UNIVERSITY OF MINDANAO
College of Engineering Education
Computer Engineering Program
Physically Distanced but Academically Engaged
Self-Instructional Manual (SIM) for Self-Directed Learning (SDL)
3rd Draft
WEEK 1-7
CEE 101- Differential Calculus for Engineering
ENGR. JOHN A. BACUS
THIS SIM/SDL MANUAL IS A DRAFT VERSION ONLY; NOT FOR
REPRODUCTION AND DISTRIBUTION OUTSIDE OF ITS INTENDED
USE. THIS IS INTENDED ONLY FOR THE USE OF THE STUDENTS
WHO ARE OFFICIALLY ENROLLED IN THE COURSE/SUBJECT.
EXPECT REVISIONS OF THE MANUAL.
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
TABLE OF CONTENTS
PAGE
Cover Page ……………………………………………………………………………………………….
1
Table of Contents……………………………………………………………………………………….
2
Course Outline…………………………………………………………………………………………...
6
Course Outline Policy…………………………………………………………………………………
6
Course Information……………………………………………………………………………………
9
Topic/ Activity
Unit Learning Outcomes- Unit 1………………………………………………………………….
10
Big Picture in Focus: ULO-1a…………………………………………………………………..…..
10
Metalanguage…………………………………………………………………………………...
10
Essential Knowledge…………………………………………………………………………
11
Self-Help…………………………………………………………………………………………..
16
In a Nutshell……………………………………………………………………………………..
19
Q & A List ……………………………………………………………………………………...…
19
Big Picture in Focus: ULO-1b…………………………………………………………………..…..
20
Metalanguage…………………………………………………………………………………..
20
Essential Knowledge…………………………………………………………………………
20
Self-Help…………………………………………………………………………………………..
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In a Nutshell……………………………………………………………………………………..
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Q & A List ………………………………………………………………………………………...
25
Big Picture in Focus: ULO-1c…………………………………………………………………..…..
25
Metalanguage…………………………………………………………………………………...
25
Essential Knowledge…………………………………………………………………………
26
Self-Help…………………………………………………………………………………………..
31
In a Nutshell……………………………………………………………………………………..
31
Q & A List ……………………………………………………………………………………...…
32
2
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
PAGE
Big Picture in Focus: ULO-1d…………………………………………………………………..…..
32
Metalanguage…………………………………………………………………………………...
32
Essential Knowledge…………………………………………………………………………
32
Self-Help…………………………………………………………………………………………..
35
In a Nutshell……………………………………………………………………………………..
37
Q & A List ………………………………………………………………………………………...
37
Big Picture in Focus: ULO-1e…………………………………………………………………..…..
38
Metalanguage…………………………………………………………………………………...
38
Essential Knowledge…………………………………………………………………………
38
Self-Help…………………………………………………………………………………………..
42
In a Nutshell……………………………………………………………………………………..
44
Q & A List ………………………………………………………………………………………...
44
Unit Learning Outcomes- Unit 2………………………………………………………………….
45
Big Picture in Focus: ULO-2a…………………………………………………………………..…..
45
Metalanguage…………………………………………………………………………………...
45
Essential Knowledge…………………………………………………………………………
45
Self-Help…………………………………………………………………………………………..
52
In a Nutshell……………………………………………………………………………………..
54
Q & A List ……………………………………………………………………………………...…
54
Big Picture in Focus: ULO-2b…………………………………………………………………..…..
55
Metalanguage…………………………………………………………………………………..
55
Essential Knowledge…………………………………………………………………………
55
Self-Help…………………………………………………………………………………………..
58
In a Nutshell……………………………………………………………………………………..
59
Q & A List ………………………………………………………………………………………...
59
Big Picture in Focus: ULO-2c…………………………………………………………………..…..
60
Metalanguage…………………………………………………………………………………...
60
3
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Essential Knowledge…………………………………………………………………………
61
Self-Help…………………………………………………………………………………………..
63
In a Nutshell……………………………………………………………………………………..
65
Q & A List ……………………………………………………………………………………...…
65
Big Picture in Focus: ULO-2d…………………………………………………………………..…..
66
Metalanguage…………………………………………………………………………………...
66
Essential Knowledge…………………………………………………………………………
67
Self-Help…………………………………………………………………………………………..
71
In a Nutshell……………………………………………………………………………………..
72
Q & A List ………………………………………………………………………………………...
73
Unit Learning Outcomes- Unit 3………………………………………………………………….
74
Big Picture in Focus: ULO-3a…………………………………………………………………..…..
74
Metalanguage…………………………………………………………………………………...
74
Essential Knowledge…………………………………………………………………………
74
Self-Help…………………………………………………………………………………………..
81
In a Nutshell……………………………………………………………………………………..
83
Q & A List ……………………………………………………………………………………...…
83
Big Picture in Focus: ULO-3b…………………………………………………………………..…..
84
Metalanguage…………………………………………………………………………………..
84
Essential Knowledge…………………………………………………………………………
84
Self-Help…………………………………………………………………………………………..
88
In a Nutshell……………………………………………………………………………………..
89
Q & A List ………………………………………………………………………………………...
89
Big Picture in Focus: ULO-3c…………………………………………………………………..…..
90
Metalanguage…………………………………………………………………………………...
90
Essential Knowledge…………………………………………………………………………
90
Self-Help…………………………………………………………………………………………..
97
In a Nutshell……………………………………………………………………………………..
99
Q & A List ……………………………………………………………………………………...…
99
4
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Big Picture in Focus: ULO-3d…………………………………………………………………..…..
100
Metalanguage…………………………………………………………………………………...
100
Essential Knowledge…………………………………………………………………………
100
Self-Help…………………………………………………………………………………………..
103
In a Nutshell……………………………………………………………………………………..
104
Q & A List ………………………………………………………………………………………...
104
Big Picture in Focus: ULO-4a…………………………………………………………………..…..
105
Metalanguage…………………………………………………………………………………...
105
Essential Knowledge…………………………………………………………………………
105
Self-Help…………………………………………………………………………………………..
110
In a Nutshell……………………………………………………………………………………..
111
Q & A List ……………………………………………………………………………………...…
111
Big Picture in Focus: ULO-4b…………………………………………………………………..…..
112
Metalanguage…………………………………………………………………………………..
112
Essential Knowledge…………………………………………………………………………
112
Self-Help…………………………………………………………………………………………..
118
In a Nutshell……………………………………………………………………………………..
119
Q & A List ………………………………………………………………………………………...
119
Big Picture in Focus: ULO-4c…………………………………………………………………..…..
120
Metalanguage…………………………………………………………………………………...
120
Essential Knowledge…………………………………………………………………………
120
Self-Help…………………………………………………………………………………………..
124
In a Nutshell……………………………………………………………………………………..
124
Q & A List ……………………………………………………………………………………...…
125
Big Picture in Focus: ULO-4d…………………………………………………………………..…..
127
Metalanguage…………………………………………………………………………………...
128
Essential Knowledge…………………………………………………………………………
127
Self-Help…………………………………………………………………………………………..
131
In a Nutshell……………………………………………………………………………………..
132
Q & A List ………………………………………………………………………………………...
132
5
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Course Outline: CEE 101 – Differential Calculus for Engineering
Course Coordinator:
Email:
Student Consultation:
Mobile:
Phone:
Effectivity Date:
Mode of Delivery:
Time Frame:
Student Workload:
Pre-requisite:
Credit:
Attendance Requirements:
John A. Bacus
johnbacus001@gmail.com
Online (LMS) or e-mail
09776420836
None
May 2020
Online Blended Delivery
108 hours
Expected Self-Directed Learning
CEE 100
5.0 units
For online (virtual/face-to-face) sessions: a minimum
of 95% attendance; for 1-day on-campus/onsite
review: 100% attendance; for 1-day oncampus/onsite final exam: 100% attendance
Course Outline Policy
Areas of Concern
Details
Contact and Non-Contact Hours
This 5-unit course self-instructional manual is intended for
blended learning mode of instructional delivery with
scheduled face-to-face or virtual sessions. The expected
number of hours will be 108, including the face-to-face or
virtual sessions which will be arranged by the course
coordinator. The face-to-face sessions shall include the
summative assessment tasks or examinations since this
course is vital in the licensure examination for engineers.
Assessment Task Submission
The first assessment task (examination) shall be given on
the 3rd week subsequent to the first day of class. The
remaining assessment tasks shall be handed every after
two (2) weeks of each examination schedule.
The
assessment paper shall be attached with a cover page
indicating the name of the course coordinator, date of
submission, and name of the student. The document should
be submitted on the same day through e-mail or Blackboard
LMS. It is also expected that you have already paid your
tuition and other fees before the submission of the
assessment task.
Since this course is included in the licensure examination
for engineers, you will be required to take the MultipleChoice Question exam inside the University as your final
exam. This should be scheduled ahead of time by your
course coordinator. This is non-negotiable for all licensurebased programs.
6
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Penalties for Late
Assignments/Assessments
The score for an assessment item submitted after the
designated time on the due date, without an approved
extension of time, will be reduced by 5% of the possible
maximum score for that assessment item for each day or
part day that the assessment item is late.
Return of Assignments/
Assessments
Assessment tasks will be returned to you two (2) weeks
after the submission with a mark and feedback. This will be
returned by email or via Blackboard portal.
For group assessment tasks, the course coordinator will
require some or few of the students for online or virtual
sessions to ask clarificatory questions to validate the
originality of the assessment task submitted and to ensure
that all the group members are involved.
Assignment Resubmission
You should request in writing addressed to the course
coordinator your intention to resubmit an assessment task.
The resubmission is premised on the student’s failure to
comply reasonable circumstances e.g. illness, accidents
financial constraints.
Re-marking of Assessment Papers
and Appeal
You should request in writing addressed to the program
coordinator your intention to appeal or contest the score
given to an assessment task. The letter should explicitly
explain the reasons/points to contest the grade. The
program coordinator shall communicate with the students
on the approval and disapproval of the request.
If disapproved by the course coordinator, you can elevate
your case to the program head or the dean with the original
letter of request. The final decision will come from the dean
of the college.
Grading System
All culled from BlackBoard sessions and traditional
contact:
Course discussions/exercise- 40%
1st formative assessment – 10%
2nd formative assessment – 10%
3rd formative assessment – 10%
All culled from on-campus/onsite sessions (TBA):
Final exam – 30%
Submission of the final grades shall follow the usual
University system and procedures.
Student Communication
You are required to have your own umindanao email
account which is a requirement to access the
BlackBoard portal. Then, the course coordinator shall
enroll the students to have access to the materials and
resources of the course. All communication formats: chat,
submission of assessment tasks, requests etc. shall be
through the portal and other university recognized
platforms.
7
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
You can also meet the course coordinator virtually
through the scheduled face-to-face sessions to raise
your issues and concerns.
For students who have not created their student email,
please contact the course coordinator or program head.
Contact Details of the Dean
Dr. Charlito L. Cañesares, PME
Email: clcanesares@umindanao.edu.ph
Phone: (082) 296-1084 local 133
Contact Details of the Program Head
Engr. Randy E. Angelia, MEP-ECE, MSCpE
Email: randy_angelia@umindanao.edu.ph
Phone: (082) 296-1084 local 133
Students with Special Needs
Students with special needs shall communicate with the
course coordinator about the nature of his/her special
needs. Depending on the nature of the need, the course
coordinator with the approval of the program head may
provide alternative assessment tasks or extension of the
deadline of submission of assessment tasks. However, the
alternative assessment tasks should still be in the service
of achieving the desired course learning outcomes.
Help Desk Contact
CEE Blackboard Administrator:
Engr. Jetron A. Adtoon, MSCpE
Email: jadtoon@umindanao.edu.ph
Phone: +63 9055 267834
CEE:
Frida Santa O. Dagatan
Email: cee@umindanao.edu.ph
Phone: +63 9055 267834
GSTC:
Ronadora E. Deala, RPsy, RPm, RGC, LPT
Email: ronadora_deala@umindanao.edu.ph
Phone: +63 921 2122846
Silvino P. Josol
Email: gstcmain@umindanao.edu.ph
Phone: +63 906 0757721
Library Contact
Brigida E. Bacani
Email: library@umindanao.edu.ph
Phone: +63 951 3766681
8
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Course Information – see/download course syllabus in the Black Board LMS
CC’s Voice: Hello forthcoming engineers! Welcome to the course CEE 101: Differential
Calculus for Engineering. You are now entering the world of Engineering
Calculus where you need to explore and understand a certain thing by looking at
its small pieces. By this time, I am positive that you really wanted to pursue your
career in the field of engineering and that you have envisioned yourself inventing,
designing, analyzing, building, and testing machines, complex systems,
structures, gadgets and materials for the benefit of humankind.
CO
Before going to the next stage of Engineering Calculus, which is the Integral
Calculus or Anti-Differentiation, you have to engage yourself first with
differentiating algebraic and transcendental functions, analyzing and
tracing transcendental curves, and applying the concepts of differentiation
in solving word problems. These are the significant course outcomes (CO) that
you need to demonstrate at the end of this course. Thus, you are now anticipated
to justify the knowledge you have acquired from your basic mathematics courses
in dealing with complex word problems and to read in advance the concepts of
differentiation.
Let us begin!
9
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Big Picture
Week 1-3: Unit Learning Outcomes 1 (ULO1): At the end of the unit, you are expected
to:
a. Evaluate the limits of a function using different techniques;
b. Determine the value/s of domain where the function discontinues and sketch
its graph;
c. Perform the long method derivation of a function;
d. Understand and apply the Differentiation Rules of Algebraic Function; and
e. Solve for the nth order derivative of a function—either implicit or explicit.
Big Picture in Focus:
ULO-a. Evaluate the limits of a function using different techniques
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to the introduction of calculus and ULO-a will be operationally defined to establish a
common frame of reference. You will encounter these terms as we delve deeper to the
study of Differential Calculus. Please refer to these definitions in case you will find it
difficult to understand mathematical concepts in relation with calculus.
1. Domain. The domain of a function is the set of all values that can be plugged
into a function and have the function exist and have a real number for a value.
1.1 It is the set of all possible x-values which will make the function “work”.
1.2 It can be determined by looking for the values of the independent variable
(usually x-values) which can be used in a function.
1.3 When finding for the value/s of domain, it is important to remember that the
denominator of the function cannot be zero and avoid negative values under
the square root sign.
2. Range. The range of a function is simply the set of all possible values that a
function can take.
2.1 It is the resulting y-values we get from substituting all the possible x-values.
2.2 It is the complete set of all possible resulting values of the dependent variable
(usually y-values).
2.3 It is the spread of all possible y-values (minimum y-value to maximum y-value).
3. Variable. It is a quantity which, during any set of mathematical operations, does
not retain the same value but is capable of assuming different values.
3.1 A variable that represents the “input numbers” for a function is called
independent variables. A variable that represents the “output numbers” is
called dependent variable because its value depends on the value of
independent variable.
4. Constant. It is a quantity which, during any set of mathematical operations, retains
the same value.
10
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
5. Polynomial. It can have constants, variables, and exponents, but never division
by a variable (e.g. 5𝑥𝑦 2 − 2𝑥 + 4𝑦 = 0, not polynomials: 5𝑥𝑦 −2 and 3𝑥⁄𝑥 + 2)
6. Conjugate. It is formed by changing the sign between two terms.
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes) for the first three (3)
weeks of this course, you need to fully understand the following essential knowledge that
will be laid down in the succeeding pages. Please note that you are not limited to
exclusively refer to these resources. Thus, you are expected to utilize other books,
research articles and other resources that are available in the university’s library (refer to
the Library Contact on page 3).
1. Functions. A function 𝑓 is a rule that assigns to every number 𝑥 in a collection D, a
number 𝑓(𝑥). The set D is called the domain of the function. And 𝑓(𝑥) is called the
value of a function at 𝑥, or commonly known as range. The set of ordered pair
(𝑥, 𝑓(𝑥)) is called the graph of 𝑓.
Example 1.1: Determine if each of the following are functions.
(a) 𝑓(𝑥) = 𝑥 2 + 2
This equation is a function. Since no matter what value of
𝑥 you put into the equation; it yields exactly one possible
value of 𝑓(𝑥).
Let x= 1;
Let x=2;
2
𝑓(𝑥) = 𝑥 + 2
𝑓(1) = 12 + 2
𝒇(𝟏) = 𝟑
(b) [𝑓(𝑥)]2 = 𝑥 + 2
Let x= 1;
𝑓(𝑥) = 𝑥 2 + 2
𝑓(2) = 22 + 2
𝒇(𝟐) = 𝟔
This equation is not a function. At a specific value
of 𝑥 , there are two (2) possible values of 𝑓(𝑥).
Let x=2;
[𝑓(𝑥)]2 = 𝑥 + 2
[𝑓(2)]2 = 2 + 2
[𝑓(2)]2 = 4
𝑓(𝑥) = √4
𝒇(𝒙) = ±𝟐
[𝑓(𝑥)]2 = 𝑥 + 2
[𝑓(1)]2 = 1 + 2
[𝑓(1)]2 = 3
𝑓(𝑥) = √3
𝒇(𝒙) = ±𝟏. 𝟕𝟑𝟐
Note:
•
“𝑓(𝑥)” can simply be written as “𝑦” or any other function notation like “𝑔(𝑥)”,
“ℎ(𝑥)”, “𝑅(𝑥)”, etc.
11
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
2. Limits. The limit of a function 𝑓(𝑥) as 𝑥 approaches 𝑎
is 𝐿, can be written as
lim 𝑓 (𝑥 ) = 𝐿
𝑥→𝑎
In other words, the value of the function 𝑓(𝑥) gets closer and closer to 𝐿 as 𝑥 gets
closer and closer to 𝑎 , without being exactly equal to 𝑎 . To understand this
concept, let’s have a simple example first.
Example 2.1: Given the function 𝑓(𝑥) = 2𝑥 + 5 as 𝑥 approaches 1, find its limit.
To denote this problem, we will write this as:
lim 2𝑥 + 5 = ?
𝑥→1
Let us now look at the function at a point 𝑥 that gets closer and closer to 𝑥 = 1.
𝑥
𝑓(𝑥)
0.89
6.78
₀
₀
0.99
6.98
1.1
7.2
1.2
7.4
The table above shows the values of 𝑥 which are very close to 1 but never equal
to 1. And if we plug in these values to the given function,
𝑓(𝑥) moves closer and closer to 7. Thus, when we evaluate the limit of the given
function above, we get 7. It can now be written as:
lim 2𝑥 + 5 = 7
𝑥→1
Please note that the above definition of “Limits” is in its informal form. Let’s stick
with this definition for the meantime since it is easier to understand, and it helps us
to have an idea of just what limits are and its relationship with functions. As we go
further in this course, you will encounter its formal and precise definition.
Why do we have to study limits?
- Because, essentially, that’s what calculus is: the study of limits. Limits
are used to examine function behavior around a specific point, and
without it, it would be very difficult to talk about rate of change. Therefore,
calculus is impossible to do without knowing what to do with limits.
12
College of Engineering Education
2nd Floor, BE Building
Matina Campus, Davao City
Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Before going to evaluate the limits of functions, it would be better if you know beforehand
the “Properties of Limits”. The proof of some these properties can be found in a separate
material (refer to Self-Help).
PROPERTIES OF LIMITS
lim 𝑐 = 𝑐; where a and c are real numbers
When taking the limit of a constant, the limit is just
that same constant.
lim 𝑥 = 𝑎; where a and c are real numbers
The limit of the function 𝑓(𝑥) = 𝑥 as 𝑥 approaches
a given point 𝑎 is 𝑎.
𝑥→𝑎
𝑥→𝑎
Let a and c be real numbers, let 𝑓(𝑥) and 𝑔(𝑥) be defined for all 𝑥 ’s that lie in some
interval around a.
lim 𝑓(𝑥) = 𝐹
lim 𝑔(𝑥) = 𝐺
𝑥→𝑎
𝑥→𝑎
exist with F and G which are real numbers. Thus, holding the following limits: `
lim (𝑓(𝑥) + 𝑔(𝑥)) = 𝐹 + 𝐺
The limit of the sum of the functions is the sum of
the limits.
lim (𝑓(𝑥) − 𝑔(𝑥)) = 𝐹 − 𝐺
The limit of the difference of the functions is the
difference of the limits.
𝑥→𝑎
𝑥→𝑎
The limit of the product of a constant and a function
is the product of their limits.
lim 𝑐𝑓(𝑥) = 𝑐𝐹
𝑥→𝑎
The limit of the product of the functions is the
product of the limits.
lim (𝑓(𝑥) ∙ 𝑔(𝑥)) = 𝐹 ∙ 𝐺
𝑥→𝑎
lim (
𝑓(𝑥)
𝐹
The limit of the quotient of the functions is the
quotient of the limits.
) = 𝐺; where 𝐺 ≠ 0
𝑥→𝑎 𝑔(𝑥)
𝑛
lim (𝑓(𝑥)) = (lim 𝑓(𝑥))𝑛 = 𝐹 𝑛 ;
𝑥→𝑎
𝑥→𝑎
The limit of a power is the power of the limit.
where 𝑛 is a positive integer
lim (𝑓(𝑥))
𝑥→𝑎
1⁄
𝑛
1⁄
𝑛
= (lim 𝑓(𝑥))
=𝐹
1⁄
𝑛;
𝑥→𝑎
where 𝑛 is an even number and 𝐹 > 0, or
an odd number and 𝐹 is any real number
𝑛 is
The limit of a power is the power of the limit.
Familiarization of the following properties will be of great help in dealing with limits in the
succeeding examples and activities.
At this point, we are now equipped with enough knowledge to start evaluating limits. Limits
of functions are computed using various techniques such as simple arithmetic,
substitution, or using algebraic simplification. These techniques are illustrated in the
following examples:
13
College of Engineering Education
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Telefax: (082)296-1084
Phone No.: (082)300-5456/300-0647 Local 131
Example 2.2: Evaluate lim 4𝑥 2 − 1
𝑥→2
lim 4𝑥 2 − 1 = (lim 4𝑥 2 ) − lim 1
𝑥→2
difference of limits
𝑥→2
2
𝑥→2
= (lim 4 ∙ lim 𝑥 ) − lim 1
𝑥→2
𝑥→2
𝑥→2
product of limits
= 4∙ (lim 𝑥 2 ) ∙ 1
limit of constant
= 4 ∙ (lim 𝑥) ∙ (lim 𝑥) − 1
product of limits
= 4 ∙ 2 ∙ 2 − 1 = 15
limit of x
𝑥→2
𝑥→2
Example 2.3: Evaluate lim
𝑥→2
𝑥 2 −9
𝑥→−3 𝑥+3
𝑥2 − 9
(−3)2 − 9 0
=
=
𝑥→−3 𝑥 + 3
(−3) + 3
0
lim
0
, which is undefined. Let’s find another way to solve this
0
one. Try algebraic factorization and simplification.
Substituting −3 for 𝑥 yields
lim
𝑥 2 −9
= lim
𝑥→−3 𝑥+3
𝑥→−3
(𝑥+3)(𝑥−3)
𝑥+3
factorization
= lim (𝑥 − 3)
simplification
= (−3 − 3)
limit of x
𝑥→−3
= −6
Example 2.4: Evaluate lim
𝑥
1
−
𝑥+3 4
𝑥→1 𝑥−1
Substituting 1 for 𝑥 yields
fraction.
𝑥
lim 𝑥+3
−
1
4
𝑥→1 𝑥−1
𝑥
−
0
,
0
1
4
= lim 𝑥+3
∙
𝑥−1
𝑥→1
which is undefined. Let us try simplifying the compound
4(𝑥+3)
find the LCD of the fraction on the numerator
4(𝑥+3)
4𝑥−(𝑥+3)
= lim 4(𝑥−1)(𝑥+3)
multiply LCD to numerator and denominator
𝑥→1
3𝑥−3
= lim 4(𝑥−1)(𝑥+3)
combine similar terms
𝑥→1
= lim
3(𝑥−1)
factor out the GCF of the numerator (3)
𝑥→1 4(𝑥−1)(𝑥+3)
= lim
3
cancel out common factor (x-1)
𝑥→1 4(𝑥+3)
3
= 16
limit of x
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𝑥
Example 2.5: Evaluate lim
𝑥→0 √1+𝑥−1
0
Plugging 0 in the numerator and denominator, it yields a value of , which is indeterminate. Since the given
0
function is not a polynomial, we must try another way. There is a lot of ways to simplify the denominator but
there’s a particular method that can eliminate its square root, and it is to multiply it by its conjugate.
𝑥
√1+𝑥−1
=
𝑥
√1+𝑥−1
=(
∙
√1+𝑥+1
√1+𝑥+1
multiply by
conjugate
conjugate
=1
𝑥 (√1+𝑥+1)
√1+𝑥−1)(√1+𝑥+1)
𝑥 (√1+𝑥+1)
√1+𝑥)2 −1∙1
=(
=
𝑥 (√1+𝑥+1)
1+𝑥−1
=
𝑥 (√1+𝑥+1)
𝑥
note:
(𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2
cancel the
𝑥
= √1 + 𝑥 + 1
lim
Now, we have
𝑥
𝑥→0 √1+𝑥−1
= lim √1 + 𝑥 + 1
𝑥→0
= √1 + 0 + 1 = 2
There are times that we need to deal with limits at infinity. Examples are illustrated:
𝒂
=𝟎
𝒙→−∞ 𝒙𝒓
𝒂
𝐥𝐢𝐦 𝒓 = 𝟎
𝒙→∞ 𝒙
𝐥𝐢𝐦
where r is a positive rational number and a is any real
where r is a positive rational number and a is any real
number
Example 2.6: Evaluate lim
number and 𝒙𝒓 is defined for 𝒙
<𝟎
2𝑥 2 +3
𝑥→+∞ 𝑥 2 −5𝑥−1
If we plug in +∞ for the value of 𝑥, we get
∞
∞
which is indeterminate. To solve this,
we need to factor the largest power of 𝑥 in the numerator from each term, same goes with
the denominator. This function will become:
lim
2𝑥 2 +3
𝑥→+∞ 𝑥 2 −5𝑥−1
3
= lim
𝑥 2 (2+ 2 )
𝑥
5𝑥
1
𝑥→+∞ 𝑥 2 (1− 2 − 2 )
𝑥
𝑥
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3
𝑥 2 (2+ 2 )
𝑥
= lim
5
1
𝑥→+∞ 𝑥 2 (1−𝑋− 2 )
𝑥
3
2+ 2
𝑥
= lim
5
1
𝑥→+∞ 1− 𝑋− 2
𝑥
3
∞2
5
1
1− − 2
∞ ∞
2+
=
=
Note:
𝑎
±∞
= 0; where a is any real number
2+0
1−0−0
=2
Example 2.7: Evaluate lim
8𝑥 2
𝑥→+∞ 𝑥+5
This problem has the same concept as the example above. Extract first the largest
power of 𝑥 in the numerator as well as in the denominator.
lim
8𝑥 2
𝑥 2 (8)
= lim
5
𝑥→−∞ 𝑥+5 𝑥→−∞ 𝑥(1+𝑥)
8
= lim (𝑥)(
= (−∞)(
= −∞ (
5
1+
𝑥
8
𝑥→−∞
)
5
−∞
1+
8
)
) = −∞
1−0
Self-Help: You can also refer to the sources below to help you further
understand the lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
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Let’s Check
Activity 1. Since you are now armed with basic knowledge of evaluating limits, it is now your turn
to prove what you have learned. Evaluate the limits of the given functions and show your solution:
1. 𝐥𝐢𝐦(𝟖 − 𝟑𝒙 + 𝟏𝟐𝒙𝟐 )
𝒙→𝟐
𝟔+ 𝟒𝒕
2. 𝐥𝐢𝐦
𝟐
𝒕→−𝟑 𝒕 +𝟏
3. 𝐥𝐢𝐦
𝟐𝒙𝟐 −𝟏𝟕𝒙+𝟖
𝟖−𝒙
𝒙→𝟖
𝒉𝟐 −𝟒𝒉−𝟐𝟏
4. 𝐥𝐢𝐦
𝟐
𝒉→𝟕 𝟑𝒉 −𝟏𝟕𝒉−𝟐𝟖
5.
𝐥𝐢𝐦
𝒚→−𝟑
6. 𝐥𝐢𝐦
√𝟐𝒚+𝟐𝟐−𝟒
𝒚+𝟑
𝒕𝟐 +𝟔
𝟐
𝒕→𝟎 𝒕 −𝟑
7. 𝐥𝐢𝐦
𝟓𝒙𝟐 −𝟏𝟔𝒙+ 𝟑
𝒙→𝟑
8.
9.
𝐥𝐢𝐦
𝟗−𝒙𝟐
𝒙𝟑 +𝟖
𝟐
𝒙→−𝟐 𝒙 +𝟖𝒙+𝟏𝟐
𝐥𝐢𝐦
𝒚→−𝟏
10. 𝐥𝐢𝐦
𝟏
𝟏
+
𝟒+𝟑𝒚 𝒚
𝒚+𝟏
𝟑− 𝒕
𝒕→𝟑 √𝒕+𝟏−√𝟓𝒕 −𝟏𝟏
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Let’s Analyze
Activity 1. Getting acquainted with the evaluation of limits is not enough. This time, you are going
to evaluate limits of functions again and explain its step-by-step procedure like the examples from
Example 2.2 to Example 2.5.
1. lim (𝑥 − 4)(𝑥 2 + 8)
𝑥→5
2(𝑡−10)2
2. lim
𝑡
𝑡→10
5+𝑥−4𝑥 2
3. lim
𝑥→∞ √1+𝑥 2 +2𝑥 4
4. lim1
𝑡→
1
1
+
3𝑡3 𝑡2 −1
2𝑡−1
2
5. lim √5𝑥 3 + 4
𝑥→1
𝑡−5
6. lim ( )
𝑡+4
𝑡→−2
7. lim
𝑥→−1
8. lim
√𝑥 2 +8 −3
𝑥+1
𝑥+1
𝑥→∞ √𝑥 2
9. lim
5+𝑥−4𝑥 2
𝑥→∞ √1+𝑥 2 +2𝑥 4
10. lim
𝑥→−∞
3𝑥 7 + 𝑥 5 −15
4𝑥 2 +32𝑥
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In a Nutshell
We are now done with the fundamentals of calculus, which are identifying functions
and computing its limits. Before proceeding to the next unit learning outcomes, be
reminded of some important points when dealing with functions and its limits.
• An equation is said to be a function if it yields only one value of 𝑦 when it is
computed at a specific value of 𝑥.
• The limit is only concerned with what is going on with around a certain point.
• Direct substitution, when evaluating the limits of a function, is a go-to method.
Use other methods or techniques, like factorization and conjugation, only when
this one fails, otherwise you are probably doing more work than you need to be.
• One cannot do Differential Calculus if he/she doesn’t have any prior knowledge
about the evaluation of limits.
Q&A List
If you have any questions regarding functions and limits, kindly write down on the
table provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-b. Determine the value/s of domain where the function discontinues and sketch
its graph
Metalanguage
You can refer to this section, and the previous one, anytime if you run across words
which you are not familiar with. This will be a common reference in terms of the definition
of mathematical concepts that we are going to discuss.
1. Graph. It is a diagram that shows the relation between variable quantities, naturally
two variables, each measured along one of a pair of axes at right angles.
2. Rational Function. It is a function that is a ratio of two polynomials. It is called
“rational” because one polynomial is divided by the other like a “ratio”.
3. Quadratic Equation. It is second-order polynomial equation in a single variable x
with a standard form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎 ≠ 0.
3.1 It always has two solutions based on the fundamental theorem of algebra.
These solutions may be both real or both complex numbers.
4. Quadratic Formula. It is a method to solve for the roots of the quadratic equation.
This method is usually used when the given quadratic equation cannot be factored
out.
4.1 To solve for the roots using this method, it is essential to determine the
values of 𝑎, 𝑏, and 𝑐 from the quadratic equation. The quadratic formula is given
by the form:
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
Essential Knowledge
In ULO-b, you will be introduced to the concept of continuity and how it is related
to limits. At the end of this ULO, you are expected to identify if the given functions are
continuous or not and at what point/s it discontinues. There will be graphs included to
deeply understand the concept of continuity.
1. Continuous Function. A function is said to be continuous at a if:
𝐥𝐢𝐦 𝒇(𝒙) = 𝒇(𝒂)
𝒙→𝒂
With that, we can say that a function, f(x), is continuous at a point a if the limit
exists at a and is equal to f(a). You can also tell that the function is continuous if
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you can draw its graph from start to finish without ever once picking up your
pencil/pen, like the graph presented below. The curve is drawn continuously.
But what we are concerned here is to find the value/s of x where the function
discontinues. A function is said to be discontinuous at a if it does not continue at a. When
graphing discontinuous function, you will have to pick up your pencil/pen at least once to
complete the sketch. To have a better understanding with this concept, here are some
graphs to look at:
There are three (3) graphs above representing different sorts of discontinuity. (1)
The first graph with the function, f(x), has a “jump discontinuity” because the function
jumps from a finite value on the left to another value on the right. (2) The graph of g(x) is
what we call the “infinite discontinuity” since the function’s limit is infinity. Lastly, (3) the
function, h(x), has a “removable discontinuity” because we can remove its discontinuity
if it will be redefined at a certain point to make it continuous at that same point.
In our case, we will mostly encounter the function with removable discontinuity and
redefine it to make it continuous. We have already done redefining functions on the
previous ULO, where we need to evaluate the limits of a function that results to numbers
0
with zero (0) as its denominator or 0. Now, let’s try identifying point/s of discontinuity in a
function.
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Example 1.1: Find the value of the domain where the function, 𝑓(𝑥) =
discontinues.
𝑥 2 −2𝑥
𝑥 2 −4
,
To solve this, let us graph the function first. Graphing this function requires values for the domain,
x, to be plugged in the given function and to compute the values of its range, f(x). Study the table shown
below.
X
-1
0
1
2
3
4
5
f(x)
-1
0
1⁄
3
0⁄
0
3⁄
5
2⁄
3
5⁄
7
Based on the table, here is its corresponding graph for reference. This way, it would be easier to
visualize the function’s behavior.
𝟎
As you can see, there is a hole at 𝒙 = 𝟐. Take note that when 𝒙 = 𝟐, the value of 𝒇(𝒙) = which
𝟎
is indeterminate. This means that at this point the function discontinues. So to answer the problem, the
value of the domain where function discontinues is 2.
Let us have another example where we will focus on the function without sketching
a graph.
4𝑥+10
Example 1.2: Find the value of the domain where the function, 𝑓(𝑥) = 𝑥 2 −2𝑥−15,
discontinues.
To easily determine the value/s of x where the function discontinues, all we need to do is to identify
where the denominator is zero since rational functions are continuous everywhere except when it is divided
by zero. To do this, set the denominator equal to zero as what is shown below:
𝑥 2 − 2𝑥 − 15 = 0
(𝑥 + 3)(𝑥 − 5) = 0
Thus, the function will not be continuous at
𝑥 = −3 and 𝑥 = 5.
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𝑥 2 −9
Example 1.3: Find the value of the domain where the function, 𝑓(𝑥) = 3𝑥 2 +2𝑥−8,
discontinues.
Again, all we need do is to isolate first the denominator, equate it to zero, and factor it out. With
this, we will be able to find the values of x where the function is discontinuous.
3𝑥 2 + 2𝑥 − 8 = 0
(3𝑥 − 4)(𝑥 + 2) = 0
Therefore, the function will be discontinuous at 𝑥 = 3⁄4 and 𝑥 = −2.
8𝑡
Example 1.4: Find the value of the domain where the function, 𝑔(𝑡) = 𝑡 2 −9𝑡−1,
discontinues.
As seen, we have a quadratic equation in the denominator of a rational function. And this quadratic
equation cannot be factored out so to find the value/s of domain where the function discontinues, we need
to use the quadratic formula.
𝑡 2 − 9𝑡 − 1
𝑡=
−(−9)±√(−9)2 −4(1)(−1)
2(1)
Thus, the function will be discontinuous at the points 𝑥
=
=
9±√85
2
= −0.10977, 9.10977
9±√85
.
2
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
Let’s Check
Activity 1. The discussion of the concept of continuity is already done. It is now time for you to
start practicing what you have learned. Determine where the given function is discontinuous.
1. 𝑓(𝑥) =
11−2𝑥
2𝑥2 −13𝑥−7
𝑡2 −1
2. 𝑅(𝑡) = 𝑡3 +6𝑡2 +𝑡
𝑥−1
3. 𝑓(𝑥) = 𝑥2 +7𝑥−4
4. 𝐺(𝑧) =
3
2𝑧 2 +3𝑧−4
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Let’s Analyze
Activity 1. Let’s try another one with a little bit of twist. Determine where the given function is
discontinuous and graph it.
1.
𝑓(𝑥) =
5𝑥+1
𝑥2+4
3𝑥
2. ℎ(𝑥) = 𝑥−5
1
3. 𝑔(𝑥) = 2𝑥 −8
4. 𝑟(𝑡) =
5. 𝑔(𝑙) =
𝑡2 +𝑡−15
𝑡−8
2𝑙2 −7𝑙
𝑙 2 +2𝑙−5
In a Nutshell
We have already dealt with the concept of continuity by sketching its graph and
identifying the value/s of domain that make/s the function discontinues. Here are some
points you should remember:
• A function is said to be continuous at a if 𝐥𝐢𝐦 𝒇(𝒙) = 𝒇(𝒂).
•
•
𝒙→𝒂
Every polynomial is continuous everywhere. In the same way, every rational
function is continuous except when its denominator is zero.
Listed below are functions that are continuous everywhere in their domain:
o Polynomials, Rational Functions
o Roots and Powers
o Trigonometric Functions and their inverses
o Exponential and the Logarithm functions
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Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
Big Picture in Focus:
ULO-c. Perform the long method derivation of a function
Metalanguage
If you run across terms you are unfamiliar with, refer to this section. This will serve
as our common reference in dealing with concepts of differential calculus.
1. Rate of Change. It is the rate at which one quantity changes in relation to another
quantity. One best example for this one is the slope of a line where there is a change
in y over the change in x.
2. Rationalization. It is a process to eliminate all the radicals that are either on the
numerator or the denominator since, sometimes, one encounters radical expressions
which are not perfect roots called irrational numbers or expressions.
2.1 To do this method, you need to multiply both the numerator and the denominator by
a radical that will get rid of the radical in either the numerator or the denominator.
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Essential Knowledge
In ULO-c, we are going to define and discuss the concept of rate of change, in
which mathematically speaking, is the derivative. At the end of this section, you need to
demonstrate how to do derivation using the long method.
1. The Derivative. A function, 𝑓, is called differentiable at 𝑎 if the difference
quotient
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
approaches to a limit as ∆𝑥 approaches 0. This limit is called the derivative of
𝑓 at 𝑥 and is denoted by 𝑓′(𝑥):
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
The concept of derivative can, actually, be demonstrated by the slope. Study
the illustrations given below.
𝑺𝒍𝒐𝒑𝒆 =
Change in Y
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒀
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑿
Change in X
This formula computes the average
slope between the two points, A and B.
A
14
B
14
average slope=
27
27
But how can we calculate for the slope
at a certain point? There’s nothing to
measure.
0
slope= =
0
????
26
That is where derivatives show up. In
derivatives, we use a small difference,
and then have it contract towards zero
to calculate its rate of change.
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We can now conclude that:
𝑺𝒍𝒐𝒑𝒆 =
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒀 𝚫𝒚
=
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑿 𝚫𝒙
Take note that “∆” is a symbol for “delta” which means “change” or a “change in”
and from the given illustration, we can see that:
𝑓(𝑥)
𝑓(𝑥 + ∆𝑥)
𝑥 transforms from 𝑥 to ∆𝑥
𝑦 transforms from 𝑓(𝑥) to 𝑓(𝑥 + ∆𝑥)
That is how the derivative and slope are related. To know the rate of change in a
certain point of a curve or a line, we need do derivation since 𝑓′(𝑥) or 𝑦′ represents the
rate of change of 𝑓(𝑥).
This time let us practice doing the long-method derivation of a function to further
understand this concept. First, study the steps of long-method derivation presented below
and look at the examples following it.
Steps in doing long-method derivation:
Δ𝑦
1. Fill in the slope-formula given by: Δ𝑥 =
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
2. Simplify it.
3. Let ∆𝑥 approach zero (0).
Example 1.1: Find the derivative of 𝑓(𝑥) = 𝑥 2 .
Let us start with transforming 𝑓(𝑥) = 𝑥 2 to 𝑓(𝑥 + ∆𝑥) = (𝑥 + ∆𝑥)2 .
Expand (𝑥 + ∆𝑥)2 : 𝑓(𝑥 + ∆𝑥) = 𝑥 2 + 2𝑥∆𝑥 + (∆𝑥)2
Use the slope formula:
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
Plug in the equivalent values of 𝑓(𝑥 + ∆𝑥) and 𝑓(𝑥):
Simplify by combining similar terms:
2𝑥∆𝑥+(∆𝑥)2
∆𝑥
27
[𝑥 2 +2𝑥∆𝑥+(∆𝑥)2 ]−𝑥 2
∆𝑥
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Simplify further by dividing the numerator by ∆𝑥: 2𝑥 + ∆𝑥
Lastly, as ∆𝑥 approaches 0, we get: 2𝑥
Thus, the derivative of 𝑥 2 is 2𝑥.
We can use some notations to simplify the symbols needed in derivation. Let us
𝑑
use "𝑑𝑥" instead of “∆𝑥 approaches 0” and “the derivative of” can be written as " 𝑑𝑥 ". So
now we can write:
𝑑 2
(𝑥 ) = 2𝑥
𝑑𝑥
Note: oftentimes, 𝑓′(𝑥) is used also for “the derivative of”.
Let’s try more examples using the notations discussed.
Example 1.2: Solve for 𝑓′(𝑥) the function, 𝑓(𝑥) = 2𝑥 2 − 16𝑥 + 35.
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
∆𝑥→0
𝑓 ′ (𝑥) = lim
2(𝑥 + ∆𝑥)2 − 16(𝑥 + ∆𝑥) + 35 − (2𝑥2 − 16𝑥 + 35)
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
2[𝑥2 + 2𝑥∆𝑥 + (∆𝑥)2 ] − 16𝑥 − 16∆𝑥 + 35 − 2𝑥2 + 16𝑥 − 35
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
2𝑥2 + 4𝑥∆𝑥 + 2(∆𝑥)2 − 16𝑥 − 16∆𝑥 + 35 − 2𝑥2 + 16𝑥 − 35
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
4𝑥∆𝑥 + 2(∆𝑥)2 − 16∆𝑥
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) = lim
∆𝑥→0
∆𝑥(4𝑥 + 2∆𝑥 − 16)
∆𝑥
𝑓 ′ (𝑥) = lim 4𝑥 + 2∆𝑥 − 16
∆𝑥→0
𝑓 ′ (𝑥) = 4𝑥 + 2(0) − 16
𝑓 ′ (𝑥) = 4𝑥 − 16
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𝑥
Example 1.3: Solve for 𝑓′(𝑥) the function, 𝑓(𝑥) = 𝑥+1.
𝑓 ′ (𝑥) = lim
∆𝑥→0
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
𝑓 ′ (𝑥) = lim (
1
𝑥 + ∆𝑥
𝑥
)(
−
)
∆𝑥 [𝑥 + ∆𝑥] + 1 𝑥 + 1
𝑓 ′ (𝑥) = lim (
[𝑥 + ∆𝑥][𝑥 + 1] − 𝑥[𝑥 + ∆𝑥 + 1]
1
)(
)
[𝑥 + ∆𝑥 + 1][𝑥 + 1]
∆𝑥
𝑓 ′ (𝑥) = lim (
1
[𝑥2 + 𝑥 + 𝑥∆𝑥 + ∆𝑥] − [𝑥2 + 𝑥∆𝑥 + 𝑥]
)(
)
[𝑥 + ∆𝑥 + 1][𝑥 + 1]
∆𝑥
∆𝑥→0
∆𝑥→0
∆𝑥→0
1
∆𝑥→0 (𝑥 + ∆𝑥 + 1)(𝑥 + 1)
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) =
1
(𝑥 + 0 + 1)(𝑥 + 1)
𝑓 ′ (𝑥) =
1
(𝑥 + 1)(𝑥 + 1)
𝑓 ′ (𝑥) =
1
(𝑥 + 1)2
Example 1.4: Solve for 𝑓′(𝑥) the function, 𝑓(𝑥) = √5𝑥 − 8.
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
√5(𝑥 + ∆𝑥) − 8 − √5𝑥 − 8
∆𝑥→0
∆𝑥
𝑓 ′ (𝑥) = lim
It looks like we are stuck here. Let us now recall the concept of rationalization (refer to the
Metalanguage) since this is what we are going to do with this problem. Probably in your previous
mathematics subject you have just tried rationalizing the denominator but in our example, we will rationalize
the numerator.
√5(𝑥 + ∆𝑥) − 8 − √5𝑥 − 8 √5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8
∙
∆𝑥→0
∆𝑥
√5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) = lim
∆𝑥→0
5(𝑥 + ∆𝑥) − 8 − (5𝑥 − 8)
∆𝑥 (√5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8)
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5𝑥 + 5∆𝑥 − 8 − 5𝑥 + 8
𝑓 ′ (𝑥) = lim
∆𝑥→0
5∆𝑥
𝑓 ′ (𝑥) = lim
∆𝑥→0
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) =
𝑓 ′ (𝑥) =
∆𝑥 (√5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8)
5
∆𝑥→0
𝑓 ′ (𝑥) =
∆𝑥 (√5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8)
(√5(𝑥 + ∆𝑥) − 8 + √5𝑥 − 8)
5
(√5(𝑥 + 0) − 8 + √5𝑥 − 8)
5
(√5𝑥 − 8 + √5𝑥 − 8)
5
2√5𝑥 − 8
In dealing with the long-method derivation, you still need to recall some of the
algebraic concepts you have encountered just like performing rationalization. The
success in doing the long-method derivation is still up to you and the process you are
going to do.
Let’s Check
Activity 1. At this point, I am positive that you will be ready to perform long-method derivation.
To prove this, find the derivative of the following functions using the long method.
1. 𝑓(𝑥) = 14
2. 𝐺(𝑡) = 2 − 27𝑡
3. ℎ(𝑥) = 5𝑥 3 − 1
4.
𝑅(𝑧) =
4
𝑧
5. 𝑘(𝑥) = 𝑥 3 − 𝑥 2 + 2𝑥 − 6
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Let’s Analyze
Activity 1. You are improving. Now, let us try doing the long-method derivation again but this
time, you need to explain the process you did.
1. 𝑓(𝑥) =
𝑥+3
𝑥+7
2. 𝑟(𝑡) = 5𝑡2 − 𝑡 + 9
3.
𝑔(𝑧) = √3z − 2
4.
𝑘(𝑚) =
4𝑚−15
3𝑚
5. ℎ(𝑥) = √𝑥2 − 5𝑥
Self-Help: You can also refer to the sources below to help you further
understand the lesson.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
In a Nutshell
The discussion of the introduction for derivatives is done. Here are some points to
consider in performing long-method derivation.
•
•
•
The process of calculating the derivative of a function is called “differentiation”.
To get a derivative, you need to do differentiation.
The formula used in performing the long-method differentiation comes from the
slope formula. That is why when doing it, you just need to fill in the formula with
appropriate values, simplify it as best as you can, and let ∆𝑥 approach zero.
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Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
Big Picture in Focus:
ULO-d. Understand and apply the Differentiation Rules of Algebraic Function
Metalanguage
This section will be your reference regarding some terms related to the rules of
differentiation of algebraic functions. This will serve as or common reference in dealing
with terms we are unfamiliar with or the words we are not quite sure of its meaning.
1. Auxiliary Variable. It is a variable found in the main function but is not necessarily
part of the solution of a function. In differential calculus, it is a variable that has an
equivalent function in terms of the independent variable. This can be usually found
in the application of chain rule.
Essential Knowledge
In ULO-d, you will run across numerous rules of differentiation which you need to
familiarize since this will be great for saving time in finding derivatives of functions. You
are already acquainted with the long-method differentiation in the previous ULO, this time,
you will be accustomed with the most efficient way of differentiating functions and you will
clearly see the difference between the two methods.
1. Differentiation of Algebraic Functions. Mathematicians developed a way to avoid
doing the long and time-consuming process of differentiation. This established method
will be essential in finding the derivatives of complex functions. Here are the standard
formulas and properties that will save you time in dealing with derivatives:
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DIFFERENTIATION RULES FOR ALGEBRAIC FUNCTIONS
𝑑
𝑑𝑥
𝑐 = 0; where 𝑐
is constant
The derivative of a constant is zero (0).
If 𝑢 and 𝑣 are functions, then the following formulas are true by the definition of
derivative:
𝑑
𝑑𝑢 𝑑𝑣
(𝑢 + 𝑣) =
+
𝑑𝑥
𝑑𝑥 𝑑𝑥
The derivative of the sum of two functions is equal
to the sum of their derivatives.
𝑑
𝑑𝑣
𝑑𝑢
(𝑢𝑣) = 𝑢
+𝑣
𝑑𝑥
𝑑𝑥
𝑑𝑥
The derivative of the product of two functions is
equal to the product of the first function and the
derivative of the second plus the product of the
second function and the derivative of the first.
𝑑𝑢
𝑑𝑣
𝑣
−𝑢
𝑑 𝑢
𝑑𝑥
𝑑𝑥
( )=
𝑑𝑥 𝑣
𝑣2
The derivative of the quotient of two functions is
equal to the product of the denominator and the
derivative of the numerator minus the product of the
numerator and the derivative of the denominator,
divided by the square of the denominator.
𝑑 𝑛
(𝑥 ) = 𝑛𝑥 𝑛−1 ;
𝑑𝑥
where 𝑛 is a positive integer
The derivative of a variable with a positive integer
exponent is the product of the exponent and the
variable raised to the integer exponent minus one.
𝑑 𝑛
𝑑𝑢
(𝑢 ) = 𝑛𝑢𝑛−1
;
𝑑𝑥
𝑑𝑥
where 𝑢 is a function and 𝑛 is a positive
integer
𝑑𝑢
𝑑
(√𝑢) = 𝑑𝑥 ;
𝑑𝑥
2√𝑢
where 𝑢 is a function and
The derivative of a function with a power is the
product of the power, the function raised to its power
minus one, and the derivative of the function. This is
called the general power formula.
This is a special case where 𝑛
1
= 2. Its derivative
is the derivative of the function divided by two times
the square root of the function.
1
𝑛=2
Let us now have examples of each of the rules for you to appreciate the method more.
The proof of the standard formulas presented above will be on separate section. You can
also prove it on your own using the long-method differentiation.
Example 1.1: Differentiate the following functions:
(a) 𝑦 = 25
𝑦′ = 0
 When differentiating a constant, it is always equal to zero (0).
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(b) 𝑓 (𝑥) = 8𝑥 − 3
𝑓 ′ (𝑥) = [8 (
𝑑
𝑑𝑥
𝑥) + 𝑥 (
𝑑
𝑑𝑥
8)] − 0
𝑓′(𝑥) = 8(1) + 𝑥(0)
𝑓′(𝑥) = 8
 The given function has a product of a constant and a variable where we need to
use the product formula for differentiation.
(c) 𝑓 (𝑥) = 𝑥 2 + 6𝑥 − 1
𝑓′(𝑥) = 2𝑥 2−1 + 6 (
𝑑
𝑑𝑥
𝑥) − 0
𝑓′(𝑥) = 2𝑥 + 6
 In this example, we used the differentiation for sum, power rule, and the product
rule.
(d)
𝑦=
𝑥
𝑥 2 −1
𝑦′ =
𝑦′ =
𝑦′ =
(𝑥 2 −1)(1)−(𝑥)(2𝑥)
(𝑥 2 −1)2
𝑥 2 −1−2𝑥 2
(𝑥 2 −1)2
−1−𝑥 2
(𝑥 2 −1)2
=−
𝑥 2 +1
(𝑥 2 −1)2
 Remember the quotient rule of differentiation, it is applied here.
(e)
𝑦 = (3𝑥 3 + 2)(𝑥 2 − 1)
𝑑
𝑑
𝑦 ′ = (3𝑥 3 + 2) 𝑑𝑥 (𝑥 2 − 1) + (𝑥 2 − 1) 𝑑𝑥 (3𝑥 3 + 2)
𝑦 ′ = (3𝑥 3 + 2)(2𝑥) + (𝑥 2 − 1)(9𝑥 2 )
𝑦 ′ = 6𝑥 4 + 4𝑥 + 9𝑥 4 − 9𝑥 2
𝑦 ′ = 15𝑥 4 − 9𝑥 2 + 4𝑥
 This example shows how the differentiation of the product of two functions work.
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(f)
𝑦 = (𝑥 2 − 4𝑥 + 8)3
𝑦 ′ = 3(𝑥 2 − 4𝑥 + 8)3−1
𝑑
𝑑𝑥
(𝑥 2 − 4𝑥 + 8)
𝑦 ′ = 3(𝑥 2 − 4𝑥 + 8)2 (2𝑥 − 4)
𝑦 ′ = (𝑥 2 − 4𝑥 + 8)2 (6𝑥 − 12)
 This is an example of the general power formula. Take note that you cannot multiply the
two functions together since the first one has a power of 2 and the other one has a power
of one 1. If you want to simplify the answer further, you may expand the first function by
multiplying it by itself and multiplying it, afterwards, with the second function.
There is also a special case called the Chain Rule where there is an auxiliary variable
in the primary function. This auxiliary variable is the function of the independent variable. This
simply means that this auxiliary variable has an equivalent function in terms of the independent
variable. The Chain Rule is given by the form:
𝒅𝒚
𝒅𝒙
where
=
𝒅𝒚
𝒅𝒖
∙
𝒅𝒖
𝒅𝒙
𝒚 = 𝒇(𝒖) and 𝒖 = ∅(𝒙)
An example is illustrated below to fully comprehend the idea of chain rule.
Example 1.2: Derive 𝑦 = 3𝑢2 + 1 if 𝑢 = √𝑥 − 1
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑢
(3𝑢2 + 1) ∙
= 6𝑢 ∙
1
2
6
2
𝑑𝑥
(√𝑥 − 1)
use Chain Rule
1
(𝑥 − 1)2−1 (1)
= 6(√𝑥 − 1) ∙
=
𝑑
differentiate √𝑥 − 1 using general power rule
1
substitute the equivalent value of 𝒖
2√𝑥−1
=3
cancel √𝑥 − 1 and simplify
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
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Let’s Check
Activity 1. Find the derivative of the following functions using the rules presented on the previous
pages. Box your final answer.
1. 𝑓(𝑥) = 𝑥 3 + 12𝑥 2 − 7𝑥 + 10
2. 𝐺(𝑡) = 4𝑡 2 ( 𝑡 − 1) 3
3. 𝑦 =
𝑥 3 −1
𝑥+2
4. ℎ(𝑧) =
4𝑧
𝑧3
5. 𝑦 = (𝑥 3 − 5𝑥 2 + 7)(𝑥 2 + 8𝑥)2
Let’s Analyze
Activity 1. Solve or the derivative of the following and label each step of the process.
1. 𝑓(𝑥) = (𝑥 + 1)2 (𝑥 − 1)(𝑥 2 + 1)
2. 𝑅(𝑡) =
(𝑡+3)3
𝑡 2 −3
3. ℎ(𝑧) = √𝑧 2 + 4
𝑚3 +2𝑚
4. 𝑙(𝑚) = √
7𝑚2
𝑥2 +3𝑥+9
5. 𝑦 = (
𝑥−2
2
)
Activity 2. The following functions are similar from the long-method differentiation activity. Since
you were done doing its long-method derivation, it’s now time for you to prove if your answers in
long-method differentiation are the same if you use the rules of differentiation.
1. 𝑓(𝑥) =
𝑥+3
𝑥+7
2. 𝑟(𝑡) = 5𝑡2 − 𝑡 + 9
3.
𝑔(𝑧) = √3z − 2
4.
𝑘(𝑚) =
4𝑚−15
3𝑚
5. ℎ(𝑥) = √𝑥2 − 5𝑥
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In a Nutshell
These are some points to remember if you are going to use the rules of
differentiation for algebraic functions:
•
•
•
The derivative of a function using the rules of differentiation is similar to the
derivative of the same function using the long-method derivation.
You need to familiarize the rules of differentiation for algebraic function to save a
great time in solving for the derivatives of functions instead of utilizing the long
method.
The standard formulas for differentiation will be your foundation in dealing with the
advanced type of functions.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-e. Solve for the nth order derivative of a function
--- either implicit or explicit
Metalanguage
In this section, you will find terms that are related to the differentiation of functions.
These terms will be your basis as to what they are used for or how they are being used
in this concept. You may go back here anytime if you run across a certain term you find
ambiguous.
1. Implicit Function. It is a function that is usually presented in terms of both
independent and dependent variables (e.g. 𝑥 2 + 𝑦 2 = 1)
2. Explicit Function. It is a function that is given in terms of the independent
variable only (e.g. 𝑦 = √1 − 𝑥 2 )
Essential Knowledge
In ULO-e, you are expected to derive functions correctly in its nth order whether it
is implicit or explicit. Equipped with the knowledge of differentiating algebraic functions
according to certain formulas or rules, you can achieve the intended learning outcome.
These are the things you need to know in addition to the rules of differentiation for
algebraic functions.
1. Higher Order Derivatives. This refers to any derivative beyond the first
derivative. In solving higher order derivatives or the nth order derivatives, you
need to repeatedly differentiate the given function. The idea of this is presented
by the examples below:
Example 1.1: Find the 2nd derivative of 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 + 5𝑥 − 8
To find the second (2nd) derivative of the given function, you need to first determine its first
derivative. To denote the first derivate of 𝑓(𝑥), we write 𝑓′(𝑥) therefore 𝑓′′(𝑥), read as “ 𝑓
double prime 𝑥”, will be written for the 2nd derivative.
(1st)
𝑓′(𝑥) = 6𝑥 2 − 6𝑥 + 5
first derivative
𝑓′′(𝑥) = 12𝑥 − 6𝑥
second derivative
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There are other notations used for representing the higher order derivatives. Study
the given table and familiarize the following:
2nd derivative
3rd derivative
4th derivative
nth derivative
most common notation
𝑓′′(𝑥)
𝑓′′′(𝑥)
𝑓 (4) (𝑥)
𝑓 (𝑛) (𝑥)
Leibniz notation
𝑑2 𝑓
𝑑𝑥 2
𝑑3 𝑓
𝑑𝑥 3
𝑑4 𝑓
𝑑𝑥 4
𝑑𝑛 𝑓
𝑑𝑥 𝑛
alternative form of
Leibniz notation
𝑑2
[𝑓(𝑥)]
𝑑𝑥 2
𝑑3
[𝑓(𝑥)]
𝑑𝑥 3
𝑑4
[𝑓(𝑥)]
𝑑𝑥 4
𝑑𝑛
[𝑓(𝑥)]
𝑑𝑥 𝑛
Euler’s notation
𝐷2 𝑓
𝐷3 𝑓
𝐷4 𝑓
𝐷𝑛 𝑓
Let us have another set of examples. This time we will use variety of notations.
Example 1.2: Find the 3rd derivative of the following functions:
(a) 𝑦 = 4𝑥 5 + 6𝑥 3 + 2𝑥 + 1
𝑦′ = 20𝑥 4 + 18𝑥 2 + 2
𝑦′′ = 80𝑥 3 + 36𝑥
𝑦′′′ = 240𝑥 2 + 36
1
(b) 𝑓(𝑥) = (𝑥+2)2 − 7𝑥2
𝑓(𝑥) = (𝑥 + 2)−2 − 7𝑥2
transform
1
to (𝑥 + 2)−2 to use general power formula
(𝑥+2)2
𝑓′(𝑥) = (−2)(𝑥 + 2)−2−1 (1) − 14𝑥
𝑓′(𝑥) = (−2)(𝑥 + 2)−3 − 14
𝑓′′(𝑥) = (6)(𝑥 + 2)−3−1 (1) − 0
𝑓′′(𝑥) = (6)(𝑥 + 2)−4
𝑓′′′(𝑥) = (−24)(𝑥 + 2)−4−1 (1)
𝑓 ′′′ (𝑥) = (−24)(𝑥 + 2)−5
−24
𝑓 ′′′ (𝑥) = (𝑥+2)5
The functions presented in our examples are explicit functions, these are usually the
functions we have dealt with already. Now, let us try differentiating functions in an implicit form.
Differentiating functions in this form is easier when not transformed into an explicit one so we will
be differentiating these functions as it is.
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Example 1.3: Differentiate the following implicit functions:
(a) 𝑥2 − 𝑥𝑦 + 𝑦2 + 𝑥 = 1
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(𝑥2 − 𝑥𝑦 + 𝑦2 + 𝑥) = 𝑑𝑥 (1)
(𝑥2 ) −
𝑑
𝑑𝑥
(𝑥𝑦) +
𝑑𝑦
𝑑
𝑑𝑥
2
(𝑦 ) +
𝑑
𝑑𝑥
(𝑥) =
𝑑
𝑑𝑥
(1)
𝑑𝑦
2𝑥 − 𝑥 𝑑𝑥 − 𝑦 + 2𝑦 𝑑𝑥 + 1 = 0
𝑑𝑦
𝑑𝑦
−𝑥 𝑑𝑥 + 2𝑦 𝑑𝑥 = −1 − 2𝑥 + 𝑦
𝑑𝑦
𝑑𝑥
(−𝑥 + 2𝑦) = −1 − 2𝑥 + 𝑦
𝑑𝑦
𝑑𝑥
=
−1−2𝑥+𝑦
−𝑥+2𝑦
𝑥
(b) 𝑦 3 = 0
𝑥𝑦 −3 = 0
Transfer 𝑦 3 to the numerator and make its exponent positive
𝑥(−3)(𝑦 −3−1 )(𝑑𝑦) + 𝑦 −3 (𝑑𝑥) = 0
−3𝑥𝑦
−4
𝑑𝑦 + 𝑦
−3
Product Rule of Differentiation
𝑑𝑥 = 0
−3𝑥𝑦 −4 𝑑𝑦 = −𝑦 −3 𝑑𝑥
−𝑦 −3
𝑑𝑦
𝑦 ′ = −3𝑥𝑦 −4
𝑦′ =
𝑑𝑥
= 𝑦′
𝑦
3𝑥
(c) (𝑥 + 𝑦)2 = 3𝑥𝑦
2(𝑥 + 𝑦)2−1 (𝑑𝑥 + 𝑑𝑦) = 3(𝑥𝑑𝑦 + 𝑦𝑑𝑥)
2(𝑥 + 𝑦)(𝑑𝑥 + 𝑑𝑦) = 3𝑥𝑑𝑦 + 3𝑦𝑑𝑥
2(𝑥𝑑𝑥 + 𝑥𝑑𝑦 + 𝑦𝑑𝑥 + 𝑦𝑑𝑦) = 3𝑥𝑑𝑦 + 3𝑦𝑑𝑥
2𝑥𝑑𝑥 + 2𝑥𝑑𝑦 + 2𝑦𝑑𝑥 + 2𝑦𝑑𝑦 = 3𝑥𝑑𝑦 + 3𝑦𝑑𝑥
2𝑥𝑑𝑦 + 2𝑦𝑑𝑦 − 3𝑥𝑑𝑦 = 3𝑦𝑑𝑥 − 2𝑦𝑑𝑥 − 2𝑥𝑑𝑥
2𝑦𝑑𝑦 − 𝑥𝑑𝑦 = 𝑦𝑑𝑥 − 2𝑥𝑑𝑥
(2𝑦 − 𝑥)𝑑𝑦 = (𝑦 − 2𝑥)𝑑𝑥
𝑑𝑦
𝑑𝑥
(𝑦−2𝑥)
= (2𝑦−𝑥)
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Example 1.4: Find the 2nd derivative of the given implicit functions:
(a) 𝑥 3 + 𝑥𝑦 = 9𝑦
3𝑥 3−1 𝑑𝑥 + 𝑥𝑑𝑦 + 𝑦𝑑𝑥 = 9 𝑑𝑦
3𝑥 2 𝑑𝑥 + 𝑥𝑑𝑦 + 𝑦𝑑𝑥 = 9 𝑑𝑦
3𝑥 2 𝑑𝑥 + 𝑦𝑑𝑥 = 9 𝑑𝑦 − 𝑥𝑑𝑦
𝑑𝑦(9 − 𝑥) = 𝑑𝑥(3𝑥 2 + 𝑦)
𝑑𝑦
𝑑𝑥
=
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
(b) 𝑦 2 =
3𝑥 2 +𝑦
=
=
=
=
9−𝑥
(9−𝑥)(6𝑥+0)−(3𝑥 2 +𝑦)(−1)
(9−𝑥)2
(54𝑥−6𝑥 2 )−(−3𝑥 2 −𝑦)
(9−𝑥)2
54𝑥−6𝑥 2 +3𝑥 2 +𝑦)
(9−𝑥)2
54𝑥−3𝑥 2 +𝑦
(9−𝑥)2
= −
3𝑥 2 −54𝑥−𝑦
(9−𝑥)2
𝑥2
𝑥+𝑦
𝑦 2 (𝑥 + 𝑦) = 𝑥 2
𝑥𝑦 2 + 𝑦 3 = 𝑥 2
[𝑥(2𝑦𝑑𝑦) + 𝑦 2 𝑑𝑥] + 3𝑦 2 𝑑𝑦 = 2𝑥𝑑𝑥
2𝑥𝑦𝑑𝑦 + 𝑦 2 𝑑𝑥 + 3𝑦 2 𝑑𝑦 = 2𝑥𝑑𝑥
(2𝑥𝑦 + 3𝑦 2 )𝑑𝑦 = (2𝑥 − 𝑦 2 )𝑑𝑥
𝑑𝑦
𝑑𝑥
2𝑥−𝑦 2
= 2𝑥𝑦+3𝑦 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
=
=
=
(2𝑥𝑦+3𝑦 2 )(2)−(2𝑥−𝑦 2 )(2𝑦)
(2𝑥𝑦+3𝑦 2 )2
(4𝑥𝑦+6𝑦 2 )−(4𝑥𝑦−2𝑦 3 )
(2𝑥𝑦+3𝑦 2 )2
4𝑥𝑦+6𝑦 2 −4𝑥𝑦+2𝑦 3
(2𝑥𝑦+3𝑦 2 )2
2𝑦 2 (3+𝑦)
= [(𝑦)(2𝑥+3𝑦)]2
2𝑦 2 (3+𝑦)
= (𝑦)2 (2𝑥+3𝑦)2 =
2(3+𝑦)
(2𝑥+3𝑦)2
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
Let’s Check
Activity 1. Find the 2nd derivative of the following functions:
1. 𝑦 = 𝑥 2 − 7𝑥 + 1
2. 𝑥 = (2 − 3𝑡)2
3. 𝑦 = √16 + 𝑥 2
Activity 2. Find the 3rd derivative of the following functions:
1. 𝑦 = 3𝑥 4 − 𝑥 3 + 5𝑥 2 + 4𝑥 + 9
2. 𝑥 = 𝑡 2 + √𝑡 + 2𝑡
3. 𝑧 = (𝑦 − 3)(𝑦 + 5)3
Activity 3. Find
𝑑𝑦
𝑑𝑥
of the following implicit functions:
1. 𝑥2 + 2𝑦 2 = 4𝑦
2. 3𝑦 2 (𝑥 + 𝑦) = 𝑥 − 𝑦
1
1
3. 𝑥 2 − 𝑥𝑦 2 = − 𝑦 2
Activity 4. Find
𝑑2 𝑦
𝑑𝑥2
of the following implicit functions:
1. 𝑥3 − 9𝑦 = 𝑥𝑦
2. 𝑦 2 − 3𝑦 2 √𝑥 = 𝑥𝑦
3. √𝑥 + 𝑦 = 𝑦 2
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Let’s Analyze
Activity 1. Find the 2nd derivative of the even-numbered functions and the 3rd derivative of the
odd-numbered ones:
1. 𝑓(𝑥) = (𝑥 + 2)4
2. 𝑦 =
𝑥2+ 2
√𝑥
3. ℎ(𝑧) = 𝑧 6 + 3𝑧 5 − 4𝑧 4 + 𝑧 3 + 2𝑧 2 + 9𝑧 + 10
4. 𝑓(𝑥) =
√𝑥 2 −3
𝑥
5. 𝑥 = (2𝑡 + 1)3
6. 𝑦 =
2
(6 + 2𝑥− 𝑥 2 )4
1
5
7. 𝑔(ℎ) = 4√ℎ3 −
8. 𝑥 =
8ℎ2
− √ℎ
1
√6𝑡+ 𝑡 4
9. 𝑦 = 6𝑥 −2 + 7𝑥 −3 − 𝑥 −4
3
10. 𝑓(𝑥) = √𝑥 2 −
32
4
√𝑥
+
1
3√𝑥 5
Activity 2. Find (a) 𝑦′ by equating the function to 𝑦 and differentiating it; (b) 𝑦′ by implicit
differentiation; and (c) 𝑦′′ using either of the solved 𝑦′.
1. 8𝑥𝑦 + 2𝑥 4 𝑦 −3 = 𝑥 3
2. 𝑥2 = 𝑦 2 − √𝑥 3 + 2𝑦
3. 3𝑥2 =
(𝑥2 +𝑦)2
𝑦
4. (𝑥 + 2𝑦)2 = 4𝑥𝑦 + √𝑥
5. 𝑥 3 − 𝑦 3 = 3
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In a Nutshell
You need to take note of these points when you run across nth order derivative of
both explicit and implicit functions:
•
•
•
To perform higher order differentiation, you just need to do derivation after another
derivation.
In differentiation implicit functions, it would be easier to derive it in its original form
than transforming it into explicit function.
Most importantly, there are many ways to find the derivative of a function. All you
have to do is find the simplest method.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture
Week 4-5: Unit Learning Outcomes (ULO): At the end of the unit, you are expected
to:
a. Calculate the tangents and normal to plane curves,
b. Determine increasing and decreasing functions and its local maximum and
minimum,
c. Solve for concavity: points of inflection using second derivative test.
d. Solve for the Application of Derivatives such as maxima and minima as well
as time rates.
Big Picture in Focus:
ULO-2a. Determine the tangents and normal to plane curves.
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to the differential calculus and ULO-a will be operationally defined to establish a common
frame of reference. You will encounter these terms as we delve deeper to the study of
Differential Calculus. Please refer to these definitions in case you will find it difficult to
understand mathematical concepts in relation with calculus. Also, you need the skill in
deriving to continue to the next pages.
We often need to find tangents and normals to curves when we are analyzing
forces acting on a moving body.
1. A tangent to a curve is a line that touches the curve at one point and has the
same slope as the curve at that point.
2. A normal to a curve is a line perpendicular to a tangent to the curve..
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Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes), you need to fully
understand the following essential knowledge that will be laid down in the succeeding
pages. Please note that you are not limited to exclusively refer to these resources. Thus,
you are expected to utilize other books, research articles and other resources that are
available in the university’s library (refer to the Library Contact on page 3).
This unit explains how differentiation can be used to calculate the equations of the
tangent and normal to a curve. The tangent is a straight line which just touches the curve
at a given point. The normal is a straight line which is perpendicular to the tangent.
To calculate the equations of these lines we shall make use of the fact that the
equation of a straight line passing through the point with coordinates (x1,y1) and having
slope m is given by
We also make use of the fact that if two lines with slope m1 and m2 respectively
are perpendicular, then m1m2 = −1. To master the techniques explained here it is vital
that you undertake plenty of practice exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able
to:
• calculate the equation of the tangent to a curve at a given point
• calculate the equation of the normal to a curve at a given point
Consider a function f(x) such as that shown in Figure 1. When we calculate the derivative,
f′, of the function at a point x = a say, we are finding the slope of the tangent to the graph
of that function at that point. Figure 1 shows the tangent drawn at x = a. The slope of this
tangent is f′(a).
Figure 1. The tangent drawn at x = a has slope f′(a).
We will use this information to calculate the equation of the tangent to a curve at a
particular point, and then the equation of the normal to a curve at a point.
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Calculating the equation of a tangent
Example 1: Find the equation of the tangent to the curve y=3x2 at the point (1, 3). Sketch
the curve and the tangent.
Step 1: Find the derivative
Use the rules of differentiation:
y = 3x2
y’ = 2(3x) = 6x
Step 2: Calculate the gradient/slope of the tangent
To determine the slope of the tangent at the point (1, 3), we substitute the x-value into
the equation for the derivative.
y’ = 6x = 6(1) = 6
Step 3: Determine the equation of the tangent
y – y1 = m(x – x1)
y – 3 = 6(x – 1)
y = 6x – 6 + 3
y = 6x – 3
Step 4: Sketch the curve and the tangent
Example 2: Find the equation of the tangent to f(x) = x3 −3x2 + x−1
Take not that we need to find f(x) to get the coordinates, at the point where x = 3
𝑓(3) = 33 − 3(32 ) + 3 − 1 = 27 − 27 + 3 − 1 = 2
So, the point of interest has coordinates (3, 2).
The next thing that we need is the slope of the curve at this point. To find this, we need
to differentiate 𝑓(𝑥):
𝑓′(𝑥) = 3𝑥 2 − 6𝑥 + 1
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We can now calculate the slope of the curve at the point where x = 3.
𝑓′(3) = 3(32 ) − 6(3) + 1 = 27 − 18 + 1 = 10
So, we have the coordinates of the required point, (3,2), and the slope of the tangent at
that point, 10.
What we want to calculate is the equation of the tangent at this point on the curve. The
tangent must pass through the point and have slope 10. The tangent is a straight line and
so we use the fact that the equation of a straight line that passes through a point (x1,y1)
and has slope m is given by the formula:
Substituting the given values,
𝑦 −2
= 10
𝑥−3
and rearranging
𝑦 − 2 = 10(𝑥 − 3)
𝑦 − 2 = 10𝑥 − 30
𝒚 = 𝟏𝟎𝒙 − 𝟐𝟖 - This is the equation of the tangent to the curve at the point (3,2).
The equation of a normal to a curve
In mathematics, the word ‘normal’ has a very specific meaning. It means ‘perpendicular’
or ‘at right angles’.
Figure 2. The normal is a line at right angles to the tangent.
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If we have a curve such as that shown in Figure 2, we can choose a point and draw in
the tangent to the curve at that point. The normal is then at right angles to the curve so it
is also at right angles (perpendicular) to the tangent.
We now find the equation of the normal to a curve. There is one further piece of
information that we need in order to do this. If two lines, having slope m1 and m2
respectively, are at right angles to each other then the product of their slope, m1m2, must
be equal to −1.
Recall that the tangent and the normal are perpendicular. Hence, from the said key
point above, the slope of the normal is just the negative reciprocal of the slope of
the tangent.
Example 1: Determine the equation of the normal to the curve xy=−4 at (−1, 4). Draw a
rough sketch.
Step 1: Find the derivative
Make y the subject of the formula and differentiate with respect to x:
4
𝑦=−
𝑥
𝑦 = −4𝑥 −1
Therefore,
𝑑𝑦
= −1(−4)𝑥 −2
𝑑𝑥
𝑦 ′ = 4𝑥 −2
4
𝑦′ = 2
𝑥
Step 2: Calculate the slope of the normal at (−1, 4)
First determine the slope of the tangent at the given point:
𝑑𝑦
4
=
𝑑𝑥 (−1)2
𝑚=4
Use the slope of the tangent to calculate the gradient of the normal:
m1m2 = -1
(4)(m2) = -1
1
m2 = − 4
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Step 3: Find the equation of the normal
Substitute the slope of the normal and the coordinates of the given point into the slopepoint form of the straight line equation.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
1
𝑦 − 4 = (𝑥 − (−1))
4
1
1
𝑦=− 𝑥− +4
4
4
1
15
𝑦=− 𝑥+
4
4
or
𝟒𝒚 = −𝒙 + 𝟏𝟓
Therefore, this is the equation of the normal
Step 4: Draw a rough sketch
Example 2: Fnd the equation of the tangent and the equation of the normal to the curve
1
𝑦 = 𝑥 + 𝑥 at the point where 𝑥 = 2.
First of all we shall calculate the y coordinate at the point on the curve where x = 2:
1
5
𝑦 =2+ =
2
2
Next, we want the gradient of the curve at the point x = 2. We need to find
Note that we can write y as 𝑦 = 𝑥 + 𝑥 −1 then
𝑑𝑦
1
−2
=
1
−
𝑥
=
1
−
𝑑𝑥
𝑥2
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𝑑𝑦
𝑑𝑥
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Furthermore, when x = 2
𝑑𝑦
1
1 − 22 =
𝑑𝑥
3
4
5
This is the gradient of the tangent to the curve at the point (2, 2). We know that the
standard equation for a straight line is
With the given values, we have
5
𝑦 −2
3
=
𝑥−2
4
Simplifying,
5 3
= (𝑥 − 2)
2 4
5
4(𝑦 − ) = 3(𝑥 − 2)
2
4𝑦 − 10 = 3𝑥 − 6
4𝑦 = 3𝑥 + 4
So, the equation of the tangent to the curve at the point where 𝑥 = 2 is 𝟒𝒚 = 𝟑𝒙 + 𝟒
𝑦−
Now we need to find the equation of the normal to the curve.
Let the gradient of the normal be m2. Suppose the slope of the tangent is m1. Recall
that the normal and the tangent are perpendicular and hence m 1m2 = −1. We know m1 =
3⁄ . So,
4
3
4
m2 = -1
m2 =
−4
3
So we know the gradient of the normal and we also know the point on the curve through
which it passes, (2, 5⁄2).
5
𝑦 −2
4
=−
𝑥−2
3
Simplifying,
5
3 (𝑦 − ) = −4(𝑥 − 2)
2
15
3𝑦 −
= −4𝑥 + 8
2
15
3𝑦 + 4𝑥 = 8 +
2
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2
𝟔𝒚 + 𝟖𝒙 = 𝟑𝟏
This is the equation of the normal to the curve at the given point.
3𝑦 + 4𝑥 =
Self-Help: Refer to the sources below to help you further understand the
lesson.
Love, C., & Rainville E., (1962). Differential Calculus Notes for Mathematics 100.
e-sources
https://www.siyavula.com/read/maths/grade-12/differential-calculus/06-differential-calculus-04
http://www.mathcentre.ac.uk/
Let’s Check
Activity 1: Determine the equations of the tangent to the curve defined by:
1. 𝑓(𝑥) = 𝑥 3 + 2𝑥 2 − 7𝑥 + 1 𝑎𝑡 𝑥 = 2
2. 𝑓(𝑥) = 3𝑥 2 − 2𝑥 + 4 𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 3
3. 𝑓(𝑥) = 𝑥𝑒 𝑥 𝑎𝑡 𝑥 = 0
Activity 2: Determine the equations of the normal at each of the points indicated.
1. 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 1 𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 4
2. 𝑓(𝑥) = 2𝑥 3 − 5𝑥 + 4 𝑎𝑡 𝑥 = −1 𝑎𝑛𝑑 1
3. 𝑓(𝑥) = 𝑡𝑎𝑛𝑥 𝑎𝑡 𝑥 =
𝜋
4
Activity 3: Determine the equations of both tangent and normal to the curve:
1. 𝑦 = 3𝑥 2 − 𝑥 + 1 𝑎𝑡 𝑥 = 1
2. 𝑦 =
𝑥−1
𝑎𝑡 𝑥 = 0
𝑥+1
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Let’s Analyze
Activity 1: For odd-numbered questions, determine the equation of the normal. For
even-numbered questions, determine the equation of the tangent.
1. 𝑦 = 3𝑥 2 − 2𝑥 + 1 𝑎𝑡 (1, 2)
2. 𝑦 = 𝑥 3 − 3𝑥 2 − 2 𝑎𝑡 (1, −4)
3. 𝑦 = 2 + 4𝑥 − 𝑥 2 𝑎𝑡 𝑥 = −1
4. 𝑥 2 − 6𝑥 + 2𝑦 − 8 = 0 𝑎𝑡 𝑥 = 3
5. 𝑥 2 + 𝑦 2 − 6𝑥 + 2𝑦 = 0 𝑎𝑡 (0, 0)
6. 𝑦 = ( 2𝑥 − 1) 3 𝑎𝑡 𝑥 = 1
Activity 2: Solve for each of the following:
1. Determine the points where the slope of the tangent to the curve:
a. 𝑓(𝑥) = 1 − 3𝑥 2 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 5
1
b. 𝑔(𝑥) = 3 𝑥 2 + 2𝑥 + 1 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 0
2. Determine the point(s) on the curve f(x)=(2x−1)2 where the tangent is:
a. parallel to the line y = 4x−2
b. perpendicular to the line 2y + x – 4 = 0.
3. Find the equation of the tangent line at the point (1, 2) with the curve 3𝑦 2 − 2𝑥 5 = 10.
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In a Nutshell
These are some points to remember in solving the equations to tangents and
normal.
•
•
•
Follow the steps in the module to avoid mistakes in getting the equation.
You need to familiarize the rules of differentiation for algebraic function to save a
great time in solving for the derivatives. With that, we can get the slope to get the
equation of either the tangent/normal.
Remember that normal slope is the negative reciprocal of the tangent slope.
Q&A List
If you have any questions regarding tangents and normals, kindly write down on
the table provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-2b. Determine increasing and decreasing functions and its local maximum
and minimum
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to differential calculus and ULO-b will be operationally defined to establish a common
frame of reference. You will encounter these terms as we delve deeper to the study of
Differential Calculus. Please refer to these definitions in case you will find it difficult to
understand mathematical concepts in relation with calculus. Also, you need the skill in
deriving to continue to the next pages.
• A maximum is a high point and a minimum is a low point:
•
A function is "increasing" when the y-value increases as the x-value increases,
like this:
•
In decreasing functions, the y-value decreases as x-value increases.
•
First Derivative Test. A method for determining whether a critical point is a
minimum, maximum, or neither.
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Essential Knowledge
In ULO-b, you are expected to be expert in determining increasing and decreasing
functions using the first derivative test.
INCREASING/DECREASING FUNCTIONS
The derivative of a function may be used to determine whether the function is
increasing or decreasing on any intervals in its domain. If f′(x) > 0 at each point in an
interval I, then the function is said to be increasing on I. f′(x) < 0 at each point in an interval
I, then the function is said to be decreasing on I. Because the derivative is zero or does
not exist only at critical points of the function, it must be positive or negative at all other
points where the function exists.
In determining intervals where a function is increasing or decreasing, you first find
domain values where all critical points will occur; then, test all intervals in the domain of
the function to the left and to the right of these values to determine if the derivative is
positive or negative. If f′(x) > 0, then f is increasing on the interval, and if f′(x) < 0, then f
is decreasing on the interval. This and other information may be used to show a
reasonably accurate sketch of the graph of the function.
Example 1: Find where the function, 𝑓(𝑥) = 𝑥 3 + 3𝑥 2 − 9𝑥 + 7 is decreasing or
increasing
Step 1: Get the first derivative
𝑓′(𝑥) = 3𝑥 2 + 6𝑥 − 9
Step 2: Now we want to find the intervals where 𝑓′ is positive or negative. This is done
using critical points, which are the points where 𝑓′ is either 0 or undefined. 𝑓′ is a
polynomial, so it's always defined. To find its zeros, we can factor it:
𝑓′(𝑥) = 3(𝑥 + 3)(𝑥 − 1)
Step 3: Find the critical points.
To find these critical points you must first take the derivative of the function.
Second, set that derivative equal to 0 and solve for x.
So, 0 = 3 (𝑥 + 3) (𝑥 − 1)
Our crucial points are x = -3 and x = 1. These points divide the number line into
three intervals.
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Step 4: Let's evaluate f’ at each interval to see if it's positive or negative on that interval.
(−∞, −3)
(−3, 1)
(1, +∞)
Step 5: Now to find its local maximum and minimum
f’(-4) = +15, f’(0) = -9, f’(2) = +15
+
-
+
max
min
To get the y-values of the max and min points, substitute to the original equation 𝑓(𝑥) =
𝑥 3 + 3𝑥 2 − 9𝑥 + 7
𝑓(−3) = (−3)3 + 3(−3)2 − 9(−3) + 7 = 34
𝑓(1) = (1)3 + 3(1)2 − 9(1) + 7 = 2
Therefore, the maximum point is at (-3, 34) and the minimum point is at (1, 2).
Try …..
For f(x) = x 4 − 8 x 2 determine all intervals where f is increasing or decreasing.
The domain of f(x) are all real numbers, and its critical points occur at x = −2, 0, and 2.
Testing all intervals to the left and right of these values for f′(x) = 4 x 3 − 16 x, you find
that.
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Hence, f is increasing on (−2,0) and (2, +∞) and decreasing on (−∞, −2) and (0,2).
Minimum points are at (-2, -16) and (2, -16) while the maximum point is at (0, 0)
Did you also get the same answer? If you did, you are a master!
Self-Help: Refer to the sources below to help you further understand the
lesson.
Love, C., & Rainville E., (1962). Differential Calculus Notes for Mathematics 100.
https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the
derivative/increasing-decreasing-functions
https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-53/a/increasing-and-decreasing-intervals-review
Let’s Check
Activity 1: Find the critical points of the following.
1. f(x)=8x3+81x2−42x−8
2. R(t)=1+80t3+5t4−2t5
3. g(w)=2w3−7w2−3w−2
Activity 2: Let us try to find where a function is increasing or decreasing given the
following.
1. f(x) = x3−4x, for x in the interval [−1,2]
2. h(x)=−x3 + 3x2 + 9
3. y = 2x – 5 on interval ( - ∞, ∞)
Activity 3: Determine the maximum and minimum points of the given functions below.
1. g(x)=x6−2x5+8x4
2. h(z)=4z3−3z2+9z+12
3. f(x) = 3x4 – 16x3 + 24x2
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Let’s Analyze
1. Find the range of values of x for which y = x3 + 5x2 - 8x + 1 is increasing.
2. For what values of x is y = 150x - 2x3 an increasing function?
3. Find the intervals in which f (x) = 2x³ + x² - 20 x is increasing or decreasing.
4. Find the intervals in which f (x) = x³ - 3 x + 1 is increasing or decreasing.
5. How will you know if a function increases and decreases and how will you locate the
maximum and minimum value? Explain.
In a Nutshell
These are some points to remember in determining where the function
increases/decreases.
•
•
•
•
You should know the first derivative of a function then equate it to 0.
You must be able to determine the critical points by factoring out the function.
Set the values of the intervals to determine where the function increases or
decreases.
To get the maximum, check the graph where it reaches its highest peak and for
the minimum, check where the graph reaches its lowest peak.
Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-2c. Solve for concavity: points of inflection using second derivative test.
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to differential calculus and ULO-c will be operationally defined to establish a common
frame of reference. You will encounter these terms as we delve deeper to the study of
Differential Calculus. Please refer to these definitions in case you will find it difficult to
understand mathematical concepts in relation with calculus. Also, you need the skill in
deriving to continue to the next pages.
• A maximum is a high point and a minimum is a low point:
•
Inflection points are points where the function changes concavity, i.e. from
being "concave up" to being "concave down" or vice versa.
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Essential Knowledge
In ULO-c, you are expected to determine the concavity and the points of inflection
of a function.
Inflection points, concavity upward and downward
A point of inflection of the graph of a function f is a point where the second derivative f′′
is 0. It’s also a point where the sense of concavity changes.
A piece of the graph of f is concave upward if the curve ‘bends’ upward. For example,
the popular parabola y=x2 is concave upward in its entirety.
A piece of the graph of f is concave downward if the curve ‘bends’ downward. For
example, a ‘flipped’ version y=−x2 of the popular parabola is concave downward in its
entirety.
The relation of points of inflection to intervals where the curve is concave up or down is
exactly the same as the relation of critical points to intervals where the function is
increasing or decreasing. That is, the points of inflection mark the boundaries of the two
different sort of behavior. Further, only one sample value of f′′ need be taken between
each pair of consecutive inflection points in order to see whether the curve bends up or
down along that interval.
In determining intervals where a function is concave upward or concave downward, you
first find domain values where f″(x) = 0 or f″(x) does not exist. Then test all intervals around
these values in the second derivative of the function. If f″(x) changes sign, then ( x, f(x))
is a point of inflection of the function. As with the First Derivative Test for Local Extrema,
there is no guarantee that the second derivative will change signs, and therefore, it is
essential to test each interval around the values for which f″(x) = 0 or does not exist.
Geometrically, a function is concave upward on an interval if its graph behaves like a
portion of a parabola that opens upward. Likewise, a function that is concave downward
on an interval looks like a portion of a parabola that opens downward. If the graph of a
function is linear on some interval in its domain, its second derivative will be zero, and it
is said to have no concavity on that interval.
Example 1: Determine the concavity of f(x) = x 3 − 6 x 2 −12 x + 2 and identify any
points of inflection of f(x).
Because f(x) is a polynomial function, its domain is all real numbers.
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Testing the intervals to the left and right of x = 2 for f″(x) = 6 x −12, you find that
If x = 1 → substitute to the 2nd derivative
f(1) = 6(1) – 12 = -6 → the negative sign shows that from -∞ to 2 the graph shows a
concavity downward
If x = 3 → substitute to the 2nd derivative
f(3) = 6(3)-12 = +6 → the positive sign shows that from 2 to +∞ the graph shows a
concavity upward.
To get the y-coordinate of the inflection point, substitute 2 to the original equation.
f(2) = 23 – 6(22) – 12(2) + 2 = -38
Therefore,
(-∞ to 2) → concave downward
(2 to +∞) → concave upward
(2, -38) → inflection point
Example 2: Find the values of x for which the curve of y=x4-4x3 has points of
inflection.
Solution: y’ = 4x3 – 12x2
y’’ = 12x2 – 24x = 12x(x-2)
Set y’’=0, 0 = 12x (x-2);
12x = 0; x-2=0
We get, x = 0 and x = 2
Testing the intervals to the left and right of x = 0 and x = 2 for y’’ = 12x(x-2), you find that
If x = -1 → substitute to 2nd derivative
f(-1) = 12(-1) 2 – 24(-1) = +36, the positive sign shows that from -∞ to 0, the graph is
concave upward
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If x = 1 → substitute to 2nd derivative
f(2) = 12(1) 2 – 24(1) = -12, the negative sign shows that from 0 to 2, the graph is
concave downward
If x = 3 → substitute to 2nd derivative
f(3) = 12(3) 2 – 24(3) = 36, the positive sign shows that from 2 to +∞, the graph is
concave upward
To get the y-coordinate of the inflection points, substitute 0 and 2 to the original
equation.
f(0) = 0
f(2) = 24 – 4(23) = -16
Therefore,
(-∞ to 0) → concave upward
(0 to 2) → concave downward
(2 to +∞) → concave upward
(0,0) and (2, -16) → inflection points
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feliciano F., and Uy, F., (1983). Differential Calculus and Integral Calculus.
https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-thederivative/concavity-and-points-of-inflection
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Let’s Check
Activity 1: Find the value or values of x for which the curve of the following functions has
a point of inflection.
1. y = (x – 1)4 (x – 6)
2. y = 2x3 – 3x2 - 36x + 25
3. y = 3x4 – 4x3 + 1
4. y = x4 – 4x3 + 4x2
5. y = 3x5 – 15x4 + 20x3 + 3
Activity 2: Find the inflection points of each of the given curve.
1. y = x3 – 3x2 + 4
2. 4y = 3x4 – 16x3 + 24x2
3. 3y = x3 + 3x2 – 9x + 3
6𝑥
4. y = 𝑥+3
5. y = 5x – x5
Let’s Analyze
1. For what values of a and b the point (−1,2) is an inflection point of the graph of the
function y(x) = ax3 + bx2?
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In a Nutshell
These are some points to remember in determining your points of inflection in your
function and where your points concave upwards/downwards.
•
•
•
•
You should be able to know the 2nd derivative of that function and then equate it to
0.
After equating it to 0, you will now get your x-coordinates’ points of inflection. Then,
test the intervals to the left and right.
Set the values of the intervals to determine the concavity.
Lastly to get the y-coordinates of your points of inflection, substitute the x-values
to the original equation.
Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-2d. Solve for the Application of Derivatives such as maxima and minima as well as
time rates
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to differential calculus and ULO-d will be operationally defined to establish a common
frame of reference. You will encounter these terms as we delve deeper to the study of
Differential Calculus. Please refer to these definitions in case you will find it difficult to
understand mathematical concepts in relation with calculus. Also, you need the skill in
deriving and your comprehension to be able to answer the following problems in the
applications of maxima and minima and time rates.
•
The terms maxima and minima refer to extreme values of a function, that is, the
maximum and minimum values that the function attains.
•
Maximum means upper bound or largest possible quantity. The absolute
maximum of a function is the largest number contained in the range of the function.
For example, the function f(x) = -16x2 + 32x + 6 has a maximum value of 22
occurring at x = 1. Every value of x produces a value of the function that is less
than or equal to 22, hence, 22 is an absolute maximum. In terms of its graph, the
absolute maximum of a function is the value of the function that corresponds to the
highest point on the graph.
•
Conversely, minimum means lower bound or least possible quantity. The absolute
minimum of a function is the smallest number in its range and corresponds to the
value of the function at the lowest point of its graph. If f(a) is less than or equal to
f(x), for all x in the domain of the function, then f(a) is an absolute minimum. As an
example, f(x) = 32x2 - 32x - 6 has an absolute minimum of -22, because every
value of x produces a value greater than or equal to -22.
•
In differential calculus, related rates problems involve finding a rate at which a
quantity changes by relating that quantity to other quantities whose rates of change
are known(Wikipedia).
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Essential Knowledge
In ULO-d, you are expected to solve the applications of derivatives. Examples are
the applications on maxima and minima and time rates.
MAXIMA AND MINIMA
Steps in Solving Maxima and Minima Problems
1. Identify the constant, say cost of fencing.
2. Identify the variable to be maximized or minimized, say area A.
3. Express this variable in terms of the other relevant variable(s), say A = f(x, y).
4. If the function shall consist of more than one variable, expressed in terms of one
variable (if possible and practical) using the conditions in the problem, say A =
f(x).
5. Differentiate and equate to zero, dA/dx = 0. Why? The roots of the resulting
equation are the critical numbers which will give the desired maximum or
minimum value of the function
The process of finding maximum or minimum values is called optimization. We are
trying to do things like maximize the profit in a company, or minimize the costs, or find
the least amount of material to make a particular object
These are very important in the world of industry(M.Bourne).
Example 1: The daily profit, P in $, of an oil refinery is given by
P = 8x − 0.02x2,
where x is the number of barrels of oil refined. How many barrels will give maximum
profit and what is the maximum profit?
Solution: In this problem, the equation is already given so, let us proceed to step 5.
𝑑𝑃
The profit is a max (or min) if 𝑑𝑥 = 0
𝑑𝑃
= 8 − 0.04𝑥
(2)
𝑑𝑥
0 = 8 − 0.04𝑥
𝑥 = 200 → 𝐼𝑠 𝑡ℎ𝑖𝑠 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚?
Remember one way of testing if it’s a maximum is to assume values to the left and to
the right of the value 200.
So, substituting x as 199 to equation (2), we have a (+) value. Substituting x as 201 to
equation 2 yields us a (–) value.
max
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We can conclude that the value x = 200 is a maximum
If x = 200, then we have the final answer as P = 8(200) – 0.02(2002) = 800$
So, if the company refines 200 barrels per day, the maximum profit of 800 is reached.
Example 2: A long strip of tin 30 cm wide is to be made into a gutter with rectangular
cross section by turning up equal widths along the edges. Find the depth of the gutter
which yields the greatest carrying capacity.
Step 1:
Solution: Let x - depth of the gutter
y - base of the rectangular cross section
A - area of the rectangular cross section
Step 2 - 4:
To ensure the greatest carrying capacity, we must make the area of the cross section
as great as possible. That is, we maximize A. Thus
A = xy → to be maximized
2x + y = 30
(2)
From equation (2), we get
y = 30 – 2x
(3)
Substituting (3) in (1), we obtain
A = x(30 – 2x) = 30x – 2x2
Step 5:
Differentiating with respect to x then equate to 0
𝑑𝐴
= 30 − 4𝑥
𝑑𝑥
0 = 30 − 4𝑥
𝒙 = 𝟕. 𝟓𝒄𝒎 →The depth of the gutter which yields the greatest carrying capacity.
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TIME RATES
 If a quantity x is a function of time t, the time rate of change of x is given by dx/dt.
 When two or more quantities, all functions of t, are related by an equation, the
relation between their rates of change may be obtained by differentiating both
sides of the equation with respect to t.
STEPS IN SOLVING TIME RATES
1. Draw a figure when necessary
2. Define the given values
3. Formulate Equations
4. Differentiate with respect to time
5. Substitute the condition/s to the equation
Example 1: Water is flowing into a vertical cylindrical tank at the rate of 24 ft 3/min. If the
radius of the tank is 4 ft, how fast is the surface rising?
Step 1: Draw a figure when necessary
Step 2: Define the given values
radius = 4 ft
𝑑𝑉
= 24 𝑓𝑡 3 /min
𝑑𝑡
𝑑ℎ
=?
𝑑𝑡
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Step 3: Formulate equations
𝑉 = 𝜋𝑟 2 ℎ = 𝜋(4𝑓𝑡)2 ℎ = 16𝑓𝑡 2 𝜋ℎ
Step 4: Differentiate with respect to time
𝑑𝑉
𝑑ℎ
= 16𝜋
𝑑𝑡
𝑑𝑡
Since
𝑑𝑉
= 24 𝑓𝑡 3 /min
𝑑𝑡
𝑑ℎ
24𝑓𝑡 3 /min = 16𝑓𝑡 2 𝜋
𝑑𝑡
𝑑ℎ
= 0.477𝑓𝑡/𝑚𝑖𝑛
𝑑𝑡
Therefore, the surface is rising at 0.477ft/min.
Example 2: Water flows into a vertical cylindrical tank at 12 ft3/min, the surface rises 6
in/min. Find the radius of the tank.
Now, what we are going to find is the radius.
Step 1: Draw a figure when necessary
Step 2: Define the given values
radius = ?
𝑑𝑉
= 12 𝑓𝑡 3 /min
𝑑𝑡
𝑑ℎ
= 6𝑖𝑛/ min = 0.5𝑓𝑡/𝑚𝑖𝑛
𝑑𝑡
Step 3: Formulate equations
𝑉 = 𝜋𝑟 2 ℎ
Step 4: Differentiate with respect to time
𝑑𝑉
𝑑ℎ
= 𝜋𝑟 2
𝑑𝑡
𝑑𝑡
Since
𝑑𝑉
= 12 𝑓𝑡 3 /min
𝑑𝑡
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12𝑓𝑡 3 /min = 𝜋𝑟 2 (0.5𝑓𝑡/𝑚𝑖𝑛)
𝑟 2 = 7.639437268𝑓𝑡 2
𝒓 = 𝟐. 𝟕𝟔𝟑𝟗𝟓 𝒇𝒕
Therefore, the radius of the tank is approximately 2.76 ft.
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feliciano F., and Uy, F., (1983). Differential Calculus and Integral Calculus.
Love, C., & Rainville E., (1962). Differential Calculus Notes for Mathematics 100.
https://www.intmath.com/applications-differentiation/7-maximum-minimum-problems.php.
Let’s Check
Activity 1: Applications of Maxima and Minima
1. Find two nonnegative numbers whose sum is 9 and so that the product of one
number and the square of the other number is a maximum.
2. A rectangular box with a square base and no top is to have a volume of 108
cubic inches. Find the dimensions for the box that require the least amount of
material.
3. An open rectangular box with square base is to be made from 48 ft.2 of material.
What dimensions will result in a box with the largest possible volume?
Activity 2: Applications of Time Rates
1. Air is being pumped into a spherical balloon such that its radius increases at a
rate of .50 in/min. Find the rate of change of its volume when the radius is 3
inches.
2. A ladder 10 meters long rests on horizontal ground and leans against a vertical
wall. The foot of the ladder is pulled away from the wall at the rate of 0.3 m/sec.
How fast is the top sliding down the wall when the foot of the ladder is 2 m from
the wall?
3. A baseball diamond is a square 90 ft on a side. A player runs from first base to
second base at 15 ft/sec. At what rate is the player's distance from third base
decreasing when she is half way from first to second base?
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Let’s Analyze
1. Find the dimensions of a rectangle with perimeter 1000 meters so that the area of the
rectangle is a maximum.
2. A closed right circular cylindrical tank is to have a capacity of 128𝜋𝑚3. Find the
dimensions of the tank that will require the least amount of material in making it.
3. The radius of a right circular cone is increasing at the rate of 6 cm/sec while its
altitude is decreasing at 3 cm/sec. Find the rate of change of its volume when its radius
is 8 cm and its altitude is 20 cm
4. Water is running out of a conical tank 3 m across the top and 4 m deep at the rate of
2 m3 /min. Find the rate at which the level of water drops when it is 1 m from the top.
In a Nutshell
Here are the important steps that you always need to remember:
Steps in Solving Maxima and Minima Problems
1. Identify the constant, say cost of fencing.
2. Identify the variable to be maximized or minimized, say area A.
3. Express this variable in terms of the other relevant variable(s), say A = f(x, y).
4. If the function shall consist of more than one variable, expressed in terms of one
variable (if possible and practical) using the conditions in the problem, say A =
f(x).
5. Differentiate and equate to zero, dA/dx = 0. Why? The roots of the resulting
equation are the critical numbers which will give the desired maximum or
minimum value of the function
Steps in solving Time Rates Problems
1. Draw a figure when necessary
2. Define the given values
3. Formulate Equations
4. Differentiate with respect to time
5. Substitute the condition/s to the equation
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Q&A List
If you have any questions regarding the applications of time rates and maxima &
minima, kindly write down on the table provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture
Week 6-7: Unit Learning Outcomes-3 (ULO-3): At the end of the unit, you are expected
to:
a. Perform differentiation of trigonometric functions for the application of maxima and
minima;
b. Apply differentiation of inverse trigonometric functions to solve for optimization
problems;
c. Solve for the derivative of exponential and logarithmic functions; and
d. Familiarize the properties of the hyperbolic functions and find its derivative.
Big Picture in Focus:
ULO-3a. Perform differentiation of trigonometric functions for the application of
maxima and minima
Metalanguage
This section will serve as your word bank where the most essential terms relevant
to ULO3-a will be operationally defined to establish a common frame of reference. You
will encounter these terms as we delve deeper to the study of Differential Calculus
regarding transcendental functions.
1. Transcendental Function. It is a function that cannot be expressed as a finite
combination of the algebraic operations, such as addition, subtraction, multiplication,
division, raising to a power, and extracting a root.
a. In general, the term transcendental means non-algebraic which is not expressible
as a solution of a polynomial equation.
2. Trigonometric Functions. These are functions that describe the relation between the
sides and angles of a right triangle.
2.1 This kind of function includes the following six (6) functions: sine, cosine, tangent,
cotangent, secant, and cosecant. For each of these functions, there is an inverse
trigonometric function.
Essential Knowledge
In ULO-3a, you will run across various formulas for the differentiation of
trigonometric functions that you need to familiarize since this will be necessary in solving
for the maxima and minima problems. It is suggested to practice solving problems
involving trigonometric functions so that it would be easier for you to recall the formulas.
For the entire two weeks, we will be dealing with the transcendental functions such
as the trigonometric, inverse trigonometric, exponential, logarithmic, and hyperbolic
functions. The differentiation of trigonometric functions will be discussed in this Unit
Learning Outcome (ULO).
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1. Differentiation of Trigonometric Functions. The formulas shown on a table on the
next page are used for differentiating trigonometric functions. The symbol 𝒖 denotes
an arbitrary differentiable function of 𝒙.
DIFFERENTIATION OF
TRIGONOMETRIC FUNCTIONS
𝑑
𝑑𝑢
(sin 𝑢) = cos 𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(cos 𝑢) = −sin 𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(tan 𝑢) = sec 2 𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(cot 𝑢) = −csc 2 𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(sec 𝑢) = sec 𝑢 tan 𝑢
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(csc 𝑢) = −csc 𝑢 cot 𝑢
𝑑𝑥
𝑑𝑥
To find the derivative of a certain trigonometric function, you must follow the
formulas shown on the table. Study the examples given below and be guided with the
table:
𝑑𝑦
Example 3a.1: Find 𝑑𝑥 if 𝑦 = sin 4𝑥.
Solution: Take note that 𝑦 = 𝑠𝑖𝑛 4𝑥 is taken from the form 𝑦 = 𝑠𝑖𝑛 𝑢 where 𝑢 = 4𝑥.
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑥
(sin 4𝑥) = cos 4𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= cos 4𝑥
𝑑𝑢
where: 𝑢 = 4𝑥
𝑑𝑥
𝑑
𝑑𝑥
(4𝑥)
= cos 4𝑥 (4)
= 4 cos 4𝑥
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𝑑𝑦
Example 3a.2: Find 𝑑𝑥 if 𝑦 = sin3 4𝑥.
Solution: The given function has a power; hence we need to use the general power formula
to find its derivative.
𝑑
𝑦 ′ = (3) (sin3−1 4𝑥) 𝑑𝑥 (sin 4𝑥)
′
general power formula:
𝑑
2
𝑑
(sin 𝑢)
𝑑𝑥
𝑦 = 3 (sin 4𝑥)(cos 4𝑥) 𝑑𝑥 ( 4𝑥)
𝑦 ′ = 3 (sin2 4𝑥)(cos 4𝑥)( 4)
𝑑
(𝑢𝑛 ) = 𝑛(𝑢𝑛−1 )
𝑑𝑥
= cos 𝑢
𝑑𝑢
𝑑𝑥
𝑑𝑢
𝑑𝑥
simplify by multiplying
𝑦 ′ = 12 sin2 4𝑥 cos 4𝑥
Example 3a.3: Find 𝑔′(𝑥) if 𝑔(𝑥) = 3sec 𝑥 − 10 cot 𝑥.
Solution: Refer to the table of the trigonometric function derivatives and differentiate each
term using the formulas there.
𝑑
𝑑
𝑑
𝑑
𝑔′ (𝑥) = (3) (𝑑𝑥 sec 𝑥) (𝑑𝑥 𝑥) − (10) (𝑑𝑥 cot 𝑥) (𝑑𝑥 𝑥)
𝑑
𝑔′ (𝑥) = (3)(sec 𝑥 tan 𝑥) (
𝑑𝑥
𝑑
𝑥) − (10)(− csc 2 𝑥) (
𝑑𝑥
𝑥)
𝑔′ (𝑥) = (3)(sec 𝑥 tan 𝑥)(1) − (10)(− csc 2 𝑥)(1)
𝑔′ (𝑥) = (3)(sec 𝑥 tan 𝑥)(1) + (10)(csc 2 𝑥)(1)
𝑔′ (𝑥) = 3 sec 𝑥 tan 𝑥 + 10 csc 2 𝑥
➔ You can isolate the constants, 3 and 10, first from each term before differentiating the function.
Be careful with the minus sign in front of the second term and make sure it is properly simplified
by combining the minus sign and the negative sign of the second term.
Example 3a.4: Find ℎ′(𝑤) if ℎ(𝑤) = 3𝑤 −4 − 𝑤 2 tan 𝑤.
Solution: The product rule of differentiation must be used on the second term of the function.
𝑑
𝑑
𝑑
ℎ′(𝑤) = (3)(−4)(𝑤 −4−1 ) (𝑑𝑤 𝑤) − [(𝑤 2 ) (𝑑𝑤 tan 𝑤) + (tan 𝑤) (𝑑𝑤 𝑤 2 )]
𝑑
𝑑
ℎ′(𝑤) = (−12)(𝑤 −5 )(1) − [(𝑤 2 )(sec 2 𝑤) (𝑑𝑤 𝑤) + (tan 𝑤)(2)(𝑤 2−1 ) (𝑑𝑤 𝑤)]
ℎ′(𝑤) = −12𝑤 −5 − [(𝑤 2 )(sec 2 𝑤)(1) + (tan 𝑤)(2𝑤)(1)]
ℎ′(𝑤) = −12𝑤 −5 − (𝑤 2 sec 2 𝑤 + 2w tan 𝑤)
ℎ′(𝑤) = −12𝑤 −5 − (𝑤 2 sec 2 𝑤 + 2w tan 𝑤)
ℎ′(𝑤) = −12𝑤 −5 − 𝑤 2 sec 2 𝑤 − 2w tan 𝑤
➔ The second term of the function requires the product rule to find its derivative. To avoid confusion
while deriving with the minus sign in front of the second term, we need parentheses to properly
group the terms since the second term will produce two terms after differentiating. After
performing differentiation to all terms, make sure you distribute the minus sign to the terms inside
the parentheses to get the simplified final answer.
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Example 3a.5: Find 𝑦′ if 𝑦 = 5 sin 𝑥 cos 𝑥 + 4 csc 𝑥.
Solution: The product rule of differentiation must be used on the first term of the function and
we will isolate the constants, 5 and 4, before differentiating the first and second terms.
𝑑
𝑑
𝑦 ′ = (5) [𝑑𝑥 (sin 𝑥 cos 𝑥)] + (4) (𝑑𝑥 csc 𝑥)
𝑑
𝑦 ′ = (5) [(sin 𝑥) (
𝑑𝑥
𝑑
cos 𝑥) + (cos 𝑥) (
𝑑𝑥
sin 𝑥) ] + (4)(− csc 𝑥 cot 𝑥) (
𝑑
𝑑𝑥
𝑑
𝑥)
𝑑
𝑦 ′ = (5) [(sin 𝑥)(− sin 𝑥) (𝑑𝑥 𝑥) + (cos 𝑥)(cos 𝑥) (𝑑𝑥 𝑥) ] + (4)(− csc 𝑥 cot 𝑥)(1)
𝑦 ′ = (5) (−sin2 𝑥 + cos 2 𝑥) − 4 csc 𝑥 cot 𝑥
𝑦 ′ = −5 sin2 𝑥 + 5 cos 2 𝑥 − 4 csc 𝑥 cot 𝑥
rearrange the terms
Or
𝑦 ′ = 5 cos2 𝑥 −5 sin2 𝑥 − 4 csc 𝑥 cot 𝑥
➔ Isolating the constants before performing differentiation will make the process easier provided that the
constant is part of the product of a term like the constants 5 and 4 in the first and the second term,
respectively.
sin 𝑡
Example 3a.5: Find 𝑃′(𝑡) if 𝑃(𝑡) = 3−2 cos 𝑡.
Solution: The quotient rule of differentiation must be used on this problem. Let 𝑢 = sin 𝑡
and 𝑣 = 3 − 2 cos 𝑡.
𝑃
′ (𝑡)
𝑃
′ (𝑡)
=
=
(3−2 cos 𝑡)(
𝑑
𝑑
sin 𝑡)−(sin 𝑡)[ (3−2 cos 𝑡)]
𝑑𝑡
𝑑𝑡
(3−2 cos 𝑡)2
𝑑
𝑑𝑡
(3−2 cos 𝑡)2
(3−2 cos 𝑡)(cos 𝑡)−(sin 𝑡)[ (3)−(2)(
𝑑
cos 𝑡)]
𝑑𝑡
𝑃′ (𝑡) =
(3−2 cos 𝑡)(cos 𝑡)−(sin 𝑡)[0−(−2 sin 𝑡)]
(3−2 cos 𝑡)2
𝑃′ (𝑡) =
(3 cos 𝑡−2 cos2 𝑡)−(sin 𝑡)(2 sin 𝑡)
(3−2 cos 𝑡)2
𝑃′ (𝑡) =
3 cos 𝑡−2 cos2 𝑡− 2 sin2 𝑡
(3−2 cos 𝑡)2
factor out (−2) from −2 cos2 𝑡 − 2 sin2 𝑡
𝑃′ (𝑡) =
3 cos 𝑡−2(cos2 𝑡+sin2 𝑡)
(3−2 cos 𝑡)2
use the identity: sin2 𝑥 + cos2 𝑥 = 1
3 cos 𝑡−2(1)
𝑃′ (𝑡) = (3−2 cos 𝑡)2
3 cos 𝑡−2
𝑃′ (𝑡) = (3−2 cos 𝑡)2
➔ A trigonometric identity is used to simplify the final answer of the problem. There are times that you need to
simplify further a trigonometric function based on various trigonometric identities.
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A table of selected trigonometric identity is laid out for you as a reference for future
trigonometric problems you will encounter. These identities are assumed to be the most
common one.
TRIGONOMETRIC IDENTITIES
RECIPROCAL IDENTITIES
1
cot 𝑥 = tan 𝑥 =
DOUBLE-ANGLE IDENTITIES
cos 𝑥
sin 2𝑥 = 2 sin 𝑥 cos 𝑥
sin 𝑥
=
1
csc 𝑥 = sin 𝑥
2 tan 𝑥
1+tan2 𝑥
cos 2𝑥 = cos 2 𝑥 − sin2 𝑥
1
sec 𝑥 = cos 𝑥
= 2 cos2 𝑥 − 1
= 1 − 2 sin2 𝑥
PYTHAGOREAN IDENTITIES
sin2 𝑥 + cos 2 𝑥 = 1
=
sin2 𝑥 = 1 − cos2 𝑥
tan 2𝑥 =
cos2 𝑥 = 1 − sin2 𝑥
cot 2𝑥 =
1 + tan2 𝑥 = sec 2 𝑥
1−tan2 𝑥
1+tan2 𝑥
2 tan 𝑥
1−tan2 𝑥
cot2 𝑥−1
2 cot 𝑥
1 + cot 2 𝑥 = csc 2 𝑥
sin2 𝑥
Example 3a.6: Find 𝑦′ if 𝑦 = cos2 𝑥.
Solution: This problem requires simplification before differentiation. Please refer to the
table of trigonometric identities to simplify the function.
sin2 𝑥
𝑦 = cos2 𝑥
sin 𝑥 2
𝑦 = (cos 𝑥)
𝑦 = (tan 𝑥)2 = tan2 𝑥
apply general power formula
𝑑
𝑦′ = 2(tan2−1 𝑥) [𝑑𝑥 (tan 𝑥)]
𝑑
𝑦′ = 2(tan 𝑥) (sec 2 𝑥) (𝑑𝑥 𝑥)
𝑦′ = 2(tan 𝑥) (sec 2 𝑥)(1)
𝑦 ′ = 2 sec 2 𝑥 tan 𝑥
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Example 3a.7: Find 𝑦′ if 𝑦 = sin 2𝑥 + cos 2 𝑥
Solution: This problem can be solved in various ways, but the most straightforward
process is used here.
𝑦′ =
𝑑
𝑑𝑥
𝑑
(sin 2𝑥) + 𝑑𝑥 (cos 2 𝑥)
𝑑
𝑑
𝑦 ′ = cos(2𝑥) [𝑑𝑥 (2𝑥)] + (2)( cos 2−1 𝑥) [𝑑𝑥 (cos 𝑥)]
𝑦 ′ = cos(2𝑥)(2) + (2)( cos 𝑥)(− sin 𝑥)
𝑦 ′ = 2 cos 2𝑥 − 2 sin 𝑥 cos 𝑥
use identity: sin 2𝑥 = 2 sin 𝑥 cos 𝑥
𝑦 ′ = 2 cos 2𝑥 − sin 2𝑥
Example 3a.8: Find 𝑦′ if 𝑦 = tan(sin 𝑥)
Solution: The given equation is not a product of functions; it is what we call composition
of functions.
𝑑
𝑦 ′ = 𝑑𝑥 [tan(sin 𝑥)]
note: 𝑢 = 𝑠𝑖𝑛 𝑥
𝑑
𝑦 ′ = sec 2 (sin 𝑥) [𝑑𝑥 (sin 𝑥)]
𝑦 ′ = sec 2 (sin 𝑥) (cos 𝑥)
Example 3a.9: Find 𝑦′ if 𝑦 = cos 3 (tan(3𝑥))
Solution: Apply series of chain rule to differentiate the function.
𝑑
𝑦′ = (3)[cos3−1 (tan(3𝑥))] [𝑑𝑥 (cos(tan(3𝑥)))]
𝑑
𝑦′ = (3)[cos2 (tan(3𝑥))][− sin(tan(3𝑥))] [𝑑𝑥 (tan(3𝑥))]
𝑑
𝑦′ = (3)[cos2 (tan(3𝑥))][−sin(tan(3𝑥))] [sec 2 (3𝑥)] [𝑑𝑥 (3𝑥)]
𝑦′ = (3)[cos2 (tan(3𝑥))][−sin(tan(3𝑥))] [sec 2 (3𝑥)](3)
𝑦′ = −9 cos2 (tan(3𝑥)) sin(tan(3𝑥)) sec 2 (3𝑥)
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After familiarizing the process of trigonometric function derivation, we need to
study how this method works when used in the application of maxima and minima. There
are different word problems to be dealt with when talking about maxima and minima, so
it is advised to practice solving these kinds of applications.
Example 3a.10: A trough for holding water is formed by taking a piece of sheet metal 60
cm wide and folding the 20 cm on either end up as shown below. Determine the angle 𝜽
that will maximize the amount of water that the trough can hold.
Solution: To solve this problem, we need to maximize the volume of the trough. But if you think about
calculating the volume of a trough, it would be the cross-sectional area multiplied by the length. So,
whatever the given length is, we can maximize the volume of the trough my maximizing its cross-sectional
area. Let us redo the sketch of the trough to get the formula of the cross-sectional area.
With this more specific sketch, we can say that the cross-sectional area of the trough has:
• a rectangle in the middle with height 𝒉 and width 𝟐𝟎
• two (2) identical right triangle with height 𝒉, base 𝒃, and hypotenuse 𝟐𝟎
Note: In basic geometry, the angle between the base 𝒃 and hypotenuse 𝟐𝟎 is equal to the angle 𝜽 of the
original sketch.
Let us now start formulating the values of 𝒃 and 𝒉 in terms of 𝜽:
𝑏
ℎ
cos 𝜃 = 20
sin 𝜃 = 20
𝑏 = 20 cos 𝜃
ℎ = 20 sin 𝜃
Now, based on the new sketch, the cross-sectional area of the trough in terms of 𝜽 is:
𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
1
1
1
𝐴 = 𝑙𝑤 + 2 𝑏ℎ + 2 𝑏ℎ = 𝑙𝑤 + 2 (2 𝑏ℎ)
1
𝐴 = 20(20 sin 𝜃) + 2 [2 (20 cos 𝜃)(20 sin 𝜃)]
𝐴 = 400 sin 𝜃 + 400 cos 𝜃 sin 𝜃
𝐴 = 400 (sin 𝜃 + sin 𝜃 cos 𝜃)
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Let us now maximize the cross-sectional area by performing differentiation:
𝐴′(𝜃) = (400) [cos 𝜃 + sin 𝜃(− sin 𝜃) + cos 𝜃(cos 𝜃) ]
𝐴′(𝜃) = (400) [cos 𝜃 − sin2 𝜃 + cos 2 𝜃 ]
use trig identity: sin2 𝜃 = 1 − cos 2 𝜃
𝐴′(𝜃) = (400) [cos 𝜃 − (1 − cos2 𝜃) + cos2 𝜃]
𝐴′(𝜃) = (400) (cos 𝜃 − 1 + cos2 𝜃 + cos 2 𝜃)
𝐴′(𝜃) = (400) (2 cos2 𝜃 + cos 𝜃 − 1)
factor out: 2 cos2 𝜃 + cos 𝜃 − 1
𝐴′(𝜃) = (400) (2 cos 𝜃 − 1)(cos 𝜃 + 1)
Let 𝐴′ (𝜃)
=0
0 = (400) (2 cos 𝜃 − 1)(cos 𝜃 + 1)
1
1
(400) 0 = [(400) (2 cos 𝜃 − 1)(cos 𝜃 + 1)] (400)
0 = (2 cos 𝜃 − 1)(cos 𝜃 + 1)
Find values of 𝜽:
2 cos 𝜃 − 1 = 0
cos 𝜃 + 1 = 0
2 cos 𝜃 = 1
cos 𝜃 = −1
1
𝜃 = cos −1(−1)
cos 𝜃 = 2
1
𝜃 = cos −1 2
𝜃1 =
𝜃2 = 𝜋
𝜋
3
We have come up with two (2) values of 𝜽, however, it won’t make sense if we consider 𝜃2 = 𝜋
𝜋
(180° 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒) since 𝜃 must be in the interval of 0 ≤ 𝜃 ≤
or in terms of degrees, 0 ≤ 𝜃 ≤
2
𝝅
𝟑
90°.Therefore, the maximum angle 𝜽 that will maximize the amount of water that the trough can hold is .
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
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Let’s Check
Activity 1. Find the derivative of the following trigonometric functions and simplify if possible. Box
your final answer.
1. 𝑦 = 3 sin 𝑥 − 4 cos 𝑥
2. 𝑦 = 𝑥 3 tan 𝑥
3. 𝑓(𝑥) =
cos 𝑥
1+sin 𝑥
4. 𝑦 = csc 𝑥 cot 𝑥
5. 𝑓(𝑥) = cos 2𝑥 + sin2 𝑥
6. 𝑦 =
sin 3𝑥
4+5 cos(2𝑥)
7. 𝑦 = 𝑥 sec 2 (𝜋𝑥)
𝑥 csc 𝑥
8. 𝑔(𝑥) = 3−csc 𝑥
Let’s Analyze
Activity 1. Solve for the derivative of the following and simplify if possible.
1. 𝑓(𝑥) = 2 cos 𝑥 − 6 sec 𝑥 + 3
2. 𝑓(𝑤) = tan 𝑤 sec 𝑤
3. ℎ(𝑡) = 𝑡 3 − 𝑡 2 sin 𝑡
4. 𝑦 = 6 + 4 √𝑥 csc 𝑥
5. 𝑅(𝑡) =
1
2 sin 𝑡−4 cos 𝑡
Activity 2. Solve the following problems and write the givens, what to maximize, and the
solution. Box your final answer.
1. Find the height of a right circular cylinder of maximum volume that can be inscribed
in a sphere of radius 15 cm.
2. Find the length of the shortest ladder which will reach from the ground level to a
high vertical wall if it must clear an 8-ft vertical fence which is 27 ft from the wall.
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3. A man in a motorboat at A receives a message at noon, calling him to B. A bus
making 40 miles per hour leaves C, bound for B, at 1:00 PM. If AC = 30 miles, what
must be the speed of the boat, to enable the man to catch the bus?
4. A gutter having a triangular cross-section is to be made by bending a strip of tin in
the middle. Find the angle between the sides when the carrying capacity is to a
maximum.
5. A trapezoidal gutter is to be made, from a strip of metal 22 inches wide by bending
up the edges. If the base is 14 inches wide, what width across the top gives the
greatest carrying capacity.
In a Nutshell
These are some points to remember if you are going to derive trigonometric
functions:
•
•
•
The derivative of the six (6) trigonometric functions can easily be determined by
using their predetermined formulas.
Trigonometric identities may be used to simplify the trigonometric function.
The rules for differentiating algebraic functions are used in performing derivation
to trigonometric functions.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-3b. Apply differentiation of inverse trigonometric functions to solve for
optimization problems
Metalanguage
In this section, you will find terms that are related to the differentiation of inverse
trigonometric functions. These terms will be your basis as to what they are used for or
how they are being used in this concept. You may go back here anytime if you run across
a certain term you find ambiguous.
1. Inverse Trigonometric Functions. These functions are used to determine the angle
measure when at least two (2) sides of a right triangle are known.
a. These functions have practical uses in navigation, physics, engineering, and
other sciences.
b. These functions give the value of the angle in degrees (°) or in radians from its
Trigonometric Function Value.
Essential Knowledge
In ULO3b, you are expected to derive inverse trigonometric functions correctly by
using certain formulas that will not require you to use the long-method differentiation.
Equipped with the knowledge of the rules for differentiating algebraic functions using its
specified rules, you can achieve the intended learning outcome. These are the things you
need to know in addition to the first transcendental function we discussed on the previous
ULO.
1. Inverse Trigonometric Functions. The six (6) inverse trigonometric functions
namely sin−1 𝑥, cos−1 𝑥, tan−1 𝑥 , csc −1 𝑥, sec −1 𝑥, and cot −1 𝑥 are used to find the
unknown measure of an angle of a right triangle when two (2) side lengths are known.
In this text, we shall denote our inverse trig functions with “Arc”.
Similar to the process of differentiating Trigonometric Functions, Inverse Trig
Functions can be derived using formulas which will be shown in the table.
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
𝑑
1
𝑑𝑢
(Arcsin 𝑢) =
2
𝑑𝑥
√1 − 𝑢 𝑑𝑥
𝑑
−1 𝑑𝑢
(Arccot 𝑢) =
𝑑𝑥
1 + 𝑢2 𝑑𝑥
𝑑
−1 𝑑𝑢
(Arccos 𝑢) =
𝑑𝑥
√1 − 𝑢2 𝑑𝑥
𝑑
1
𝑑𝑢
(Arcsec 𝑢) =
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
𝑑
1
𝑑𝑢
(Arctan 𝑢) =
2
𝑑𝑥
1 + 𝑢 𝑑𝑥
𝑑
−1
𝑑𝑢
(Arccsc 𝑢) =
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
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Examples on how to use the formulas for differentiating the inverse trigonometric
functions will be presented on the succeeding pages.
𝑑𝑦
Example 3b.1: Find 𝑑𝑥 if 𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛 3𝑥
Solution: The function 𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛 3𝑥 takes the form 𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛 𝑢 where 𝑢 = 3𝑥.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
1
(
𝑑
√1−(3𝑥)2 𝑑𝑥
1
√1−(3𝑥)2
3𝑥)
(3)
3
= √1−9𝑥 2
𝑑𝑦
Example 3b.2: Find 𝑑𝑥 if 𝑦 = 𝐴𝑟𝑐𝑡𝑎𝑛
𝑥
4
Solution:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
=
1
𝑑
1
1
𝑥2
1+ 2
4
1
𝑑
[4 (𝑑𝑥 𝑥)]
1
𝑥2
1+
16
=[
𝑥
[𝑑𝑥 (4)]
𝑥 2
1+( )
4
1
[4 (1)] = (
𝑥2
1+
16
1
]=
𝑥2
(1+ )4
16
1
) (4 )
1
(4+
1
4𝑥2
)
16
1
= [4(16)+4𝑥 2] = 4(16+𝑥 2 )
4
= 16+𝑥 2
Example 3b.3: Find 𝑦′ if 𝑦 = 𝐴𝑟𝑐𝑠𝑖𝑛 (𝑥 − 1)
Solution:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
1
𝑑
√1−(𝑥−1)2
[𝑑𝑥 (𝑥 − 1)]
1
√1−(𝑥 2 −2𝑥+1)
1
= √1−𝑥 2
+2𝑥−1
(1)
1
= √−𝑥 2
+2𝑥
1
= √2𝑥−𝑥 2
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𝑥+1
Example 3b.4: Find 𝑦′ if 𝑦 = 𝐴𝑟𝑐𝑡𝑎𝑛 (𝑥−1)
Solution:
𝑦′ =
𝑦′ =
𝑦′ =
1
𝑑
𝑥+1 2
1+(
)
𝑥−1
1
(𝑥+1)2
1+
(𝑥−1)2
𝑥+1
[𝑑𝑥 (𝑥−1)]
[
(𝑥−1)(1)−(𝑥+1)(1)
(𝑥−1)2
]
(𝑥−1)(1)−(𝑥+1)(1)
(1+
(𝑥+1)2
)(𝑥−1)2
(𝑥−1)2
𝑥−1−𝑥−1
𝑦′ = (𝑥−1)2 +(𝑥+1)2
−2
𝑦′ = 𝑥 2 −2𝑥+1+𝑥 2+2𝑥+1
−2
𝑦′ = 2𝑥 2 +2
−1
𝑦′ = 𝑥 2 +1
2
Example 3b.5: Find 𝑦′ if 𝑦 =
2𝑥+1
√3
𝐴𝑟𝑐𝑐𝑜𝑡 (
√3
)
Solution:
𝑦′ =
2
−1
[
𝑑
2
√3 1+(2𝑥+1)
2𝑥+1
] [𝑑𝑥 (
)]
√3
√3
𝑦′ = −
2
1
[
1
√3 1+(2𝑥+1)
2
2
𝑑
] [( ) 𝑑𝑥 (2𝑥 + 1)]
√3
(√3)
𝑦′ = −
2
√3
1
(
4
2
)( )
4𝑥2 +4𝑥+1
1+
3
√3
1
𝑦 ′ = − 3 ( 3+4𝑥2 +4𝑥+1 )
3
4
3
𝑦 ′ = − 3 (4𝑥 2+4𝑥+4)
𝑦′ =
−1
𝑥 2 +𝑥+1
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At this point, we will solve a problem using the differentiation of inverse
trigonometric function.
Example 3b.6: A ladder 25 ft long leans against a vertical wall. If the lower end is pulled
away at the rate of 6 ft/sec, how fast is the angle between the ladder and the floor
changing when the lower end is 7ft from the wall?
Solution: A sketch must be done first to deeply understand the problem. A label based on the givens from
the problem is also necessary.
Note: Let 𝜃 be the angle
between the ladder and the
floor.
𝑑𝑥
= 6𝑓𝑡/𝑠𝑒𝑐
𝑑𝑡
𝜃
x
7 ft
We need to find
𝑑𝜃
𝑑𝑡
when
𝑑𝑥
𝑑𝑡
= 6𝑓𝑡/𝑠𝑒𝑐 and 𝑥 = 7 𝑓𝑡. To do this, we shall use the angle 𝜃,
𝒙
hypotenuse, and 𝑥 of the right triangle made by the ladder, floor, and wall. Use 𝐜𝐨𝐬 𝜽 = . Find 𝜃 by
𝟐𝟓
isolating it on the left side of the equal sign (=). Thus, we have:
𝜃 = arccos
𝑥
25
Differentiate the equation with respect to 𝑡 to compute for
𝑑𝜃
𝑑𝑡
𝑑𝜃
𝑑𝑡
=
−1
−1
=
√1−
𝑑𝑥
Now, substitute the values of 𝑥 = 7 and
𝑑𝜃
𝑑𝑡
𝑑𝜃
𝑑𝑡
=
−1
2
√1−(7)
625
𝑑
2
√1−( 𝑥 )
25
𝑑𝑡
1
𝑥2
𝑥
[𝑑𝑡 (25)]
𝑑𝑥
(25) ( 𝑑𝑡 )
625
= 6.
1
( ) (6)
25
1
= − 4 𝑟𝑎𝑑/𝑠𝑒𝑐
The negative sign tells us that the angle
𝜃 is decreasing.
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𝑑𝜃
𝑑𝑡
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
Let’s Check
Activity 1. Find the derivative of the following inverse trigonometric functions and simply it
whenever possible.
1. 𝑦 = 𝐴𝑟𝑐𝑐𝑜𝑠
2. 𝑦 = 𝐴𝑟𝑐𝑡𝑎𝑛
𝑥
1−𝑥
4
𝑥
3. 𝑦 = 𝐴𝑟𝑐𝑐𝑜𝑡 (tan 2𝑥)
4. 𝑦 = 𝐴𝑟𝑐𝑠𝑒𝑐 √4𝑥 + 1
5. 𝑦 = 𝐴𝑟𝑐𝑐𝑠𝑐
𝑥
2
Let’s Analyze
Activity 1. Differentiate the following functions and simplify if possible.
1. 𝑦 =
1
4
𝐴𝑟𝑐𝑡𝑎𝑛
2. 𝑦 = 𝐴𝑟𝑐𝑐𝑜𝑠
𝑥
4 sin 𝑥
3+5 cos 𝑥
−
2
√4−𝑥 2
3. 𝑦 = 𝑥√1 − 4𝑥 2 +
𝑥
1
2
𝐴𝑟𝑐𝑠𝑖𝑛 2𝑥
4. 𝑦 = 𝐴𝑟𝑐𝑐𝑜𝑡 𝑥 + 𝐴𝑟𝑐𝑡𝑎𝑛
2+𝑥
1−2𝑥
5. 𝑦 = (𝑥 − 1)√2𝑥 − 𝑥 2 − 𝐴𝑟𝑐𝑐𝑜𝑠 (𝑥 − 1)
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Activity 2. Solve the following problems using inverse trigonometric functions.
1. The lower edge of a picture is 4 ft., the upper edge is 9 ft above the eye of an observer.
At what horizontal distance should he stand if the angle subtended by the picture is a
maximum?
2. An isosceles triangle has legs 10 cm. The base decreases at the rate of 4 cm/sec. Find
the rate of change of the angle at the apex when the base is 16 cm.
3. A ladder 14 ft long is leaning against a fence 8 ft high, with the upper end projecting over
the fence. If the lower end slides away from the fence at the rate of 2 ft/sec, find the rate
at which the angle between the ladder and the ground is changing when the upper end is
just at the top of the fence.
In a Nutshell
You need to take note of these points when dealing with the differentiation of
inverse trigonometric functions.
•
•
•
A set of formulas for differentiating inverse trigonometric functions exist to avoid
using the long method.
Inverse trigonometric functions are necessary to determine the angle of a triangle
with known 2 sides.
Inverse trigonometric function is the opposite of the trigonometric functions. These
functions are namely sin−1 𝑥 , cos −1 𝑥 , tan−1 𝑥 , csc −1 𝑥 , sec −1 𝑥 , and cot −1 𝑥.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-3c. Solve for the derivative of exponential and logarithmic functions
Metalanguage
The terms that will be useful in this unit learning outcome are defined in this
section. These terms are all related as basis of logarithmic and exponential functions.
You may go back here anytime if you run across a certain term you find ambiguous.
1. Logarithmic Function. This function is the inverse of exponential function, where all
exponential functions can be expressed as logarithmic functions.
1.1 This function is necessary in permitting us to work with very large numbers while
manipulating numbers of a much more manageable size.
2. Exponential Function. This function is the opposite of the logarithmic functions. All
logarithmic functions can be expressed as exponential functions.
2.1 This function has an equation where the variable appears as the exponent.
Essential Knowledge
In ULO3c, you are expected to recall your knowledge about the properties of the
logarithmic and exponential functions. These properties will be of great help when dealing
with the differentiation of these functions. The logarithmic and exponential functions are
the 3rd and 4th transcendental functions to be differentiated.
1. The Logarithmic and Exponential Functions. Logarithms and exponents work well
together because they “undo” each other, as long as the base is the same. They can
be called as “Inverse Functions.”
A logarithmic function is defined by:
𝒚 = 𝐥𝐨𝐠 𝒃 𝒙
On the other hand, an exponential function is defined by the equation:
𝒙=𝒃
𝒚
Where in both forms, 𝑥 > 0, 𝑏 > 0 and 𝑏 ≠ 1. Putting them side by side, we can
conclude that they are equivalent to one another.
𝒙=𝒃
𝒚 = 𝐥𝐨𝐠 𝒃 𝒙
the base remains the base
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The laws of exponents, radicals, and logarithms have been part of your study in
Algebra and Trigonometry. Some of these laws are necessary to aid in our present
topic and it is listed down here for easy reference.
Laws of Exponents:
E1. 𝒂𝒎 𝒂𝒏 = 𝒂𝒎+𝒏
𝒂𝒎
E2.
= 𝒂𝒎−𝒏
𝒂𝒏
where 𝒎 > 𝒏, 𝒂 ≠ 𝟎
E3. (𝒂𝒎 )𝒏 = 𝒂𝒎𝒏
E4. (𝒂𝒃)𝒏 = 𝒂𝒏 𝒃𝒏
𝒂 𝒏
𝒂𝒏
(𝒃) = 𝒃𝒏
E5.
where 𝒃 ≠ 𝟎
Laws of Radicals:
𝒏
R1. √𝒂𝒏 = 𝒂
𝒏
𝒎
𝒏
𝒎
√𝒂𝒎 = ( √𝒂) = 𝒂 𝒏
R2.
𝒏
𝒏
𝒏
R3. √𝒂 ∙ √𝒃 = √𝒂𝒃
𝒏
𝒂
R4. √𝒃 =
R5.
𝒏
√𝒂
√𝒃
𝒏
𝒎 𝒏
√ √𝒂 = 𝒎𝒏√𝒂
Laws of Logarithms:
L1. 𝐥𝐨𝐠 𝒃 𝑴𝑵 = 𝐥𝐨𝐠 𝒃 𝑴 + 𝐥𝐨𝐠 𝒃 𝑵
L2.
𝑴
𝐥𝐨𝐠 𝒃 𝑵 = 𝐥𝐨𝐠 𝒃 𝑴 − 𝐥𝐨𝐠 𝒃 𝑵
L3. 𝐥𝐨𝐠 𝒃 𝑵𝒑 = 𝐩 𝐥𝐨𝐠 𝒃 𝑵
L4. 𝐥𝐨𝐠 𝒃 𝒃 = 𝟏
L5.
𝒃𝐥𝐨𝐠𝒃 𝑵 = 𝑵
Aside from the laws stated above, we need to introduce special logarithms that
occur on a very usual basis. Here are the definitions and notations of the logarithms:
common logarithm: log 𝑥 = log10 𝑥
natural logarithm: ln 𝑥 = log 𝑒 𝑥
91
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2. Differentiation of Logarithmic Functions. The formulas given below provide
rules for finding the derivatives of logarithmic functions. In these formulas, 𝑢 is a
function of 𝑥.
DIFFERENTIATION OF LOGARITHMIC FUNCTIONS
𝒅
𝟏
𝒅𝒖
(𝐥𝐨𝐠 𝒃 𝒖) = (𝐥𝐨𝐠 𝒃 𝒆)
𝒅𝒙
𝒖
𝒅𝒙
𝒅
𝟏 𝒅𝒖
(𝐥𝐧 𝒖) =
𝒅𝒙
𝒖 𝒅𝒙
To use these formulas, study the following examples for differentiating logarithmic
functions.
𝑑𝑦
Example 3c.1: Find 𝑑𝑥 if 𝑦 = log 5 (4𝑥 + 3)
Solution: Let 𝑢 = 4𝑥 + 3
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
1
𝑑
= 4𝑥+3 (log 5 𝑒) [𝑑𝑥 (4𝑥 + 3)]
1
= 4𝑥+3 (log 5 𝑒)(4)
=
4 log5 𝑒
4𝑥+3
Example 3c.2: Find
𝑑𝑦
𝑑𝑥
if 𝑦 = ln (2𝑥 + 1)4
Solution: Let 𝑢 = (2𝑥 + 1)4
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
1
(2𝑥+1)4
𝑑
[𝑑𝑥 (2𝑥 + 1)4 ]
1
𝑑
= (2𝑥+1)4 (4)(2𝑥 + 1)4−1 [𝑑𝑥 (2𝑥 + 1)]
1
= (2𝑥+1)4 (4)(2𝑥 + 1)3 (2)
1
= (2𝑥+1)4 (8)(2𝑥 + 1)3
8
= 2𝑥+1
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Alternative Solution for Example 3c.2:
𝑦 = ln (2𝑥 + 1)4
𝑦 = 4 ln (2𝑥 + 1)
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑
=4∙
𝑑𝑥
use Law L3 to simplify the equation
[ ln (2𝑥 + 1)]
1
𝑑
= 4 (2𝑥+1) [𝑑𝑥 (2𝑥 + 1)]
1
= 4 (2𝑥+1) (2)
8
= 2𝑥+1
➔ You can choose to do either method since it will still result to a same answer.
𝑥+4
Example 3c.3: Find 𝑦′ if 𝑦 = ln √𝑥−4
Solution: We can apply Law L3 to simplify the equation
1
𝑦=
𝑥+4 2
ln (𝑥−4)
1
𝑥+4
𝑦 = 2 ln (𝑥−4)
1
𝑦 = 2 [ln(𝑥 + 4) − ln(𝑥 − 4)]
1
1
𝑑
1
use Law L2
𝑑
𝑦′ = 2 {𝑥+4 [𝑑𝑥 (𝑥 + 4)] − 𝑥−4 [𝑑𝑥 (𝑥 − 4)]}
𝑦′ =
1
2
[
1
𝑥+4
1
(1) −
1
1
𝑥−4
(1)]
1
𝑦′ = 2 (𝑥+4 − 𝑥−4)
1
(𝑥−4)−(𝑥+4)
𝑦 ′ = 2 [ (𝑥+4)(𝑥−4) ]
1
𝑥−4−𝑥−4
𝑦′ = 2 [(𝑥+4)(𝑥−4)]
1
−8
𝑦′ = 2 (𝑥 2 −16)
−4
𝑦′ = 𝑥 2 −16
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2
Example 3c.4: Find 𝑦′ if 𝑦 =
𝐴𝑟𝑐𝑐𝑜𝑡 (
√3
2𝑥+1
√3
)
Solution:
𝑦′ =
2
−1
[
𝑑
2
√3 1+(2𝑥+1)
2𝑥+1
] [𝑑𝑥 (
)]
√3
√3
𝑦′ = −
𝑦′ = −
2
√3
2
√3
[
1
1
(2𝑥+1)2
1+
2
(√3)
] [( )
√3 𝑑𝑥
1
(
(2𝑥 + 1)]
2
)( )
4𝑥2 +4𝑥+1
1+
3
4
𝑑
√3
1
𝑦 ′ = − 3 ( 3+4𝑥2 +4𝑥+1 )
3
4
3
𝑦 ′ = − 3 (4𝑥 2+4𝑥+4)
−1
𝑦 ′ = 𝑥 2 +𝑥+1
𝑥
Example 3c.5: Find 𝑦′ if 𝑦 = ln (tan 2)
Solution: This problem involves a trigonometric function, thus recall the derivative of the
trigonometric function.
𝑦′ =
1
𝑑
tan
𝑥
2
𝑥
[𝑑𝑥 (tan 2)]
𝑥
𝑥
𝑑
𝑥
𝑦 ′ = cot 2 (sec 2 2) [𝑑𝑥 (2)]
𝑥
𝑥
1
𝑥
𝑥
1
note:
1
tan u
= cot u
𝑑
𝑦 ′ = cot 2 (sec 2 2) [(2) 𝑑𝑥 (𝑥)]
𝑦 ′ = cot 2 (sec 2 2) [(2) (1)]
1
𝑥
𝑥
𝑦 ′ = 2 (cot 2) (sec 2 2)
simplify all trig functions by using sine and cosine
𝑥
′
𝑦 =
1 cos2
1
( 𝑥 ) ( 2 𝑥)
2 sin
cos
2
1
𝑦′ = 2 (
2
1
𝑥
2
sin cos
𝑥
2
)=
1
𝑥
2
2 sin cos
𝑥
𝑥
𝑥
2
2
2
use double-angle formula: 2 sin cos = 𝑠𝑖𝑛 [2 ( )]
𝑥
2
1
𝑦 ′ = sin 𝑥 = csc 𝑥
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3. Logarithmic Differentiation. This technique is used when the function to be
derived is expressed as a product, quotient, power, or root of two (2) or more
differentiable functions of 𝑥. This process is done by the following steps:
i. Take the natural logarithm (ln) of both sides of the equations which defines
the function.
ii. Make use of the properties and laws of logarithms to simplify the right-side
member of the equation.
𝑑𝑦
iii. Differentiate both sides of the equation with respect to 𝑥 and solve for 𝑑𝑥 .
𝒅𝒚
Example 3c.6: Find 𝒅𝒙 if 𝒚 = (𝟐𝒙 + 𝟏)√𝟑𝒙 + 𝟓, using logarithmic differentiation.
Solution: The equation given is a product of two differentiable functions of x, thus we will use
logarithmic differentiation to find 𝑑𝑦/𝑑𝑥.
𝒍𝒏 𝒚 = 𝒍𝒏 (𝟐𝒙 + 𝟏)√𝟑𝒙 + 𝟓
take 𝒍𝒏 of both sides
𝒍𝒏 𝒚 = 𝒍𝒏 (𝟐𝒙 + 𝟏) + 𝒍𝒏 √𝟑𝒙 + 𝟓
use Law L1
𝟏
𝒍𝒏 𝒚 = 𝒍𝒏 (𝟐𝒙 + 𝟏) + 𝟐 𝒍𝒏 (𝟑𝒙 + 𝟓)
𝒅
𝒅𝒙
𝒅
(𝒍𝒏 𝒚) =
𝟏 𝒅𝒚
𝒅𝒙
𝟏
use Law L3
𝟏
[𝒍𝒏 (𝟐𝒙 + 𝟏) + 𝟐 𝒍𝒏 (𝟑𝒙 + 𝟓)]
𝒅
𝟏
𝟏
𝒅
( ) = (𝟐𝒙+𝟏) [𝒅𝒙 (𝟐𝒙 + 𝟏)] + (𝟐) (𝟑𝒙+𝟓) [𝒅𝒙 (𝟑𝒙 + 𝟏)]
𝒚 𝒅𝒙
𝟏 𝒅𝒚
𝟏
𝟏
𝟏
( ) = (𝟐𝒙+𝟏) (𝟐) + (𝟐) (𝟑𝒙+𝟓) (𝟑)
𝒚 𝒅𝒙
𝟏 𝒅𝒚
𝟐
𝟏
𝟑
( ) = 𝟐𝒙+𝟏 + (𝟐) (𝟑𝒙+𝟓)
𝒚 𝒅𝒙
𝒅𝒚
𝒅𝒙
𝒅𝒚
𝒅𝒙
𝟐
= [(𝟐𝒙 + 𝟏)√𝟑𝒙 + 𝟓] [
𝒅𝒚
=
𝒅𝒙
𝒅𝒚
𝒅𝒙
𝟑
= 𝒚 [𝟐𝒙+𝟏 + 𝟐(𝟑𝒙+𝟓)]
=
transpose 𝑦 to the right side of the equal sign
𝟐(𝟐)(𝟑𝒙+𝟓)+(𝟑)(𝟐𝒙+𝟏)
𝟐(𝟐𝒙+𝟏)(𝟑𝒙+𝟓)
[(𝟐𝒙+𝟏)√𝟑𝒙+𝟓] (𝟏𝟐𝒙+𝟐𝟎+𝟔𝒙+𝟑)
]
cancel similar terms from numerator and denominator
𝟐(𝟐𝒙+𝟏)(𝟑𝒙+𝟓)
𝟏𝟖𝒙+𝟐𝟑
𝟐√𝟑𝒙+𝟓
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Example 3c.7: Find 𝒚′ if 𝒚 = 𝒙𝒙 using logarithmic differentiation.
Solution: The equation given is a power of two differentiable functions of x so we can use
logarithmic differentiation here.
𝐥𝐧 𝒚 = 𝐥𝐧 𝒙𝒙
Take natural logarithms (𝑙𝑛) of both sides
𝐥𝐧 𝒚 = 𝒙 𝐥𝐧 𝒙
use L3
𝒅
𝒅𝒙
(𝐥𝐧 𝒚) =
𝟏 𝒅
𝒚
𝒅
𝒅𝒙
(𝒙 𝐥𝐧 𝒙)
differentiate both sides
𝟏
(𝒚) = 𝒙 ( ) (𝟏) + 𝐥𝐧 𝒙 (𝟏)
𝒅𝒙
𝒙
𝒅𝒚
𝒅𝒙
𝒅𝒚
𝒅𝒙
use product rule for
𝑑
𝑑𝑥
(𝑥 ln 𝑥)
= 𝒚(𝟏 + 𝐥𝐧 𝒙 )
simplify and transpose 𝑦 to the right side of “=”
= 𝒙𝒙 (𝟏 + 𝐥𝐧 𝒙 )
substitute the value of 𝑦 which is 𝑥 𝑥
By using the logarithmic differentiation, you can prove the general power formula
𝑑𝑢
= 𝑛𝑢𝑛−1 𝑑𝑥 . Do it on your own and you will see how amazing it is that differentiation
𝑑𝑥
formulas are connected.
𝑑
(𝑢𝑛 )
4. Differentiation of Exponential Functions. The given formulas below are used to
derive another type of transcendental function, the exponential function.
DIFFERENTIATION OF EXPONENTIAL FUNCTIONS
𝒅 𝒖
𝒅𝒖
(𝒂 ) = 𝒂𝒖 (𝐥𝐧 𝒂)
𝒅𝒙
𝒅𝒙
𝒅 𝒖
𝒅𝒖
(𝒆 ) = 𝒆𝒖
𝒅𝒙
𝒅𝒙
Take note that the variable 𝑢 in the formulas of the above table is a differentiable
function of x. See the examples on the next page to deeply understand the point.
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Example 3c.8: Find 𝒚′ if 𝒚 = 𝟒𝟐𝒙 .
Solution: The base of the given function is a constant while its power has the variable 𝑥. This
means we need to use the differentiation of exponential function.
𝒚 = 𝟒𝟐𝒙
where 𝑢 = 2𝑥
𝒅
𝒚′ = 𝟒𝟐𝒙 (𝐥𝐧 𝟒) (𝒅𝒙 𝟐𝒙)
𝒚′ = 𝟒𝟐𝒙 (𝐥𝐧 𝟒)(𝟐)
𝒚′ = 𝟒𝟐𝒙 (𝟐 𝐥𝐧 𝟒)
𝒚′ = 𝟒𝟐𝒙 (𝐥𝐧 𝟒𝟐 )
use L3
𝒚′ = 𝟒𝟐𝒙 (𝐥𝐧 𝟏𝟔)
𝟑
Example 3c.9: Find 𝒚′ if 𝒚 = 𝒆−𝒙 .
Solution:
𝟑
𝒅
𝒚′ = 𝒆−𝒙 [𝒅𝒙 (−𝒙𝟑 )]
𝒅
𝟑
𝒚′ = 𝒆−𝒙 [(−𝟑)(𝒙𝟑−𝟏 ) (𝒅𝒙 𝒙)]
𝟑
𝒚′ = 𝒆−𝒙 [(−𝟑)(𝒙𝟑−𝟏 )(𝟏)]
𝟑
𝒚′ = 𝒆−𝒙 (−𝟑𝒙𝟐 )
𝒚′ = −𝟑𝒙𝟐 𝒆−𝒙
𝟑
Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Lax, P. D., & Terell, M. S. (2014). Calculus with Applications. New York City: Springer.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
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Let’s Check
Activity 1. Find the derivative of the following functions and simply it whenever possible.
1. 𝑦 = 34𝑥
6. 𝑦 = ln(ln sec 𝑥)
2. 𝑦 = log √2𝑥 + 5
7. 𝑦 = 𝑥 𝑒
𝑥
8. 𝑦 = ln(𝑥 + 3)4
𝑥
2
3. 𝑦 = 1+
1− 2𝑥
4. 𝑦 = log (sin2 4𝑥)
9. 𝑦 = 𝑒 −4𝑥
5. 𝑦 = 4𝑥 ln 4𝑥
10. 𝑦 = 𝑥 4 (1 − ln 𝑥 4 )
Let’s Analyze
Activity 1. Differentiate the following functions using logarithmic differentiation.
2
1
1. 𝑦 = (2𝑥 + 1)3 (3𝑥 − 4)2
2. 𝑦 =
(𝑥−2)3 √𝑥+1
(𝑥−3)4
3
(4𝑥−1)(𝑥+2)
3. 𝑦 = √
(𝑥−5)2
Activity 2. Find the derivative of the following functions and simplify whenever possible.
𝑥
1.
𝑦 = 𝑥2
6.
𝑦 = ln(𝑥 + √𝑥 2 + 1)
2.
𝑦 = 𝑒 𝑥 𝑒 ln 𝑥
7.
𝑦 = ln2 (𝑥 + 3)
3.
𝑦 = ln(𝑒 𝑥 𝑥 2 )
8. 𝑦 = ln (1+sin 𝑥)
4. 𝑦 = ln
1−sin 𝑥 3
𝑒 2𝑥 −1
9. 𝑦 = 𝑥 𝐴𝑟𝑐𝑡𝑎𝑛 𝑥 − ln √1 + 𝑥 2
𝑒 2𝑥 +1
5. 𝑒 𝑥+𝑦 = ln
10. 𝑦 = ln(sec 𝑥 + tan 𝑥)
𝑥
𝑦
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In a Nutshell
You need to take note of these points when dealing with the differentiation of
logarithmic and exponential functions.
•
•
•
•
The graph of a logarithmic function is always on the positive side of the y-axis.
The graph of an exponential function never crosses the x-axis and always
intersects the y-axis at 𝑦 = 1.
There are properties and laws of logarithms and exponents that will help simplify
the functions process of differentiation.
Logarithmic Differentiation is a method used when an equation to be derived is
expressed as a product, quotient, root, or power of two (2) or more differentiable
functions of x.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-3d. Familiarize the properties of the hyperbolic functions and find its derivative
Metalanguage
In this section, you will find terms that can be used when dealing with hyperbolic
functions. You may go back here anytime if you run across a certain term you find
ambiguous.
1. Hyperbolic Functions. These functions are analogs of the standard trigonometric
functions which are defined for the hyperbola rather than on the circle.
1.1 These functions are sometimes called the “hyperbolic trigonometric functions”
since it has many connections between them and the standard trigonometric
functions.
1.2 These functions relate to the hyperbola same as the trigonometric functions
relate to the circle.
Essential Knowledge
In ULO3d, you are expected to familiarize the properties of hyperbolic functions
for its easy derivation. This function is the last transcendental function which will be
discussed.
1. The Hyperbolic Functions. These functions are a combination of certain exponential
functions, 𝑒 𝑥 and 𝑒 −𝑥 . Hyperbolic functions are defined as follows:
DEFINITIONS OF HYPERBOLIC FUNCTIONS
𝑒 𝑥 − 𝑒 −𝑥
sinh 𝑥 =
2
𝑥
𝑒 + 𝑒 −𝑥
cosh 𝑥 =
2
tanh 𝑥 =
cosh 𝑥
sinh 𝑥
1
sech 𝑥 =
cosh 𝑥
coth 𝑥 =
sinh 𝑥
cosh 𝑥
csch 𝑥 =
1
sinh 𝑥
The notation sinh 𝑥 which is pronounced as “shine” can be read “hyperbolic sine
of x”. The others can also be read in the same manner.
Like the trigonometric functions, hyperbolic functions also have its identities which
are illustrated on a table on the next page. These identities can be directly deduced from
the definitions of hyperbolic functions.
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HYPERBOLIC IDENTITIES
H1. cosh2 𝑥 − sinh2 𝑥 = 1
H2. tanh2 𝑥 + sech2 𝑥 = 1
H3. coth2 𝑥 − csch2 𝑥 = 1
H4. sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥
H5. cosh 2𝑥 = cosh2 𝑥 + sinh2 𝑥
= 1 + 2 sinh2 𝑥
= 2 cosh2 𝑥 − 1
The hyperbolic identities would come in handy later as we start differentiating the
hyperbolic functions.
2. Differentiation of Hyperbolic Functions. The rules for differentiating the hyperbolic
functions are given by the following formulas where 𝑢 is a function of 𝑥.
DIFFERENTIATION OF HYPERBOLIC FUNCTIONS
𝒅
𝒅𝒖
(𝒔𝒊𝒏𝒉 𝒖) = 𝒄𝒐𝒔𝒉 𝒖
𝒅𝒙
𝒅𝒙
𝒅
𝒅𝒖
(𝒄𝒐𝒕𝒉 𝒖) = −𝒄𝒔𝒄𝒉𝟐 𝒖
𝒅𝒙
𝒅𝒙
𝒅
𝒅𝒖
(𝒄𝒐𝒔𝒉 𝒖) = 𝒔𝒊𝒏𝒉 𝒖
𝒅𝒙
𝒅𝒙
𝒅
𝒅𝒖
(𝒔𝒆𝒄𝒉 𝒖) = −𝒔𝒆𝒄𝒉 𝒖 𝒕𝒂𝒏𝒉 𝒖
𝒅𝒙
𝒅𝒙
𝒅
𝒅𝒖
(𝒕𝒂𝒏𝒉 𝒖) = 𝒔𝒆𝒄𝒉𝟐 𝒖
𝒅𝒙
𝒅𝒙
𝒅
𝒅𝒖
(𝒄𝒔𝒄𝒉 𝒖) = −𝒄𝒔𝒄𝒉 𝒖 𝒄𝒐𝒕𝒉 𝒖
𝒅𝒙
𝒅𝒙
Study the examples on the succeeding page for you to know how to use the
formulas in differentiating hyperbolic functions.
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𝑑𝑦
Example 3d.1: Find 𝑑𝑥 if 𝑦 = sinh(4𝑥 + 3).
Solution:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑
= cosh(4𝑥 + 3) [𝑑𝑥 (4𝑥 + 3)]
= cosh(4𝑥 + 3)(4)
= 4 cosh(4𝑥 + 3)
Example 3d.2: Find 𝑦’ if 𝑦 = 3 cosh2 4𝑥
Solution:
𝑑
𝑦 ′ = (3)(2)(cosh2−1 4𝑥) [𝑑𝑥 (cosh 4𝑥)]
𝑑
𝑦 ′ = 6 cosh 4𝑥 sinh 4𝑥 (𝑑𝑥 4𝑥)
𝑦 ′ = 24 cosh 4𝑥 sinh 4𝑥
This is not yet the final answer since we can still simplify it using the hyperbolic identity
sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥.
𝑦 ′ = 12 (2 sinh4𝑥 cosh 4𝑥)
where
𝑥 = 4𝑥
𝑦 ′ = 12 sinh 2(4𝑥)
𝑦 ′ = 12 sinh 8𝑥
Example 3d.3: Find 𝑦’ if 𝑦 = sinh(ln 𝑥)
Solution:
𝑑
𝑦 ′ = (cosh ln 𝑥) [𝑑𝑥 (ln 𝑥)]
1
𝑦 ′ = (cosh ln 𝑥) (𝑥)
We can simplify the equation further by applying one of the definitions of hyperbolic function-𝑒 𝑥 + 𝑒 −𝑥
cosh 𝑥 =
2
𝑦′ = (
𝑦′ =
𝑒 ln 𝑥 +𝑒 − ln 𝑥
2
𝑒 ln 𝑥 +𝑒 − ln 𝑥
2𝑥
1
) (𝑥 )
where 𝑥 = ln 𝑥
.
Use the logarithmic identity: 𝑒 ln 𝑥 = 𝑥 and 𝑒 − ln 𝑥 = 𝑒 ln 𝑥
′
𝑦 =
𝑥+
1
𝑥
2𝑥
=
𝑥 2 +1
2𝑥 2
.
102
−1
= 𝑥 −1 =
1
𝑥
to simplify further..
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Feldman, J., & Rechnitzer, A. (2015). Differential Calculus Notes for Mathematics 100.
Terano, H. J. (2015). Calculus 1: A simplified Text in Differential Calculus.
Let’s Check
Activity 1. Find the derivative of the following hyperbolic functions and simplify whenever
possible.
1. 𝑦 = sinh2 5𝑥
2. 𝑦 = 𝐴𝑟𝑐𝑡𝑎𝑛 (sinh 𝑥)
3. 𝑦 = ln(tanh 2𝑥)
4. 𝑦 =
sinh 𝑥
1+cosh 𝑥
5. 𝑦 = tanh 𝑥 2
Let’s Analyze
Activity 1. Differentiate the following functions and simplify if possible.
1
1. 𝑦 = coth 𝑥
2. 𝑦 = csch2 3𝑥
3. 𝑦 = sech2 ln 𝑥
4. 𝑦 = sinh 𝑥 cosh 𝑥 − 𝑥
5. 𝑦 =
𝑥
[cosh(ln 𝑥)
2
+ sinh(ln 𝑥)]
6. 𝑦 = 𝐴𝑟𝑐𝑡𝑎𝑛 (tanh 4𝑥)
7. 𝑦 = 𝑒 𝑥 ln(sinh 𝑥)
8. 𝑦 = cosh2 6𝑥 +
1
cosh 12 𝑥
2
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In a Nutshell
You need to take note of these points when dealing with the differentiation of
hyperbolic functions.
•
•
•
Hyperbolic function is a combination of the exponential functions 𝑒 𝑥 and 𝑒 −𝑥 .
The definitions and identities of hyperbolic functions are necessary to simplify the
equation.
Differentiation of hyperbolic functions have rules, which are expressed in formulas,
that should be followed to avoid using the long method.
Q&A List
If you have any questions regarding continuity, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture
Week 8-9: Unit Learning Outcomes (ULO): At the end of the unit, you are expected
to:
a. Solve Parametric Equations: curve tracing,
b. Solve for Partial derivatives,
c. Learn Approximate limits that produce indeterminate form and use L`Hospitals
Rule to evaluate limits
d. Learn and solve for solutions of equations: Newton’s Method.
Big Picture in Focus:
ULO-4a. Solve Parametric Equations: curve tracing
Metalanguage
Instead of a function y(x) being defined explicitly in terms of the independent
variable x, it is sometimes useful to define both x and y in terms of a third variable, t say,
known as a parameter. In this unit we explain how such functions can be differentiated
using a process known as parametric differentiation.
In order to master the techniques explained here, it is vital that you undertake
plenty of practice exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should
be able to:
• differentiate a function defined parametrically
• find the second derivative of such a function
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes), you need to fully
understand the following essential knowledge that will be laid down in the succeeding
pages. Please note that you are not limited to exclusively refer to these resources. Thus,
you are expected to utilize other books, research articles and other resources that are
available in the university’s library (refer to the Library Contact on page 3).
It is often necessary to find the rate of change of a function defined parametrically; that
𝑑𝑦
is, we want to calculate 𝑑𝑥 . The following example will show how this is achieved.
𝑑𝑦
Example: Suppose we wish to find 𝑑𝑥 when x = cost and y = sint.
We differentiate both x and y with respect to the parameter, t:
𝑑𝑥
𝑑𝑦
= −𝑠𝑖𝑛𝑡
= 𝑐𝑜𝑠𝑡
𝑑𝑡
𝑑𝑡
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From the chain rule we know that,
𝑑𝑦 𝑑𝑦 𝑑𝑥
=
.
𝑑𝑡 𝑑𝑥 𝑑𝑡
So that, by rearrangement
𝑑𝑦
𝑑𝑦 𝑑𝑡
𝑐𝑜𝑠𝑡
=
=
= −𝑐𝑜𝑡 𝑡
𝑑𝑥 𝑑𝑥
−𝑠𝑖𝑛𝑡
𝑑𝑡
𝑑𝑦
EXAMPLE 1: Suppose we wish to find 𝑑𝑥 when 𝑥 = 𝑡 3 − 𝑡 𝑎𝑛𝑑 𝑦 = 4 − 𝑡 2 .
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𝑑𝑦
EXAMPLE 2: Suppose we wish to find 𝑑𝑥 when 𝑥 = 𝑡 3 𝑎𝑛𝑑 𝑦 = 𝑡 2 − 𝑡.
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SECOND DERIVATIVES
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Love, C., & Rainville E., (1962). Differential Calculus Notes for Mathematics 100.
Feliciano F., and Uy, F., (1983). Differential Calculus and Integral Calculus.
e-source
http://www.mathcentre.ac.uk/
Let’s Check
1.
2.
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Let’s Analyze
1.
In a Nutshell
For second derivative,
Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-4b. Solve for Partial derivatives
Metalanguage
Before proceeding to “Essential Knowledge”, you should already master the
principle of differentiating a function of one variable. I believe you have already learned
that in the previous units. To be able to catch up to the lesson, please refer to the following
definition of terms.
A partial derivative of a function of several variables is its derivative with respect to one
of those variables, with the others held constant (as opposed to the total derivative, in
which all variables are allowed to vary). The symbol used to denote partial derivatives is
𝝏.
An independent variable is a variable that stands alone and isn't changed by the other
variables you are trying to measure. For equations such as y = 3x – 2, the independent
variable is x. The variable y is not independent since it depends on the number chosen
for x.
A dependent variable is a variable that depends on one or more other variables. For
equations such as y = 3x – 2, the dependent variable is y.
Essential Knowledge
When a function of more than one independent input variable changes because of
changes in one or more of the input variables, it is important to calculate the change in
the function itself. This can be investigated by holding all but one of the variables constant
and finding the rate of change of the function with respect to the one remaining variable.
This process is called partial differentiation. In this section, we show how to carry out the
process.
FIRST PARTIAL DERIVATIVES
The x partial derivative
For a function of a single variable, y=f(x), changing the independent variable x leads to a
corresponding change in the dependent variable y. The rate of change of y with respect
𝑑𝑓
to x is given by the derivative, written 𝑑𝑥 . A similar situation occurs with functions of more
than one variable. For clarity we shall concentrate on functions of just two variables.
In the relation z=f(x, y) the independent variables are x and y and the dependent
variable z. Now, both variables x and y may change simultaneously inducing a change
in z. However, rather than consider this general situation, to begin with we shall hold one
of the independent variables fixed. This is equivalent to moving along a curve obtained
by intersecting the surface by one of the coordinate planes.
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Consider f(x, y) =x3 + 2x2y+ y2+ 2x+ 1.
Suppose we keep y constant and vary x; then what is the rate of change of the function
f?
Suppose we hold y at the value 3 then
f(x,3) =x3+ 6x2+ 9 + 2x + 1 = x3 + 6x2+ 2x+ 10
In effect, we now have a function of x only. If we differentiate it with respect to x we
obtain the expression:
3x2 + 12x+ 2.
We say that f has been partially differentiated with respect tox. We denote the partial
𝜕𝑓
derivative of f with respect to x by 𝜕𝑥 (to be read as ‘partial dee f by dee x’). In this
example when y= 3:
𝜕𝑓
= 3x2+ 12x+ 2
𝜕𝑥
In the same way if y is held at the value 4 then f(x,4) =x3 + 8x2 + 16 + 2x + 1 = x3 + 8x2 +
2x + 17and so, for this value of y
𝜕𝑓
= 3x2+ 16x+ 2
𝜕𝑥
Now if we return to the original formulation
f(x, y) = x3 + 2x2y + y2 + 2x + 1
and treat y as a constant then the process of partial differentiation with respect to x
gives
𝜕𝑓
= 3x2+ 4xy + 0 + 2
𝜕𝑥
𝜕𝑓
𝜕𝑥
= 3x2+ 4xy + 2
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EXAMPLE:
a.)
Here, we treat y as constant. Therefore,
𝜕𝑓
= 3𝑥 2 + 1 + 0 + 0
𝜕𝑥
𝝏𝒇
= 𝟑𝒙𝟐 + 𝟏
𝝏𝒙
b.)
Variable y is constant here. Therefore,
𝜕𝑓
= 2𝑥(0) + 𝑦(2𝑥) + 𝑥(0) + 𝑦 3 (1)
𝜕𝑥
𝝏𝒇
= 𝟐𝒙𝒚 + 𝒚𝟑
𝝏𝒙
EXAMPLE:
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𝜕𝑓
We can calculate the partial derivative of f with respect to x and the value of 𝜕𝑥 at a
specific point e.g. x= 1, y=−2.
EXAMPLE:
Functions of several variables
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Second partial derivatives
EXAMPLE:
We can use the alternative notation when evaluating derivatives.
EXAMPLE:
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Mixed second derivatives
EXAMPLE 1:
EXAMPLE 2:
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Helm (2008). Workbook 18: Functions of Several Variables
Feliciano F., and Uy, F., (1983). Differential Calculus and Integral Calculus.
Miller, Jeff (2009-06-14). "Earliest Uses of Symbols of Calculus". Earliest Uses of Various Mathematical
Symbols. Retrieved 2009-02-20
Let’s Check
Let’s Analyze
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In a Nutshell
Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-4c. Learn Approximate limits that produce indeterminate form and use
L`Hospitals Rule to evaluate limits
Metalanguage
To be able to catch up to the lesson, please familiarize yourself to the following definition
of terms:
A mathematical expression can also be said to be indeterminate if it is not definitively
or precisely determined. Certain forms of limits are said to be indeterminate when merely
knowing the limiting behavior of individual parts of the expression is not sufficient to
actually determine the overall limit.
An indeterminate form is an expression involving two functions whose limit cannot be
determined solely from the limits of the individual functions. These forms are common in
calculus; indeed, the limit definition of the derivative is the limit of an indeterminate form.
Types if indeterminate forms:
1. 0/0
3. 0x∞
5. 00
7. ∞0
∞
2. ∞/∞
4. ∞ − ∞
6. 1
Essential Knowledge
In the previous unit, you studied limits such as
In those sections, you discovered that direct substitution can produce an indeterminate
form such as 0/0 or ∞/∞. For instance, if you substitute x=1 into the first limit, you
obtain
which tells you nothing about the limit. To find the limit, you can factor and divide out
like factors, as shown.
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For the second limit, direct substitution produces the indeterminate form ∞/∞, which
again tells you nothing about the limit. To evaluate this limit, you can divide the
numerator and denominator by x. Then you can use the fact that the limit of 1/x, as x
approaches ∞ is 0.
Algebraic techniques such as these tend to work well as long as the function itself is
algebraic. To find the limits of other types of functions, such as exponential functions or
trigonometric functions, you generally need to use a different approach.
L’Hôpital’s Rule
L’Hôpital’s Rule, which is named after the French mathematician Guillaume Francois
Antoine de L’Hôpital (1661–1704), describes an analytic approach for evaluating limits.
EXAMPLE 1:
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Because direct substitution produces the indeterminate form 0/0, you can apply
L’Hôpital’s Rule to obtain the same result.
EXAMPLE 2:
Again, because direct substitution produces the indeterminate form 0/0, you can apply
L’Hôpital’s Rule to obtain the same result.
EXAMPLE 3:
Because direct substitution produces the indeterminate form ∞/∞.
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Sometimes it is necessary to apply L’Hôpital’s Rule more than once to remove an
indeterminate form. This is shown in Example 4.
EXAMPLE 4:
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Self-Help: Refer to the sources below to help you further understand the
lesson.
Feliciano F., and Uy, F., (1983). Differential Calculus and Integral Calculus.
https://www.math.ucdavis.edu/~marx/Sec.%208.6.pdf
Let’s Check
Decide whether the limit produces an indeterminate form.
Let’s Analyze
Use L’Hôpital’s Rule to find the limit. You mayneed to use L’Hôpital’s Rule repeatedly.
1.
2.
3.
4.
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5.
6.
7.
8
9.
10.
11.
12.
In a Nutshell
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Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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Big Picture in Focus:
ULO-4d. Learn and solve for solutions of equations: Newton’s Method
Metalanguage
To be able to catch up to the lesson, please familiarize yourself to the following definition
of terms:
Newton's Method (also called the Newton-Raphson method) is a recursive algorithm for
approximating the root of a differentiable function.
Iteration is the repetition of a mathematical or computational procedure applied to the
result of a previous application, typically as a means of obtaining successively closer
approximations to the solution of a problem.
Essential Knowledge
The next application that we’ll take a look at in this unit is an important application that is
used in many areas. If you’ve been following along to this point it’s quite possible that
you’ve gotten the impression that many of the applications that we’ve looked at are just
made up by us to make you work. This is unfortunate because all of the applications that
we’ve looked at to this point are real applications that really are used in real situations.
The problem is often that in order to work more meaningful examples of the applications
we would need more knowledge than we generally have about the science and/or physics
behind the problem. Without that knowledge we’re stuck doing some fairly simplistic
examples that often don’t seem very realistic at all and that makes it hard to understand
that the application we’re looking at is a real application.
That is going to change in this section. This is an application that we can all understand,
and we can all understand needs to be done on occasion even if we don’t understand the
physics/science behind an actual application.
In this section, we are going to look at a method for approximating solutions to equations.
We all know that equations need to be solved on occasion and in fact we’ve solved quite
a few equations ourselves to this point. In all the examples we’ve looked at to this point
we were able to actually find the solutions, but it’s not always possible to do that exactly
and/or do the work by hand. That is where this application comes into play. So, let’s see
what this application is all about.
Let’s suppose that we want to approximate the solution to 𝑓(𝑥) = 0 and let’s also suppose
that we have somehow found an initial approximation to this solution say, 𝑥0 . This initial
approximation is probably not all that good, in fact it may be nothing more than a quick
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guess we made, and so we’d like to find a better approximation. This is easy enough to
do. First, we will get the tangent line to 𝑓(𝑥) at 𝑥0 .
Now, take a look at the graph below.
The blue line is the tangent line at 𝑥0 . We can see that this line will cross the x-axis much
closer to the actual solution to the equation than 𝑥0 does. Let’s call this point where the
tangent at 𝑥0 crosses the x-axis 𝑥1 and we’ll use this point as our new approximation to
the solution.
So, how do we find this point? Well we know it’s coordinates, (𝑥1 , 0), and we know that
it’s on the tangent line so plug this point into the tangent line and solve for 𝑥1 as follows,
0 = 𝑓(𝑥0 ) + 𝑓′(𝑥0 )(𝑥1 − 𝑥0 )
𝑓(𝑥0 )
𝑥1 − 𝑥0 = −
𝑓′(𝑥0 )
𝑓(𝑥0 )
𝑥1 = 𝑥0 −
𝑓′(𝑥0 )
So, we can find the new approximation provided the derivative isn’t zero at the original
approximation.
Now we repeat the whole process to find an even better approximation. We form up the
tangent line to f(x) at 𝑥1 and use its root, which we’ll call 𝑥2 , as a new approximation to
the actual solution. If we do this, we will arrive at the following formula.
𝑓(𝑥1 )
𝑥2 = 𝑥1 −
𝑓′(𝑥1 )
This point is also shown on the graph above and we can see from this graph that if we
continue following this process will get a sequence of numbers that are getting very close
the actual solution. This process is called Newton’s Method.
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Newton’s Method
This should lead to the question of when do we stop? How many times do we go through
this process? One of the more common stopping points in the process is to continue until
two successive approximations agree to a given number of decimal places.
Before working any examples, we should address two issues. First, we really do need to
be solving f(x)=0 in order for Newton’s Method to be applied. This isn’t really all that much
of an issue, but we do need to make sure that the equation is in this form prior to using
the method.
Secondly, we do need to somehow get our hands on an initial approximation to the
solution (i.e. we need 𝑥0 somehow). One of the more common ways of getting our hands
on 𝑥0 is to sketch the graph of the function and use that to get an estimate of the solution
which we then use as 𝑥0 . Another common method is if we know that there is a solution
to a function in an interval then we can use the midpoint of the interval as 𝑥0 .
Let’s work an example of Newton’s Method.
EXAMPLE 1: Use Newton’s Method to determine x2 for f(x)=x3−7x2+8x−3 if x0=5.
𝑓(𝑥𝑛 )
𝑥𝑛+1 = 𝑥𝑛 −
𝑓′(𝑥𝑛 )
So all we need to do is run through this twice.
Here is the derivative of the function since we’ll need that.
𝑓’(𝑥) = 3𝑥 2 − 14𝑥 + 8
The first iteration through the formula for x1 is,
𝑓(𝑥0 )
𝑓(5)
−13
𝑥1 = 𝑥0 −
=5 −
=5 −
= 6
𝑓′(𝑥0 )
𝑓′(5)
13
The second iteration through the formula for x2 is,
𝑓(𝑥1 )
𝑓(6)
9
𝑥2 = 𝑥1 −
=6 −
=6 −
= 5.71875
𝑓′(𝑥1 )
𝑓′(6)
32
So, the answer for this problem is 𝟓. 𝟕𝟏𝟖𝟕𝟓.
EXAMPLE 2: Use Newton’s Method to determine x2 for f(x) = xcos(x) − x2 if x0 = 1.
𝑓(𝑥𝑛 )
𝑥𝑛+1 = 𝑥𝑛 −
𝑓′(𝑥𝑛 )
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Here is the derivative of the function since we’ll need that.
𝑓′(𝑥) = 𝑐𝑜𝑠(𝑥) − 𝑥𝑠𝑖𝑛(𝑥) − 2𝑥
The first iteration through the formula for x1 is,
𝑓(𝑥0 )
𝑓(1)
−0.4596976941
𝑥1 = 𝑥0 −
=1 −
=1 −
= 0.8002329432
𝑓′(𝑥0 )
𝑓′(1)
−2.301168679
Important!!! Don’t forget that for us angles are always in radians so make sure
your calculator is set to compute in radians.
The second iteration through the formula for x2 is,
𝑓(𝑥1 )
𝑓(0.8002329432)
𝑥2 = 𝑥1 −
= 0.8002329432 −
= 0.7440943985
𝑓′(𝑥1 )
𝑓′(0.8002329432)
So, the answer for this problem is 𝟎. 𝟕𝟒𝟒𝟎𝟗𝟒𝟑𝟗𝟖𝟓.
EXAMPLE 3: Use Newton’s Method to determine an approximation to the solution to
𝑐𝑜𝑠𝑥 = 𝑥 that lies in the interval [0,2]. Find the approximation to six decimal places.
First note that we weren’t given an initial guess. We were however, given an interval in
which to look. We will use this to get our initial guess. As noted above the general rule of
thumb in these cases is to take the initial approximation to be the midpoint of the interval.
So, we’ll use 𝑥0 =1 as our initial guess.
Next, recall that we must have the function in the form 𝑓(𝑥) = 0. Therefore, we first
rewrite the equation as,
𝑐𝑜𝑠𝑥 − 𝑥 = 0
We can now write down the general formula for Newton’s Method. Doing this will often
simplify up the work a little so it’s generally not a bad idea to do this.
Let’s now get the first approximation.
At this point we should point out that the phrase “six decimal places” does not mean just
get 𝑥1 to six decimal places and then stop. Instead it means that we continue until two
successive approximations agree to six decimal places.
Given that stopping condition we clearly need to go at least one step farther.
Alright, we’re making progress. We’ve got the approximation to 1 decimal place. Let’s
do another one, leaving the details of the computation to you.
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We’ve got it to three decimal places. We’ll need another one.
And now we’ve got two approximations that agree to 9 decimal places and so we can
stop. We will assume that the solution is approximately 𝒙𝟒 = 𝟎. 𝟕𝟑𝟗𝟎𝟖𝟓𝟏𝟑𝟑𝟐.
Self-Help: Refer to the sources below to help you further understand the
lesson.
Dawkins, P., Paul's Online Notes. Copyright 2003-2020
https://tutorial.math.lamar.edu/Solutions/CalcI/NewtonsMethod/Prob5.aspx
Let’s Check
For problems 1 – 3 use Newton’s Method to determine x2 for the given function and
given value of x0.
For problems 4 – 8 use Newton’s Method to find the root of the given equation, accurate
to six decimal places, that lies in the given interval.
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Let’s Analyze
For problems 9 – 12 use Newton’s Method to find all the roots of the given equation
accurate to six decimal places.
13. Suppose that we want to find the root to x3−7x2+8x−3=0. Is it possible to use x0=4
as the initial point? What can you conclude about using Newton’s Method to
approximate roots from this example?
In a Nutshell
The approximation formula is given by,
𝑓(𝑥𝑛 )
𝑓′(𝑥𝑛 )
Where n is the no. of iterations. We need to be a little careful with Newton’s method. It
will usually quickly find an approximation to an equation. However, there are times when
it will take a lot of work or when it won’t work at all.
𝑥𝑛+1 = 𝑥𝑛 −
Q&A List
If you have any questions regarding this topic, kindly write down on the table
provided.
QUESTIONS
ANSWERS
1.
2.
3.
4.
5.
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