + 1 − t
(3) sin3 (cos2 t)
4. Find the equation of the tangent line to the given equation at the specified point.
(1) x2 + y2 = 1 at x = 21
(2) x2 − 2x + 4y2 = 3 at x = 0
(3) sin y + x2 = 0 at x = √12
√
(4) tan(s2 ) + t 3 = 9 at s = 2π
(5) x2 = cos(3t) at t = π4
5. For a positive number n, the function f is given by:
(
sin(xn ) cos(1/x) if x ̸= 0,
f (x) =
0
if x = 0.
Find the smallest value of n such that f is differentiable at x = 0.
6. The values of differentiable functions f and g at each point are given below.
x -3 -2 -1 0 1 2 3 4
f (x) 4 1 0 1 4 9 16 25
g(x) -9 - 83 - 31 0 13 83 9 64
3
(1) Compute the average slope of f , g over each interval [i, i + 1].
(2) Use the table and the Mean Value Theorem to estimate f ′ ( 12 ).
(3) Use the table and the Mean Value Theorem to estimate (g ◦ f )′ ( 12 ).
Lecture 5
Integration & Fundamental Theorem of Calculus
In Lectures 3 and 4, we studied differentiation and differentiation techniques. Now, we
will study integration. Integration is the process of reversing differentiation. Performing
integration requires techniques, which can be developed through practice. The techniques
of integration will be discussed in more detail in Part III.
5.1 Antiderivative
The antiderivative of a function f (x) is a function whose derivative results in f (x). That
is,
F ′ (x) = f (x)
A function F(x) that satisfies this condition is called the antiderivative of the function f (x). Finding the antiderivative is essentially reversing the differentiation process.
However, in practice, finding antiderivatives is not always straightforward. Problem 3.5
presents the rules of differentiation. By applying these rules in reverse, we can sometimes
determine antiderivatives.
The first important point to note is that antiderivatives are not unique. Since differentiation eliminates constant terms, the process of integration introduces a generic constant.
Problem 5.1. Find the antiderivative of the following functions.
(1) f (x) = 0.
(2) f (x) = 1.
(3) f (x) = xα , α ̸= −1.
(4) f (x) = x−1 .
Solution 5.1 Even if the problem simply asks to ”find,” it is understood that all possible
solutions should be found.
(1) Since the derivative of any constant function is 0, any constant function is an antiderivative of 0. That is, F(x) = C for any constant C ∈ R. Here, C is an arbitrary constant,
often referred to as the ”generic constant.” If we multiply or add another constant to C, it
still remains a generic constant. Therefore, sometimes, if we have 3C, we simply rewrite
it as C.
37
38
5 Integration & Fundamental Theorem of Calculus
(2) If we set F(x) = x, then F ′ (x) = 1. Thus, F(x) = x is an antiderivative of the constant function 1. However, since the derivative of a constant is 0, for any constant C ∈ R,
the function F(x) = x +C is also an antiderivative of f (x) = 1.
(3) We know that differentiating a power function decreases its exponent by one, i.e.,
(xα )′ = αxα−1 . Consequently, the exponent of the antiderivative should increase by one,
and we must adjust the coefficient accordingly. Adding a constant, we obtain:
F(x) =
1 α+1
x
+C.
α +1
However, there is a restriction: the denominator in the coefficient cannot be zero, meaning
that α ̸= −1. This is why the problem specifies the condition α ̸= −1. Additionally, this
antiderivative may not be valid for all values of x. As discussed earlier, depending on the
values of α and α + 1, the domain may be all real numbers, x ≥ 0, or x > 0.
(4) What about the case when α = −1, i.e., f (x) = x−1 ? For all other values of α except
−1, we found an antiderivative easily, but this specific case is different. It should be noted
that the function f (x) = x−1 holds special significance. In fact, it plays a crucial role in
mathematics. The antiderivative of this function and its inverse function are among the
most important functions in mathematics. This will be discussed in Lecture 7. ⊔
⊓
Beyond the cases above, more practice is needed to find antiderivatives of various functions.
The antiderivatives found in Problem 5.1 always include a generic constant C. This
constant appears automatically when determining the general antiderivative.
Question 5.1 (Antiderivative of 0). Other than the antiderivatives found above, are there
truly no other antiderivatives? In other words, can we obtain all possible antiderivatives
simply by adding a generic constant C?
This question is equivalent to asking whether the only antiderivatives of 0 are constant
functions. The reasoning is as follows: Suppose F and G are different antiderivatives of
the same function f . Then,
(F − G)′ = F ′ − G′ = f − f = 0.
This means that the difference F − G is an antiderivative of 0. If the only antiderivatives
of 0 are constant functions, then adding a generic constant C accounts for all possible
antiderivatives.
But is there truly nothing other than a constant that satisfies this? Could there exist
some strange function that, when differentiated, results in 0? This is a somewhat philosophical and mysterious question. It can be answered using the Mean Value Theorem.
Since antiderivatives are differentiable functions, the theorem applies. Suppose F is a
differentiable function satisfying F ′ (x) = 0. If F were not a constant function, then there
would exist two points x ̸= y where F(x) ̸= F(y). Consequently, the average rate of change
would be:
5.2 Area Function as an Integral
39
F(x) − F(y)
̸= 0.
x−y
By the Mean Value Theorem, there must exist some c between x and y such that:
F ′ (c) =
F(x) − F(y)
̸= 0.
x−y
However, this contradicts the assumption that F is an antiderivative of 0. Thus, the only
antiderivatives of 0 are constants.
The derivative of a composite function f (g(s)) is given by the chain rule as f ′ (g(s))g′ (s).
That is, the antiderivative of f ′ (g(s))g′ (s) is f (g(s)). The substitution method is an integration technique that reverses the chain rule. To apply this method effectively, one must
carefully choose the function g(s). Using different variable names for f (x) and g(s) helps
to avoid confusion.
Problem 5.2. Use substitution to find the antiderivative of the following functions.
(1) 2s sin(s2 )
(2) x(x2 + 1)α
(3) (x + 1)100
Solution 5.2 (1) Let g(s) = s2 , then g′ (s) = 2s. Thus, the given function can be rewritten
as g′ (s) sin(g(s)). Since sin(x) is the derivative of − cos(x), the antiderivative of 2s sin(s2 )
is − cos(s2 ). Adding the generic constant C, we get:
− cos(s2 ) +C.
(2) Using the same approach, let g(x) = x2 + 1. Then, the given function can be written
as:
1 α ′
g g.
2
Thus, its antiderivative is:
1
1
gα+1 +C =
(x2 + 1)α+1 +C.
2(α + 1)
2(α + 1)
(3) Setting u = x + 1, we have u′ = 1. The function becomes u100 u′ , whose antiderivative
is:
1 101
1
u +C =
(x + 1)101 +C.
101
101
⊔
⊓
5.2 Area Function as an Integral
Let f : R → R be a given function, and define F(t) as the area between the x-axis and
the graph of f over the interval from x = a to x = t > a, as shown in the figure. If the
region lies below the x-axis (i.e., when f < 0), the area is considered negative. The value
a is treated as a fixed constant, while t is considered a variable. This area function is
commonly represented using the integral notation as follows:
40
5 Integration & Fundamental Theorem of Calculus
Z t
F(t) =
f (x)dx.
a
Now, let us determine the derivative of the area function F(t), which is defined in terms
of the area.
Problem 5.3. If F(t) is the area function corresponding to the given function f (t), prove
the following equation:
F ′ (t) = f (t).
(5.1)
However, this equation does not always hold. Provide the necessary conditions on f for
this equation to be satisfied.
Solution 5.3 The derivative of the area function is given by the following limit:
F ′ (t) = lim
h→0
F(t + h) − F(t)
.
h
The numerator F(t + h) − F(t) represents the area of the purple region on the left side of
the figure. Dividing this area by the width h gives an approximation of the function value
f (t), which becomes more accurate as h approaches zero. To rigorously establish this
convergence, the function f must be continuous at t. If f is discontinuous, the left-hand
and right-hand limits will differ. Thus, continuity of f at t is a necessary condition. The
explanation provided here is an intuitive reasoning. In Problem 5.4, we will use the ε-δ
definition to rigorously prove this result. ⊔
⊓
Question 5.2 (Definition of the Area Function). A natural question arises as we proceed
with our discussion: Given a function f (x), is the area function always well-defined?
To determine whether the area function is well-defined, we must first specify how the
area is computed. This is fundamentally a question of how integration is performed. We
define the area using the Riemann sum approach. However, there exist functions for which
the area function cannot be properly defined. This happens mainly for two reasons: (1)
the area diverges to infinity, or (2) the function oscillates too wildly to allow meaningful
calculation.
In calculus, we typically consider functions that are continuous everywhere except at a
finite number of points and that take finite values over any finite interval. If these conditions are met, the function is bounded, ensuring that the area does not diverge to infinity,
and the continuity allows for proper computation of the area.
5.3 Riemann Sum and Area
41
Problem 5.4 (Differentiability of the Area Function). Assume that the area function
F(t) is well-defined. Prove using the ε-δ definition that if f is continuous, then equation
(5.1) holds.
Solution 5.4 Let ε > 0 be given. Since the function f is continuous at x = t, there exists
δ > 0 such that for all |h| < δ , we have | f (t) − f (t + h)| < ε. Consequently, for all |h| < δ ,
we obtain:
f (t + h) < f (t) + ε,
which implies
F(t + h) − F(t) ( f (t) + ε)h
<
< f (t) + ε.
h
h
Similarly, for all |h| < δ , we also have:
f (t + h) > f (t) − ε,
which leads to
F(t + h) − F(t) ( f (t) − ε)h
>
> f (t) − ε.
h
h
Thus, for all |h| ≤ δ , we obtain:
F(t + h) − F(t)
− f (t) ≤ ε.
h
By the definition of the derivative, this shows that F is differentiable at t and that its
derivative is f (t). ⊔
⊓
5.3 Riemann Sum and Area
In this section, we define
the Riemann integral using the partitioning method and use it to
R
define the integral ab f (x)dx. Let f : [a, b] → R be a continuous function defined on the
closed interval [a, b]. First, consider a set of points,
π = {x0 , x1 , · · · , x p }
s.t.
a = x0 < x1 < x2 < · · · < x p = b,
which we call a partition of the closed interval [a, b]. The symbol π is often used for
the mathematical constant pi, but here it denotes a partition. The number of points in the
partition of [a, b] can vary, but it always starts at x0 = a and ends at x p = b, with all points
ordered in increasing sequence. The length of the k-th subinterval is given by:
△xk := xk − xk−1 ,
k = 1, · · · , p.
The gauge of the partition π is denoted as ∥π∥ and is defined as the length of the largest
subinterval:
∥π∥ := max △xk .
1≤k≤p
Finally, the Riemann sum is defined as:
42
5 Integration & Fundamental Theorem of Calculus
p
ck ∈ [xk−1 , xk ].
∑ f (ck )△xk ,
Riemann Sum
k=1
A point ck ∈ [xk−1 , xk ] is chosen within each subinterval, and different choices of ck can
result in different Riemann sums, even when using the same partition π. Let us compute
Riemann sums for some specific cases to understand their properties.
Problem 5.5. Consider the interval [0, 1] and the partition π = nk : k = 0, 1, · · · , n . (1)
Compute the Riemann sum for the constant function f (x) = c and compare it with the
area. (2) Compute the Riemann sum for the function f (x) = x and compare it with the
area. Find the limit of the Riemann sum as n → ∞. (3) Compute the Riemann sum for
the function f (x) = x2 and compare it with the area.
Solution 5.5 (1) Since f (x) = c is a constant function, the choice of ck does not affect the
result:
n
n
n
∑ f (ck )△xk = ∑ c△xk = c ∑ △xk = c.
k=1
k=1
k=1
This matches the area between the graph of the constant function and the x-axis.
(2) Choosing ck ∈ [xk−1 , xk ] as the left endpoint ck = k−1
n , we obtain:
n
n
1
∑ f (ck )△xk = n2 ∑ (k − 1) =
k=1
k=1
n(n − 1)
.
2n2
This result is slightly less than the actual area 21 , but taking the limit as n → ∞, we get 12 .
(3) Choosing ck ∈ [xk−1 , xk ] as the right endpoint ck = nk , we obtain:
n
n
1
∑ f (ck )△xk = n3 ∑ k2 =
k=1
k=1
n(n + 1)(2n + 1)
.
6n3
This result is slightly greater than the actual area 31 , but taking the limit as n → ∞, we get
1
⊓
3. ⊔
Definition 5.1. (1) A number I is called the integral of f over [a, b] if
p
∑ f (ck )△xk = I
∥π∥→0
lim
k=1
for any choice of ck ∈ [xk−1 , xk ]. We denote this as I = ab f (x)dx. (2) If this limit exists,
R
R
we say that f is integrable. (3) We define ba f (x)dx = − ab f (x)dx.
R
Theorem 5.1. (1) If a function f : [a, b] → R is continuous, then it is integrable. (2) If
a function f : [a, b] → R is continuous except at a finite number of points and is bounded,
then it is integrable.
The condition that f is continuous except at a finite number of points is necessary for
integrability. The following problem illustrates why this condition is important.
5.3 Riemann Sum and Area
43
Problem 5.6. Define the function f : R → R as follows:
(
1, if x is a rational number,
f (x) =
0, if x is an irrational number.
Show that this function f is not integrable over any finite interval [a, b].
Solution 5.6 Every subinterval [xk−1 , xk ] contains both rational and irrational numbers.
If we always choose ck as a rational number, the Riemann sum becomes 1, while if we
always choose ck as an irrational number, the Riemann sum becomes 0. According to the
definition of the integral, the Riemann sums must converge to the same value regardless
of the choice of ck . However, since different choices yield different limits, the function is
not integrable. ⊔
⊓
Problem 5.7. Let f and g be integrable functions defined on the interval [a, b]. Prove the
following:
R
R
(1) ab k f (x)dx = k ab f (x)dx for any k ∈ R
R
R
R
(2) ab ( f + g)dx = ab f (x)dx + ab g(x)dx
Rb
Rc
Rb
(3) a f (x)dx = a f (x)dx + c f (x)dx for any c ∈ [a, b]
R
R
(4) If f (x) ≤ g(x) for all x in [a, b], then ab f (x)dx ≤ ab g(x)dx.
Solution 5.7
⊔
⊓
Now, we introduce the Fundamental Theorem of Calculus (FTC) in two forms.
Theorem 5.2 (Fundamental Theorem of Calculus). RLet f be a continuous function on
the closed interval [a, b]. Then, the function F(x) := ax f (t)dt is differentiable, and its
derivative satisfies:
F ′ (x) = f (x).
Proof. To prove this theorem, we need to utilize the properties of integration. However,
we accept this result without proof. ⊔
⊓
The Fundamental Theorem of Calculus states that a continuous function is integrable
and that the function obtained by integrating a continuous function is differentiable, with
its derivative returning to the original function. In other words, differentiation and integration are inverse processes.
Theorem 5.3 (Cauchy’s Fundamental Theorem of Calculus). Let f be a continuous
function on the closed interval [a, b] and let F(x) satisfy F ′ (x) = f (x). Then, the integral
satisfies:
Z b
f (x)dx = F(b) − F(a).
a
Proof. We prove this using the first version of the Fundamental Theorem. Define G(x) =
a f (t)dt. Then, G(x) is also an antiderivative of f . Therefore,
Rx
(F − G)′ (x) = F ′ (x) − G′ (x) = f (x) − f (x) = 0.
44
5 Integration & Fundamental Theorem of Calculus
Since F − G is an antiderivative of 0, it must be a constant. That is, there exists a constant
C such that F − G = C. Thus, F = G +C, and
F(b) − F(a) = G(b) − G(a) =
Z b
f (x)dx.
a
⊔
⊓
Cauchy’s version provides a method for computing integrals or areas. One first
finds an
R
antiderivative F and then calculates F(b)−F(a) to determine the integral value ab f (x)dx.
If an antiderivative can be found, this approach is preferable to computing the integral
using Riemann sums. However, finding an antiderivative is not always easy, and when
using computers, numerical integration via Riemann sums is often more practical.
Problem 5.8. Using the Fundamental Theorem of Calculus (FTC), compute the following
integrals:
Z a
(1)
Z a
x2 dx
(2)
0
Z a
xα dx
(3)
0
(x2 + x4 )dx
Z 2
(4)
0
x−1 dx
1
⊔
⊓
Solution 5.8
Exercises
1. Let π be a partition of the interval [a, b], and let ck ∈ [xk−1 , xk ]. Compute the limit of
the following Riemann sums and determine the conditions on [a, b] for convergence.
p
p
∑ 2c2k ∆ xk
∥π∥→0
∑ sin(ck )∆ xk
∥π∥→0
(1) lim
(2) lim
k=1
p
k=1
∑ (3ck + c2k )∆ xk
∥π∥→0
(4) lim
k=1
p
p
∑ c−1
k ∆ xk
(5) lim
(6) lim
∥π∥→0 k=1
p
∑ tan(ck )∆ xk
∥π∥→0
(3) lim
k=1
∑ c0.5
k ∆ xk
∥π∥→0 k=1
2. Find the antiderivatives of the following functions:
(2) f (y) = y−2
(3) f (t) = −|t|
(1) f (x) = x2
(4) f (t) = 3t − 1
(5) f (y) = sin(y)
(6) f (x) = x−1
3. Compute the following integrals:
Z 2
(1)
Z0 3
(3)
(x − x − 1) dx
|t| dt
Z−2
2π
(5)
0
Z 2
2
(4)
(2)
1
Z 3 2
y − y4
y2
1
sin(2x)
dx
sin(x)
Z π
(6)
0
x100 dx
dy
(cos(x) + | cos(x)|) dx
5.3 Riemann Sum and Area
45
4. Compute the following derivatives:
Z
Z 2
d x 2
d t
(1)
t dt
(2)
f (x) dx
dx 0
dt 2
Z 1
Z √t
d
d
x−2/3 dx
(4)
(3)
√ sin(t) dt
dx x
dt 1
5. Compute the area of the region bounded by y = x2 and y = 1.
√
6. Compute the area of the region bounded by y = x, x = 1, and y = 0.
7. A particle moves along the x-axis with velocity v(t) = 3t 2 − 6t + 2. Compute the total
distance traveled over the interval [0, 3].
8. The population of a city at time t changes at a rate given by P′ (t) = 1000 − 2t. If the
initial population is 50,000, determine the population function P(t).
Z x
9. Use the Chain Rule to differentiate the given functions: (1) f (x) =
Z cos(x)
sin(x)
1
dt (3) h(x) =
1 − t2
Z √x−π
√
− x
1
1/x t
dt (2) g(x) =
cos(t 2 ) dt
10. Given a continuous function f on the interval [0, 3], compute the following:
R 2
(1) Given G(x) = 0x f (t) dt, find G′ (1).
R 3
(2) Given H(x) = xx f (t) dt, find H ′ (3).
R 2
(3) Given F(x) = 1x f (t 2 ) dt, find F ′′ (2).
11. Let the function f be defined as follows:
(
−1, for x < 0,
f (x) =
2,
for x ≥ 0.
Prove the following:
(1) Compute the Riemann sum of f for the partition π = { ki : k = 0, ±1, ±2, . . . , ±n}
over the interval [−1, 1], and find its limit as n → ∞.
(2) Show that the function f is integrable over every interval.
12. Define the function g : R → R as follows:
Z x
g(x) =
sin(x − t) dt.
0
Show that the function y = g(x) satisfies the differential equation:
y′′ + 4y = 0,
y(0) = 0.
13. Let the function f be defined on the interval [−2, 2] as follows:
(
x2 − 1, if − 2 ≤ x < 0,
f (x) =
x + 2, if 0 ≤ x ≤ 2.
46
5 Integration & Fundamental Theorem of Calculus
Rx
Define the area function as F(x) = −2 f (t) dt. Prove the following:
(1) Is the function F continuous on the interval [−2, 2]?
(2) Show that there exists a constant c in the interval [−2, 2] such that F(c) = 0.
(3) Does the function f satisfy the Fundamental Theorem of Calculus for all x ∈
[−2, 2]? If not, provide a proof.
(4) Does the function f satisfy the Fundamental Theorem of Calculus for all x ∈ [0, 2]?
If so, compute F ′ (1).
14. (Mean Value Theorem for Integration) Suppose the function f is continuous on the
interval [a, b]. Prove that there exists a constant c in [a, b] such that:
1
b−a
Z b
f (x) dx = f (c).
a
Lecture 6
Inverse Functions and Their Derivatives
This chapter introduces the definition of inverse functions, the conditions under which
they exist, and how to compute their derivatives using the chain rule. We will also explore
integration techniques using variable substitution.
6.1 Bijection (One-to-One and Onto Function)
When denoting a function f as f : A → B, the set A is called the domain, and the set B is
called the co-domain. The subset of B containing all the values taken by the function is
denoted as
R( f ) := {y ∈ B | there exists x ∈ A such that y = f (x)},
and is called the range of f .
A function must assign exactly one value to each element of its domain. If an element
is assigned more than one value, the mapping is not a function. While this definition has
been implicitly used, it is formally stated as follows:
Definition 6.1. The mapping f : A → B is called a function if for each x ∈ A, there exists
a unique y ∈ B such that f (x) = y.
The definition of an inverse function is straightforward. However, since an inverse function must also be a function, we must be careful about the conditions under which it exists.
Definition 6.2. A function f : A → B is called a one-to-one (or injection) if no value y ∈ B
is taken more than once. In other words, if f (x1 ) = f (x2 ), then it must be that x1 = x2 .
A function f : A → B is called an onto (or surjection) if every element y ∈ B is taken at
least once, meaning the range R( f ) = B. A function f is called a bijection if it is both
one-to-one and onto.
We can define an inverse function only for functions that are both one-to-one and onto.
Definition 6.3. If f : A → B is a bijection, we can define a function g : B → A such that
g(y) is the preimage of y under f , meaning that f (g(y)) = y. This function g is called the
inverse function of f . The inverse function of f is typically denoted by f −1 .
47
48
6 Inverse Functions and Their Derivatives
The notation f −1 for inverse functions can sometimes be inconvenient. We often use the
notation f α to represent the composition of a power function with f , such that ( f )α = f α .
This notation coincides with inverse functions when α = −1. However, these two cases
are different, so we must determine from the context whether f −1 represents the inverse
′
function of f or simply 1f . In particular, writing derivatives as f −1 (x) or ( f −1 )′ (x) can
be cumbersome. Therefore, it is often more convenient to explicitly define the inverse
function as g : B → A, as done in the definition above.
Problem 6.1. Show that if a function f : A → B is a bijection, then its inverse function is
well-defined.
Solution 6.1 To show that the inverse function is well-defined in this case, we must
demonstrate that the reverse mapping defines a function. Specifically, we need to show
that g : B → A is a function, meaning that it assigns a unique value g(y) for every y ∈ B.
Since f is onto, for every y ∈ B, there exists at least one x ∈ A such that f (x) = y,
meaning that g(y) is always defined.
Furthermore, if g(y) assigned two different values x1 ̸= x2 , then f (x1 ) = f (x2 ) = y,
which contradicts the assumption that f is one-to-one. Therefore, g(y) is uniquely defined
for every y ∈ B, proving that g is a function. ⊔
⊓
Question 6.1. What is the relationship between the graphs of a bijective function f (x)
and its inverse function f −1 (x)? The two graphs are symmetric about the line y = x. Can
you explain why?
The graph of a function f consists of the points in the xy-plane satisfying y = f (x).
These same points also satisfy the inverse relationship x = f −1 (y). Thus, the graphs of
f (x) and f −1 (y) are the same. The only difference is whether we consider the independent
variable to be x or y.
At a point (x0 , y0 ) on the graph of f (x), the slope of the tangent line is f ′ (x0 ). This
same tangent line is also the tangent to the graph of x = f −1 (y) at (x0 , y0 ). The derivative
′
f −1 (y0 ) represents the slope of this tangent line when viewed with y as the independent
△y
variable. Since the slope △x
△y is the reciprocal of △x , the following relationship holds:
′
f −1 (y0 ) =
1
f ′ (x0 )
if
y0 = f (x0 ),
f ′ (x0 ) ̸= 0.
6.1 Bijection (One-to-One and Onto Function)
49
If we wish to place y on the horizontal axis, we must reflect the graph across the line
y = x, producing the second diagram. If we then relabel the variable as x, we obtain the
third diagram. However, sometimes it is preferable to keep the original variable notation
to avoid confusion. The above reasoning leads to the following theorem.
Theorem 6.1 (Derivative of the Inverse Function). Let f : A → B be a bijective function
that is continuously differentiable at x0 ∈ A, with f ′ (x0 ) ̸= 0. Then, its inverse function
g : B → A is continuously differentiable at f (x0 ) ∈ B, and its derivative is given by:
g′ ( f (x0 )) =
1
.
f ′ (x0 )
(6.1)
What if a function f : A → B is not bijective? In that case, we cannot define an inverse
function for the entire domain. Instead, we can restrict A and B to create a smaller one-toone and onto function. This process is known as choosing a branch, and different branches
lead to different inverse functions. When choosing a branch, we aim to include as much
of the domain as possible. For commonly used functions, standard branch choices exist
and should be followed.
Problem 6.2. Find sets A, B ⊂ R that make the following functions f : A → B bijections.
Find their inverse functions.
(1) f (x) = 12 x + 1 (2) f (x) = x2
Solution 6.2 Let g denote the inverse function of f .
(1) If we set A = R and B = R, then f is one-to-one and onto. To find the inverse function,
we start by setting f (x) = y and solving for x in terms of y:
1
x + 1 = y ⇒ x = 2(y − 1) ⇒ g(y) = 2(y − 1).
2
Thus, the inverse function is g(y) = 2(y − 1). Of course, writing it as g(x) = 2(x − 1) is
also fine, but unnecessary.
(2) The function f is not one-to-one over R. To restrict it to a one-to-one domain, we can
choose either A = [0, ∞) or A = (−∞, 0]. Most people prefer A = [0, ∞), so we follow that
convention. Additionally, setting B = [0, ∞) ensures f : A → B is a bijection. In this case,
y = x2 ⇒ x =
Thus, the inverse function is g(y) =
√
√
y ⇒ g(y) = y,
y ≥ 0.
√
y. ⊔
⊓
To differentiate an inverse function, we can either explicitly compute the inverse and
then differentiate it, or use the inverse function differentiation formula. The following
problem examines both approaches and compares the results.
Problem 6.3. Find the inverse function and its derivative, and compare the result with the
inverse function differentiation formula.
(1) f (x) = x2
(2) f (x) = x3 − 2, find ( f −1 )′ (6).
Solution 6.3 (1) Since f (x) is not a bijection over all real numbers, we must restrict its
domain to ensure bijectivity. The most natural choice is the branch where x, y ≥ 0. Setting
50
6 Inverse Functions and Their Derivatives
y = x2 , we solve for x:
√
y, x, y ≥ 0.
√
Thus, the inverse function is f −1 (y) = y = y1/2 . Differentiating directly gives:
x=
1
1
( f −1 )′ (y) = y−1/2 = √ .
2
2 y
Now, using the inverse function differentiation formula:
( f −1 )′ (y) = ( f −1 )′ ( f (x)) =
Do these two results match? Yes, since x =
1
1
= .
f ′ (x) 2x
√
y, they are equivalent.
(2) This function is a bijection over all real numbers. Setting y = x3 − 2, we solve for x:
x = (y + 2)1/3 .
Thus, the inverse function is f −1 (y) = (y + 2)1/3 , and differentiating directly:
1
( f −1 )′ (y) = (y + 2)−2/3 .
3
Using the inverse function differentiation formula:
( f −1 )′ (y) = ( f −1 )′ ( f (x)) =
1
f ′ (x)
=
1
.
3x2
Again, the two results match. Now, we compute ( f −1 )′ (6). Substituting into our first
expression:
1
1
( f −1 )′ (6) = 8−2/3 = .
3
12
For the second formula, we first determine x when y = 6, finding x = 2, then computing:
( f −1 )′ (6) =
1
f ′ (2)
=
1
1
= .
2
3×2
12
Both results agree. ⊔
⊓
In many cases, explicitly computing an inverse function requires substantial effort or
may not be possible at all. Even when feasible, it is often more convenient to apply the
inverse function differentiation formula rather than first finding the inverse and then differentiating.
6.2 Inverse Functions and Variable-Oriented Notation
The inverse function differentiation formula can be derived using the chain rule. Suppose
f : A → B and g : B → A are inverse functions, with f (x0 ) = y0 and f ′ (x0 ) ̸= 0. Then, we
6.2 Inverse Functions and Variable-Oriented Notation
51
have g( f (x)) = x, so differentiating both sides gives:
(g ◦ f )(x) = x
⇒
(g ◦ f )′ (x) = 1.
Applying the derivative rule for composite functions, assuming g is differentiable at f (x0 ),
we obtain:
1
g′ ( f (x0 )) f ′ (x0 ) = 1 ⇒ g′ ( f (x0 )) = ′
.
f (x0 )
This derivation assumes that g is differentiable at f (x0 ).
Now, let us consider inverse functions and their derivatives using variable-oriented notation. Suppose we have two variables, x and y, where x can be expressed as a function
of y, meaning that for a given y, there is a unique corresponding x. In variable-oriented
notation, we do not introduce a separate function name f , but instead simply write x(y).
Similarly, we can express y as a function of x. Under these assumptions, if A is the range
of y and B is the range of x, then x : A → B is a bijection, and x(y) and y(x) are inverse
functions of each other.
The relationships x(y(x)) = x and y(x(y)) = y are trivial identities. Using Leibniz notation, the inverse function differentiation formula is written as:
dx
−1
dy
−1
dy
dx
=
,
=
.
(6.2)
dx at (x0 ,y0 )
dy at (x0 ,y0 )
dy at (x0 ,y0 )
dx at (x0 ,y0 )
Here, y takes the value y0 when x = x0 , and the derivatives are computed at that location.
Problem 6.4. The two variables x and y are dependent on each other, meaning that once
one is determined, the other is also uniquely determined. Thus, they can be understood as
functions of each other.
dy
(1) When the relationship is x + y = 1, compute dx
dy and dx at x = 1.
dy
(2) When the relationship is x2 + y2 + y = 1, compute dx
dy and dx at x = 1.
(The calculations require implicit differentiation, but understanding the meaning in relation to the differentiation of inverse functions is crucial.)
dy
= 0, so we
Solution 6.4 (1) Differentiating the equation with respect to x gives 1 + dx
dy
dx
find that dx is the constant −1. Differentiating with respect to y gives dy + 1 = 0, so dx
dy is
also −1. This satisfies equation (6.2).
(2) Differentiating x2 + y2 + y = 1 with respect to x gives:
2x + (2y + 1)
dy
Solving for dx
, we obtain:
dy
= 0.
dx
dy
2x
=−
.
dx
2y + 1
Differentiating with respect to y gives:
2x
dx
+ 2y + 1 = 0.
dy
52
6 Inverse Functions and Their Derivatives
Solving for dx
dy , we obtain:
dx
2y + 1
=−
.
dy
2x
This satisfies equation (6.2). At x = 1, we have y = 0 or y = −1, so:
2x
dy
=−
= 2 or − 2.
dx
2y + 1
⊔
⊓
6.3 Integration by Change of Variables
In Problem 5.2, the substitution technique was used to find antiderivatives. However, applying substitution in definite integration requires an additional condition: the function
g(x) must be one-to-one. For this reason, substitution in definite integration is more appropriately called a change of variables.
Consider the definite integral of the function f : [a, b] → R and compute it using the
R
integral of the composite function f ◦ g : [c, d] → R. To evaluate the integral ab f (u)du,
we first partition the interval [a, b] using π = {u0 , u1 , . . . , u p } and choose representative
points ui within each subinterval [ui−1 , ui ]. The integral can then be approximated as:
Z b
a
p
f (u)du ≈ ∑ f (ui )∆ ui ,
∆ ui = ui − ui−1 .
i=1
Now, consider integrating the composite function f (g(x)). Since g : [c, d] → [a, b] is a
one-to-one function, it must be either increasing or decreasing. First, consider the case
where g(x) is increasing. Define a partition π = {x0 , x1 , . . . , x p } such that g(xi ) = ui and
choose the representative points xi = g−1 (ui ) for each subinterval [xi−1 , xi ]. Then, we have
f (g(xi )) = f (ui ), and the integral approximation becomes:
Z d
c
p
p
f (g(x))dx ≈ ∑ f (g(xi ))∆ xi = ∑ f (ui )∆ xi ,
i=1
∆ xi = xi − xi−1 .
i=1
The difference between these two integrals arises not from the function values but from
the difference between ∆ x and ∆ u, i.e., the partition sizes or domain transformations. In
6.3 Integration by Change of Variables
53
the limit, as the partition is refined, their ratio converges to:
∆ ui
g(xi ) − g(xi−1 )
→ g′ (xi ) as ∥π∥ → 0.
=
∆ xi
xi − xi−1
Thus, this ratio converges to the derivative of the transformation function g. The relationship between the two definite integrals is then:
Z b
a
p
p
f (u)du ≈ ∑ f (ui )∆ ui = ∑ f (g(xi ))
i=1
i=1
∆ ui
∆ xi ≈
∆ xi
Z d
f (g(x))g′ (x)dx.
c
When g(x) is an increasing function, the change of variables formula is:
Z b
Z d
f (u)du =
a
f (g(x))g′ (x)dx,
g′ (x) ≥ 0.
c
Problem 6.5. Derive the change of variables formula when the transformation function
g(x) is decreasing.
Solution 6.5 In this case, the partition π = {x0 , x1 , . . . , x p } of [c, d] satisfies g(xi ) = u p−i ,
and the representative points satisfy g(xi ) = u p−i+1 . Then, the integral approximation is:
Z d
c
p
p
p
f (g(x))dx ≈ ∑ f (g(xi ))∆ xi = ∑ f (u p−i+1 )∆ xi = ∑ f (ui )∆ x p−i+1 .
i=1
i=1
i=1
The ratio of subinterval sizes is:
g(x p−i+1 ) − g(x p−i )
ui − ui−1
∆ ui
=
=−
≈ −g′ (x p−i+1 ).
∆ x p−i+1
x p−i+1 − x p−i
x p−i+1 − x p−i
Thus, we obtain:
Z b
f (u)du = −
a
Z d
f (g(x))g′ (x)dx,
g′ (x) ≤ 0.
⊔
⊓
c
The formula for change of variables in definite integrals holds regardless of whether
the transformation function g is increasing or decreasing. It can be expressed as:
Z b
Z d
f (u)du =
a
f (g(x))|g′ (x)|dx.
(6.3)
c
The absolute value |g′ (x)| inside the integral ensures that this formula remains valid even
when g is a decreasing function.
Problem 6.6. If we write the change of variables formula (6.3) without the absolute value,
we obtain:
Z
Z
g−1 (b)
b
f (u)du =
a
g−1 (a)
Explain why these two formulas are equivalent.
f (g(x))g′ (x)dx.
(6.4)
54
6 Inverse Functions and Their Derivatives
Solution 6.6 When the transformation function g is increasing, the two formulas are
identical. When g is decreasing, we have g−1 (a) = d, g−1 (b) = c, and g′ (s) ≤ 0. This
gives:
Z g−1 (b)
g−1 (a)
f (g(s))g′ (s)ds =
Z c
f (g(s))g′ (s)ds =
Z d
d
f (g(s))|g′ (s)|ds.
c
Thus, equation (6.4) is equivalent to (6.3). ⊔
⊓
Problem 6.7. Evaluate the following definite integrals using change of variables.
Z 2
(1)
x sin(x2 )dx
Z 1
(2)
0
Z 1
s sin(1 − s2 )ds
(3)
2
xex dx
−1
0
Solution 6.7 The problems are designed to be solved easily using the change of variables
technique. While the method is the same, multiple approaches can be used to find the
answers.
(1) This integral involves a composite function. Setting g(x) = x2 , we see that g :
[0, 2] → [0, 4] is a one-to-one function. Since g′ (x) = 2x, we rewrite the integral as:
Z 2
x sin(x2 )dx =
0
Z x=2
1
1 ′
g (x) sin(g(x))dx =
2
x=0 2
Z g=4
4
1
sin(g)dg = − cos(g) .
2
0
g=0
(2) The variable s plays the same role as x. Instead of using g, we introduce u = 1 − s2 .
(Using u or g makes no difference.) Over the interval [0, 1], the function is one-to-one,
and since du = −2sds, we rewrite the integral as:
Z s=1
s sin(1 − s2 )ds = −
s=0
1
2
Z u=0
sin udu =
u=1
1
2
Z u=1
1
1
sin udu = − cos u .
2
0
u=0
(3) Setting u = x2 , we have du = 2xdx, and the integral transforms into:
Z x=1
2
xex dx =
x=−1
1
2
Z u=1
eu du = 0.
u=1
However, this result is incorrect because the function g(x) = x2 is not one-to-one over the
interval [−1, 1]. Instead, we must split the integral into two parts:
Z 1
−1
2
xex dx =
Z 0
−1
2
xex dx +
Z 1
2
xex dx.
0
With this approach, g is one-to-one over each subinterval. ⊔
⊓
Exercises
1. Find the inverse function of the following functions along with their domains.
(1) f (x) = x5
(2) f (x) = 2x + 1
(3) f (x) = (1 − x)3
6.3 Integration by Change of Variables
55
2. Compute the derivative of the inverse function ( f −1 )′ (2), i.e., the derivative at y = 2.
x
(1) f (x) = x5
(2) f (x) = 2√
+1
(3) f (x) = x3 + 1
x
x+3
(4) f (x) =
(5) f (x) = √
(6) f (x) = (x2 − 1)0.3
x−1
x−3
3. Given the function f (x) = −x3 − 4x2 − x + 3, do the following:
(1) Find the largest interval [a, b] containing −1 such that f is one-to-one.
(2) If g is the inverse function of f on the interval [a, b], find g′ (−3).
4. Given the following relationships between two variables x and y, compute the derivady
dx
and
at the given points and verify whether they satisfy equation (6.2).
tives
dy
dx
(1) x2 + y2 = 25, (x0 , y0 ) = (3, 4)
(2) xy + y2 = 5, (x0 , y0 ) = (4, 1)
2
x
(3) y + 2 = 8, (x0 , y0 ) = (2, 2)
(4) cos x + sin y = 0, (x0 , y0 ) = (π, π/2)
5. Compute
integrals.
Z √ the following indefinite
Z
2
(1)
3 − xdx
(2) √
dx
2x + 4
Z
Z
(4)
Z
(7)
3x(x2 + 1)3 dx (5) sin x cos xdx
Z p
√
(8) x x2 − 1dx
x x − 1dx
1
x
√
dx (11)
dx
(4x + 1)3/2
4x + 1
6. Compute the following definite integrals.
Z 1
Z 3√
3 − xdx
(2)
3x(x2 + 1)3 dx
(1)
Z
0
0
Z π
Z π/2
sin x cos xdx
0
(7)
Z 1 p
x
0
Z √
7s − 1ds
Z
x2 + 1dx
(5)
sin x cos3 xdx
√
x2 x − 1dx
(6)
Z
(9)
Z
(10)
(4)
(3)
Z
(12)
x3 (x2 − 1)2/3 dx
Z 1
(3)
3x(x2 + 1)3 dx
Z−12π
sin x cos xdx
−π/2
Z 1
(8)
0
x
−1
(6)
p
x2 + 1dx
sin x cos3 xdx
Lecture 7
Logarithm, Exponential, & Implicit Differentiation
In this lecture, we will explore the logarithm function, one of the most important functions, and its inverse, the exponential function.
7.1 Natural Logarithm
In the previous lecture, we were able to find the antiderivative of the power function xα
for all α except α = −1. So, what is the antiderivative of x−1 ? We define the area function
of f (x) = x−1 and call this function the natural logarithm, denoted and defined as:
Z x
ln x =
1
dt.
1 t
(7.1)
This area function calculates the area starting from a = 1. Thus, for x > 1, we have ln x > 0,
and for x < 1, we have ln x < 0, with ln 1 = 0.
Question 7.1. Why is the natural logarithm ln x defined by integrating from 1 to x?
Wouldn’t it be more natural to integrate from 0 to x? Or perhaps from x to ∞? Is there
a specific reason why we must integrate from 1?
The most natural choice might seem to be integrating from 0. However, we cannot
integrate x−1 from 0 because the integral diverges to infinity:
57
58
7 Logarithm, Exponential, & Implicit Differentiation
Z x
1
dt = ∞.
c→0 c t
lim
The next natural choice would be integrating from x to ∞, but that is also not possible
because this integral also diverges to infinity:
Z ∞
x
c1
1
dt = lim
dt = ∞.
c→∞ x t
t
Z
Thus, the only possible choice is to integrate from a nonzero real number, and people
chose 1. As a result, equation (7.1) became the definition of the natural logarithm ln x.
In summary, (1) the natural logarithm ln x is only defined for x > 0, and (2) for x > 0, it
d
ln x = x−1 .
satisfies dx
Problem 7.1. Find the antiderivative of x−1 .
d
Solution 7.1 (Ignoring the general constant C for now.) Since dx
ln x = 1x , the function
ln x is only a partial answer because it is only valid for x > 0. However, x−1 is defined
for all x ̸= 0, so we need a function that also holds for x < 0. The answer is ln |x|. This
function is defined for all x ̸= 0. We already know that its derivative is x−1 for x > 0, so
we only need to check for x < 0. Since x < 0, we have ln |x| = ln(−x), and differentiating
gives:
d
d
1
1
ln(|x|) =
ln(−x) =
× (−1) = , x < 0.
dx
dx
−x
x
d
ln |x| = x−1 , meaning the antiderivative of x−1 is ln |x| +C.
Thus, for all x ̸= 0, we have dx
⊔
⊓
Now, we have found the antiderivative for all power functions xα :
(
Z
1
xα+1 +C, α ̸= −1,
xα dx = α+1
ln |x| +C,
α = −1.
The following problem explores some useful properties of the natural logarithm.
Problem 7.2 (Properties of the Natural Logarithm). Prove the following properties.
(1) ln(xy) = ln x + ln y.
(2) ln xk = k ln x for x > 0 and k ∈ R.
′ (x)
d
(3) ln(x/y) = ln x − ln y.
(4) dx
ln | f (x)| = ff (x)
for all x such that f (x) ̸= 0.
Solution 7.2 Differentiation is very useful in proving these rules.
To prove (1), we differentiate both sides. We must differentiate both sides in the same
manner. Differentiating with respect to x, and applying the chain rule, we obtain:
1
1
d
ln(xy) = y = ,
dx
xy
x
d
1
(ln x + ln y) = .
dx
x
Since both derivatives are equal, the original functions must differ by a constant. That is,
ln(xy) = ln x + ln y +C.
To find the constant C, we substitute x = 1, yielding C = 0. This completes the proof.
7.1 Natural Logarithm
59
To prove (2), we differentiate both sides:
1
k
d
ln xk = k kxk−1 = ,
dx
x
x
d
k
k ln x = .
dx
x
Since both derivatives are equal, the original functions must differ by a constant:
ln xk = k ln x +C.
Substituting x = 1, we obtain C = 0. The proof is complete.
(3) follows directly from the properties already proven:
ln(x/y) = ln(xy−1 ) = ln x + ln y−1 = ln x − ln y.
(4) follows from the chain rule and the fact that ln |x| is the antiderivative of x−1 . ⊔
⊓
Problem 7.3. Let a > 0. Using Problem 7.2, prove the following:
d x
a = ax ln a.
dx
Solution 7.3 We prove this by differentiating ln ax in two different ways. Using Problem
7.2(2),
d
d
ln(ax ) = (x ln a) = ln a.
dx
dx
Now, using Problem 7.2(4), we note that since a > 0, we have ax = |ax |. Thus,
d
(ax )′
ln(ax ) = x .
dx
a
Since both expressions must be equal, we obtain (ax )′ = ax ln a. ⊔
⊓
Using the natural logarithm, we can simplify the differentiation of complex fractional
functions, as shown in the following problem.
2
1/2
Problem 7.4. Differentiate the function (x +1)(x+3)
x−1
.
Solution 7.4 Attempting to differentiate this function directly using the quotient and
product rules leads to cumbersome calculations. However, taking the logarithm simplifies
multiplication, division, and exponentiation, making differentiation easier. Observe the
process carefully. Define the function as:
y=
(x2 + 1)(x + 3)1/2
.
x−1
Taking the natural logarithm of both sides and using logarithmic properties, we obtain:
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1).
2
Differentiating both sides with respect to x, we get:
60
7 Logarithm, Exponential, & Implicit Differentiation
y′
2x
1
1
= 2
+
−
.
y
x + 1 2(x + 3) x − 1
Multiplying both sides by y and substituting y back, we obtain:
!
2x
1
1
(x2 + 1)(x + 3)1/2
y′ =
+
−
.
x2 + 1 2(x + 3) x − 1
x−1
For simple fractions, direct computation is manageable, but for more complex functions,
using logarithmic differentiation is advantageous. ⊔
⊓
Problem
7.5. Find theZfollowing indefinite
Z
Z integrals.
2 cos x
dx
(1) cot x dx
(2) tan x dx
(3)
3 + 2 sin x
Z
Z
(4) sec x dx
(5) csc x dx
Solution 7.5 Finding antiderivatives is generally difficult. However, these problems use
the logarithmic property from Problem 7.2(4) to determine the antiderivative. If a given
function can be expressed as
f ′ (x)
F(x) =
,
f (x)
then the antiderivative follows directly from Problem 7.2(4):
Z
F(x)dx = ln | f (x)| +C.
Be sure to include the absolute value. ⊔
⊓
7.2 Exponential Function
The special number e, also known as the Euler number or the natural constant, is defined
as the number satisfying:
Z e
1
dt = 1.
1 t
That is, it is the number for which ln e = 1. Like π, this number is an irrational number.
Now, let’s consider the derivative of the exponential function with base e, given by f (x) =
ex . According to Problem 7.3,
d x
e = ex ln e = ex .
dx
This means that the function is unique in that its derivative is the same as the function
itself.
Now, let’s consider the inverse function of the natural logarithm. The inverse function
of the natural logarithm ln x is called the exponential function and is denoted by exp x. If
we write the natural logarithm function as ln : A → B, its domain is A = (0, ∞) and its range
is B = (−∞, ∞) = R. Therefore, the exponential function is defined as exp : R → (0, ∞).
7.3 Implicit Differentiation
61
Now, let’s compute the derivative of exp x. If we set y = exp(x), then x = ln(y). Using
the differentiation rule for inverse functions, we obtain:
exp′ (x) =
1
1
=
= y = exp(x).
ln′ (y) 1/y
In other words, like the function ex , the exponential function also has the property that its
derivative is itself. Could these two functions be the same? What would we need to show
to prove that they are the same function? Consider the following problem.
Problem 7.6. Suppose two functions f (x) and g(x) are both positive and satisfy the following:
f ′ (x) = f (x), g′ (x) = g(x), f (c) = g(c) for some c.
Show that the two functions are identical.
f (x)
is well-defined. Using the
Solution 7.6 Since both functions are positive, the fraction g(x)
given conditions, we can show that the derivative of this fraction is zero, which implies
that the fraction is a constant function. Since f and g are equal at x = c, this constant must
be 1. That is, for all x,
f (x)
= 1.
g(x)
Thus, we conclude that f (x) = g(x). ⊔
⊓
Both functions ex and exp x are positive and satisfy the differential equation f ′ (x) =
f (x). Moreover, since e0 = 1 and ln 1 = 0, we have exp 0 = 1. Therefore, these two functions must be identical, meaning:
exp x = ex ,
x ∈ R.
Both notations are used, but ex is more commonly used. Now, let’s summarize some
important properties of the exponential function.
Theorem 7.1 (Exponential Function). The exponential function ex satisfies the following
properties:
d x
d f (x)
(1) dx
e = ex .
(2) dx
e
= f ′ (x)e f (x) .
(3) ex ey = ex+y .
Proof. (1) has already been shown. (2) follows directly from (1) by applying the chain
rule. To prove (3), we take the natural logarithm of both sides:
ln(ex ey ) = ln ex + ln ey = x + y = (x + y) ln e = ln(ex+y ).
Since the natural logarithm is a one-to-one function, we conclude that ex ey = ex+y . ⊔
⊓
7.3 Implicit Differentiation
(This section would logically follow Section 4.2, but for better content distribution, it
has been moved here.) When a function is not explicitly given, but rather a relationship
62
7 Logarithm, Exponential, & Implicit Differentiation
between two variables is provided, the value of one variable may determine the value
of the other (either uniquely or with multiple possibilities). In some cases, these two
variables can locally form a functional relationship, allowing differentiation. The process
of computing derivatives in such cases is called implicit differentiation.1
Suppose the two variables x and y satisfy the equation:
x2 + y2 − 25 = 0.
Then, we can consider one of the variables as independent and express the other as its
function. For instance, if we treat x as the independent variable, we obtain:
p
p
y2 = 25 − x2 ⇒ y = 25 − x2 or y = − 25 − x2 .
Thus, the derivative is given by:
dy
x
x
=√
or − √
dx
25 − x2
25 − x2
for
− 5 < x < 5.
p
Similarly, if we express x as a function of y, we get x = ± 25 − y2 .
However, explicitly solving for y in terms of x before differentiating is cumbersome.
Moreover, in many cases, expressing y as a function of x is extremely difficult or even
impossible. Implicit differentiation allows us to compute derivatives without solving
explicitly for y.
For example, consider the equation:
x3 + 2y3 − 9xy = 1.
(7.2)
In this case, solving for y explicitly as a function of x is not straightforward. Instead, we
treat y as an implicit function of x, imagining:
x3 + 2(y(x))3 − 9xy(x) = 1.
The second term, 2(y(x))3 , is a composition of functions. Differentiating both sides with
respect to x, we obtain:
d 3
dy
dy
(x + 2y3 − 9xy) = 3x2 + 6y2 − 9y − 9x = 0.
dx
dx
dx
dy
Here, we used the chain rule and the product rule. Solving for dx
, we get:
dy 9y − 3x2
=
.
dx 6y2 − 9x
(7.3)
dy
Problem 7.7. Given that x and y satisfy equation (7.2), compute dx
at x = 1.
1 The term ”implicit function” does not refer to a specific type of function but rather indicates that the
function is not explicitly given and is instead implicitly defined by a relationship. A more accurate translation of ”implicit differentiation” might be ”암묵적 미분” (indirect differentiation).
7.3 Implicit Differentiation
63
Solution 7.7 We substitute x = 1 into equation (7.3), but we first need to determine the
corresponding y-values. Substituting x = 1 into (7.2), we obtain the equation 2y3 − 9y = 0,
which factors into y(2y2 − 9) = 0. Solving, we find y = 0 and y = ± 92 . Substituting these
values into (7.3) gives three possible derivative values. ⊔
⊓
Let’s think about this for a moment. A function must assign a unique output for each input. However, here we find three possible y-values for x = 1, indicating that y is not truly a
function of x. It is locally a function but not globally. Nevertheless, implicit differentiation
successfully finds all possible derivatives.
Problem 7.8. Given that x and y satisfy x2 + y2 = 25, use implicit differentiation to find
dx
dy . What is its value at the point (5, 0)?
Solution 7.8 This time, we treat y as the independent variable and differentiate both sides:
d 2
dx
(x + y2 ) = 2x + 2y = 0.
dy
dy
Thus,
dx
y
y
=− = p
.
dy
x ± 25 − y2
Substituting (5, 0), we find:
dx
= 0.
dy x=5,y=0
The given equation is symmetric in x and y (interchanging them leaves the equation unchanged), so the derivative follows the same functional form. ⊔
⊓
Problem 7.9. A circle has its center at the origin and passes through the point (3, −4).
Find the slope of the tangent line to the circle at this point.
p
Solution 7.9 The radius of the circle is 32 + (−4)2 = 5, so its equation is x2 + y2 = 25.
Differentiating implicitly with respect to x, we obtain:
2x + 2y
dy
= 0.
dx
Thus, the slope of the tangent line at (3, −4) is:
dy
x
3
3
=− =−
= .
dx
y
−4 4
⊔
⊓
Problem 7.10. A circle is centered at the origin and passes through the point (5, 0). Find
the slope of the tangent line to the circle at this point.
Solution 7.10 The radius of the circle is 5, so it satisfies the equation x2 + y2 = 25.
Differentiating implicitly with respect to x, we obtain:
2x + 2y
dy
= 0.
dx
64
7 Logarithm, Exponential, & Implicit Differentiation
Thus, the slope of the tangent line at (5, 0) is:
dy
x 5
=− = .
dx
y 0
However, something is wrong here—the result is undefined. Ah, in this case, near the point
dy
. Geometrically,
(5, 0), y cannot be considered a function of x, so we cannot compute dx
the tangent line at this point is a vertical line. While we might informally say the slope is
”infinity,” it is more precise to state that y is not differentiable at this point. ⊔
⊓
Question 7.2. So far, we have considered cases where a single equation describes the
relationship between two variables. Now, consider the following cases: (1) What happens
if we have a single equation involving three variables? (2) What happens if we have two
equations involving three variables?
These cases are studied in multivariable calculus. However, consider case (1): Can we
select one variable and express the other two as functions of this variable? The answer
is no—why? On the other hand, in case (2), it is generally possible to express two of
the variables as functions of the remaining variable. Why is that? Note that ”generally
possible” does not mean ”always possible.” As seen in Problem 7.10, there are cases
where differentiation is not possible, or the variables cannot be expressed as functions.
Exercises
1. Compute the derivatives of the following functions.
2
2
(1) y = ln(x3 + x)
(2) y =
(3) y = ex +3
rln(ln x )
1 1+t
1+t
(5) y =
(6) y = ln
(4) y =
1−t
1−t
t(1 + t)
2. Compute the following integrals.
Z −1
Z π
Z 2
1
cos x
(1)
dx
(2)
dx
(3)
ex dx
−2 x
0 10 + sin x
0
Z 2
Z
Z
dx
7t
(4)
(5) 10 tan 2x dx
(6)
dt
2 − 10
x
ln
x
t
1
3. The inverse trigonometric functions are defined as follows:
f (x) = sin x
g(x) = cos x
h(x) = tan x
if x ∈ [−π, π],
if x ∈ [0, π],
π π
if x ∈ [− , ],
2 2
arcsin(x) = f −1 (x)
if x ∈ [−1, 1],
−1
arccos(x) = g (x) if x ∈ [−1, 1],
arctan(x) = h−1 (x)
if x ∈ [−∞, ∞].
Use the properties of inverse functions to compute the derivatives of the following
functions.
(a) arcsin(x)
(b) arccos(x)
(c) arctan(x)
2
2
(d) arccos(ln x)
(e) arctan(ln(x + 1))
(f) arcsin(e−x )
7.3 Implicit Differentiation
65
4. Differentiate
the following using logarithmic properties.
s
√
2+t
(x + 1)(x − 1)2
(2)
(3)
tan
θ
(1)
2θ − 1
(x2 + 1)x2
t sint
dy
5. Compute dx
for the following implicit functions.
x
(1) y = y
(2) ex = x2y+2x
(3) ln y = x3 y
x
2x
(4) y = x
(5) y = (tan x)
(6) ysin x = xcos y
6. Compute the following integrals.
Z
Z
dx
2 ln(10x)
dx
(2)
(1)
x
−
1
x
ln x
Z
Z
√
3x 2 dx
(4)
Z ln x
(3)
√
(5)
Z 0
x 2−1 dx
(6)
1
1
dt
t
5−x dx
−2
7. Compute f ′ ( π4 ) for the function f (x) = ln(tan(x)).
2
8. Find the antiderivative of f (x) = xex and compute the integral 02 f (x)dx.
R
9. Compute f ′ (π/4) for the function f (x) = arcsin(ln(sin x)).
10. Find the area enclosed by the function y = ln(x2 ), the x-axis, and the line y = 2 ln 3.
11. Using the product rule ( f g)′ = f ′ g + g′ f , prove the following.
Z
(a)
Z
(b)
ln x dx = x ln x − x +C.
d
ln(x2 + 1) dx = x ln(x2 + 1) + 2 arctan(x) − 2x +C. (Hint: dx
(arctan(x)) = x21+1 )
dy
12. Use implicit differentiation to find dx
.
2
2
2
(1) x + y = 4
(2) x + y = 1
(3) sin x + y2 = 1
2
2
(4) xy + x = 3
(5) xy = sin(xy)
(6) (2xy + y2 )2 = x2 − y2
dy
13. Use implicit differentiation to find dx
and dx
dy at the given points.
2
2
(1) x + y = 4 at x = 1
(2) y − x2 + x = 4 at x = 1
(3) xy + x2 + y2 = 1 at
2
x=1
(4) x + y − 2y = 4 at x = 1
dy
14. Use the results above to verify whether the product of dx
and dx
dy at the same point
equals 1.
Index
antiderivative, 37
area function, 40
bijection, 47
Cauchy’s Fundamental Theorem of Calculus, 43
Cauchy’s Mean Value Theorem, 25
chain rule, 29
closed interval, 4
co-domain, 47
continuous, 4
converge, 4
inverse function, 47
left limit, 4
Leibniz notation, 33
limit, 4
mean growth rate, 20
Mean Growth Rate Theorem, 26
Mean Value Theorem, 26
natural logarithm, 57
derivative, 20
domain, 47
one-to-one function, 47
onto function, 47
open interval, 4
exponential function, 60
partition, 41
function, 47
fundamental theorem of calculus, 43
range, 47
Riemann sum, 41
right limit, 4
Rolle’s Theorem, 25
gauge, 41
Implicit differentiation, 62
implicit differentiation, 62
injection, 47
instantaneous growth rate, 21
integrability, 42
integral, 42
Intermediate Value Theorem, 25
surjection, 47
variable-centered notation, 33
순간 증가율, 21
평균 증가율, 20
67