SENIOR HIGH SCHOOL
BASIC CALCULUS
QUARTER 3 - Module 1
Lessons 1-9
(DO_Q3_BASICCALCULUS_MODULE1_LESSONS1-9)
i
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
RESOURCE TITLE: Basic Calculus
Alternative Delivery Mode
Quarter 3 – Weeks 1-9
Second Edition, 2022
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of
the Government of the Philippines. However, prior approval of the government agency or office
wherein the work is created shall be necessary for exploitation of such work for profit. Such
agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names,
trademarks, etc.) included in this module are owned by their respective copyright holders.
Every effort has been exerted to locate and seek permission to use these materials from their
respective copyright owners. The publisher and authors do not represent nor claim ownership
over them.
Published by the Department of Education
Secretary: Sara Z. Duterte-Carpio
Development Team of the Module
Writers:
SETIEL P. GO; JOHN ARVIE B. CARINAN
Reviewer: REBECCA M. BIÑAS
Content Editor: SILVERIO M. AGUSTIN
Language Editor: MARICION T. RUMBAUA-SABUG, ED.D.
Illustrator: NATHANIEL D.C. DEL MUNDO
Layout Artist: RAPHAEL A. LOPEZ
Management Team:
MELITON P. ZURBANO, Schools Division Superintendent
FILMORE R. CABALLERO, CID Chief
MELVIN WILLY B. ROQUE, PSDS, OIC LRMS
EDNA LLANERA, Ed. D – Division SHS Focal Person
MARILYN B. SORIANO- EPS in Mathematics
Printed in the Philippines by ________________________
Department of Education – National Capital Region – SDO VALENZUELA
Office Address:
Telefax:
E-mail Address:
Pio Valenzuela St., Marulas, Valenzuela City
(02) 292 – 3247
sdovalenzuela@deped.gov.ph
ii
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
SENIOR HIGH
BASIC
CALCULUS
(LEARNING
AREA)
QUARTER
- Module 1
(QUARTER3 NUMBER)
Lesson
1:
(MODULE
NUMBER)
The Limit and Continuity of
a Function
iii
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners, can
continue your studies and learn while at home. Activities, questions, directions,
exercises, and discussions are carefully stated for you to understand each lesson.
Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you.
Pre-tests are provided to measure your prior knowledge on lessons in each
SLM. This will tell you if you need to proceed on completing this module or if you
need to ask your facilitator or your teacher’s assistance for better understanding of
the lesson. At the end of each module, you need to answer the post-test to self-check
your learning. Answer keys are provided for each activity and test. We trust that
you will be honest in using these.
In addition to the material in the main text. Notes to the Teacher are also
provided to our facilitators and parents for strategies and reminders on how they can
best help you on your home-based learning.
Please use this module with care. Do not Put Unnecessary marks on any part
of this SLM. Use a separate sheet of paper in answering the exercises and tests. And
read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the
tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
iv
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Targets:
1. Illustrate the limit of a function using a table of values and the graph
of the function (STEM_BC11LC-IIIa-1);
2. Distinguish between
and
(STEM_BC11LC-IIIa-2);
3. Illustrate the limit laws (STEM_BC11LC-IIIa-3) ; and
4. Apply the limit laws in evaluating the limit of algebraic functions
(polynomial, rational and radical) (STEM_BC11LC-IIIa-4).
What I Know
Directions: Choose the best answer in the following question base on the
idea about limits and continuity of a functions.
1−x2
1. How should the function f(x) = 1+x be defined so that it will be continuous
at 𝑥 = −1?
a. f(−1) = ∞
b. f(−1) = −2
c. f(−1) = 0
d. f(−1) = 2
2. Evaluate the lim
𝑥→2
4𝑥 2 −8𝑥
𝑥−2
a. 4
b. 8
c. 2
d. undefined
2x + 5 if x ≤ 2
3. Find k so the function f(x) = {
can be defined so that it will
kx + 1 if x > 2
be a continuous function.
a. 8
b. 4
c. 2
d. 0
4. What is the average velocity of 𝑥 over [2, 2 + h]?
2
a. 4 + h
b. 2h
c. 4h + ℎ2
d.
4+4ℎ+ℎ2
ℎ
5. A cliff diver plunges 42 m into the crashing Pacific, landing in a 3-metre
deep inlet. The position of the diver at any time t is given by 𝑠(𝑡) = −4.9𝑡 2 +
42. What is the average velocity of the diver over the interval [0, 2]?
a. 19.6 m/s
b. -9.8 m/s
c. -11.2m/s
1
d. 9.8 m/s
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Lesson The Limit and Continuity of a
1
Function
The study of different types of functions, limits associated with these functions
and how these functions change, together with the ability to graphically illustrate
basic concepts associated with these functions, is fundamental to the understanding
of calculus. These important issues are presented along with the development of
some additional elementary concepts which will aid in our later studies of more
advanced concepts. In this module and throughout this text be aware that definitions
and their consequences are the keys to success for the understanding of calculus
and its many applications and extensions.
The idea of the limit of a function is what connects algebra and geometry
to the mathematics of calculus. In working with the limit of function, we
encounter notation of the form
(DO_Q3_BasicCalculus
_Lesson1)
This is read as “the limit of
as approaches c equals
the number N.”
Here is a function defined on some open interval containing the number c;
needs not be
defined at c, however.
We may describe the meaning of
as follows:
For all x approximately equal to c, with
, the corresponding value of
is approximately equal to N.
Another description of
is
As x gets closer to c, but remains unequal to c, the corresponding value
of
gets closer to N.
Example 1: Finding a Limit Using a Table
When choosing the values of in a table, the number to start with and the
subsequent entries are arbitrary. However, the entries should be chosen so
2
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
that the table makes clear what the corresponding values of
close to.
are getting
Find: 𝐥𝐢𝐦(𝟓𝒙𝟐 )
𝒙→𝟑
Solution: Here
and
. We choose values of
close to 3,
arbitrarily starting with 2.99. Then, we select additional numbers that get
closer to 3, but remain less than 3. Next, we choose values of greater than
3, starting with 3.01, that gets closer to 3. Finally, we evaluate at each
choice to obtain Table 1.
2.99
2.999
𝑥
44.701
44.97
𝑓(𝑥)
From Table 1, we infer that as
closer to 45. That is,
2.9999
3.0001
3.001
44.997
45.003
45.30
gets closer to 3 the value of
3.01
45.301
gets
You can visit the website below for more detailed explanation:
https://www.youtube.com/watch?v=RzDKt.
Example 2: Finding a Limit Using a Table (Trigonometric Functions)
Find:
Solution:
First, we observe the domain of the function
Table 4, where is measured in radians.
is
-0.03
-0.02
-0.01
0.01
𝑥
0.99985 0.99993 0.99998 0.99998
𝑓(𝑥)
We infer from the table above that
= 1.
Distinguish between
The graph of a function
figure below.
and
0.02
0.99993
0.03
0.99985
.
can also be of help in finding limits. Refer to the
In each graph, notice that, as
the number
. We create
gets closer to , the value of
. We conclude that
gets closer to
. This is the conclusion
3
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
regardless of the value of at . In Figure 1(a),
, and in Figure 1(b),
. Figure 1(c) illustrated that
, even if is not defined at .
Algebra Techniques for Finding Limits: Laws of Limits
We mentioned in the previous section that algebra can sometimes be used to
find the exact value of a limit. This is accomplished by developing two
formulas involving limits and several properties of limits.
Two Formula: lim 𝑏 as 𝑥 → 𝑐.
𝑥→𝑐
Limit of a Constant
For the constant function 𝑏
lim 𝑏 = 𝑏,
𝑥→𝑐
where 𝑐 is any number.
Limit of 𝒙
For the identity function 𝑓(𝑥) = 𝑥,
lim 𝑓(𝑥) = lim 𝑥 = 𝑐,
where 𝑐 is any number.
𝑥→𝑐
𝑥→𝑐
Example 3:
Formula (1) in words means, the limit of a constant is the constant, while
the formula (2) states that the limit of as approaches is .
(a)
(b)
(c)
(d)
Find the Limit of Sum, a Difference, and a Product
In the following properties, we assume that
are two functions for
which both
and
exist.
LIMIT OF A SUM
lim[𝑓(𝑥) + 𝑔(𝑥)] = lim 𝑓(𝑥) + lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
Formula in words means that the limit of the sum of two functions equals
the sum of their limits.
Example 4: Find:
Solution: The limit we seek is the sum of two functions
From formulas (1) and (2), we know that
4
.
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
and
from formula (3), it follows that
.
LIMIT OF A DIFFERENCE
[𝑓(𝑥)
lim
− 𝑔(𝑥)] = lim 𝑓(𝑥) − lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
Formula in words means that the limit of the difference of two functions equals the
difference of their limits.
Example 5: Find:
Solution: The limit we seek is the difference of two functions
. From formulas (1) and (2), we know that
and
from formula above, it follows that
LIMIT OF A PRODUCT
lim[𝑓(𝑥) ∙ 𝑔(𝑥)] = [lim 𝑓(𝑥)][lim 𝑔(𝑥)]
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
The limit of the product of two functions equals the product of their
limits.
Example 6: Find:
Solution: The limit we seek is the product of two functions
. From formulas (1) and (2), we know that
and
(DO_Q3_BasicCalculus _Lesson1)
from formula above, it follows that
LIMIT OF A MONOMIAL
If 𝑛 ≥ 1 is a positive integer and 𝑎 is a constant,
lim(𝑎𝑥 𝑛 ) = 𝑎𝑐 𝑛
for any number 𝑐.
𝑥→𝑐
5
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Example 7: Find:
Solution:
LIMIT OF A POLYNOMIAL
If 𝑃 is a polynomial function, then
lim 𝑃(𝑥) = 𝑃(𝑐)
𝑥→𝑐
for any number 𝑐.
Formula means that the limit of a polynomial function as 𝑥 approaches 𝑐, all
we need to do is evaluate the polynomial at 𝑐.
Example 8: Find:
Solution:
LIMIT OF A POWER OR ROOT
𝑓(𝑥) exist and if 𝑛 ≥ 2 is a positive integer,
If lim
𝑥→𝑐
lim[𝑓(𝑥)]𝑛 = [lim 𝑃(𝑥)]𝑛
𝑥→𝑐
and
𝑥→𝑐
lim 𝑛√𝑓(𝑥) = 𝑛√lim 𝑓(𝑥).
𝑥→𝑐
𝑥→𝑐
In the formula above, we require that both 𝑛√𝑓(𝑥) and 𝑛√lim 𝑓(𝑥) be defined.
𝑥→𝑐
Example 9:
Find: (a)
Solution:
(b)
(c)
(a)
(b)
(c)
.
LIMIT OF A QUOTIENT
lim 𝑓(𝑥)
𝑓(𝑥)
lim [
] = 𝑥→𝑐
𝑋→𝐶 𝑔(𝑥)
lim 𝑔(𝑥)
𝑥→𝑐
provided that lim 𝑔(𝑥) ≠ 0.
𝑥→𝑐
6
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Example 10:
Find:
Solution: The Limit we seek involves the quotient of two functions:
and
First, we find the limit of the denominator
.
.
Since the limit of the denominator is not zero, we can proceed to use
formula (10).
.
I. Directions: Use the concepts of finding the limit using a table, find the
limit of the following.
(a)
(b)
II. Directions: Use the Laws of Limits in finding the limits of the following
algebraic terms.
(a)
(b)
In this lesson, I have learned that
_________________________________________________
_________________________________________________
_________________________________________________
7
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Directions: Use the different Laws of Limits to perform the following.
1. Evaluate:
2. Given the function:
Evaluate the following limits, if they exist.
(a)
(b)
Directions: Supply the correct answer.
1. The limit of the product of two functions equal the ______________ of their
limits.
2.
= __________ .
3.
= __________ .
4. True or False: The limit of a polynomial function as approaches 5
equals the value of the polynomial at 𝑥 = 5.
5. True or False: The limit of a rational function at 5 equals the value of the
rational function at 𝑥 = 5.
Evaluate the following limits:
1. lim
𝑥→2
𝑥 2 +4𝑥−12
𝑥 2 −2𝑥
2. lim
2(−3+ℎ)2 −18
ℎ
ℎ→0
8
3. lim
𝑡→4
𝑡−√3𝑡+4
4−𝑡
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
9
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
References:
•
•
•
•
•
•
•
•
•
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines:
Anvil Publishing Inc.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of
America: Pearson Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second
Edition. United States of America: Alpha Books
Teaching Guide for Senior High School Basic Calculus (With permission
to use the concepts and examples)
https://tutorial.math.lamar.edu/Classes /Calc/ComputingLImits.aspx
https://www.analyzemath.com/calculus/limits/find_limits_functions.ht
ml
https://www.onlinemathlearning.com/limits-calculus.html
https://archives.math.utk.edu/visual.casual/1/limits.15/index.html
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondi
rectory/LimitConstant.html
10
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
SENIOR HIGH SCHOOL
BASIC
CALCULUS
(LEARNING
AREA)
(QUARTER3 NUMBER)
QUARTER
- Module 1
(MODULE
NUMBER)
Lesson
2:
The Limits of Exponential,
Logarithmic and
Trigonometric Functions
11
(DO_Q3_BASICCALCULUS_MODULE1_LESSON1)
Targets:
1. Compute the limits of exponential, logarithmic, and trigonometric
functions using tables of values and graphs of the functions
(STEM_BC11LC-IIIb-1); and
2. Illustrate limits involving the expressions
tables of values (STEM_BC11LC-IIIb-2).
and using
What I Know
Directions: Choose the best answer in the following question base on the
idea about limits of trigonometric functions.
𝑠𝑖𝑛𝑥
1. What is the value of lim
a. 2/𝜋
𝑦→𝜋/2 𝑥
?
b. 𝜋/2
c. 1
2. What is the value of the limit 𝑓(𝑥) =
a. 1/√2
b. -1/√2
1−𝑐𝑜𝑠𝑥
3. The value of the limit lim 𝑥 is:
𝑥→0
a. 2
b. 0
4. What is the value of the lim 𝑥𝑠𝑒𝑐𝑥?
𝑠𝑖𝑛2 𝑥+√2𝑠𝑖𝑛𝑥
𝑥 2 −4𝑥
d. 0
if x approaches 0?
c. -1/2√2
d. 1/2√2
c. ∞
d. 1
𝑥→0
a. Not defined b. 1
c. 0
d. -1
sin 3𝑥−3 𝑠𝑖𝑛𝑥
5. If you are going to evaluate the lim (𝜋−𝑥)3 , what is its value?
a. 4
Lesson
2
b. ¼
𝑥→𝜋
c. -1/4
d. -4
The Limits of Exponential,
Logarithmic and Trigonometric
Functions
Real-world situations can be expressed in terms of functional
relationships. These functional relationships are called mathematical models.
In applications of calculus, it is quite important that one can generate these
mathematical models. They sometimes use functions that you encountered in
12
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
pre-calculus, like the exponential, logarithmic, and trigonometric functions.
Hence, we start this lesson by recalling these functions and their
corresponding graphs.
(a) If
, the exponential function with base b is defined by
.
(b) Let
. If
then is called the logarithm of x to the base b,
denoted
.
EVALUATING LIMITS OF EXPONENTIAL FUNCTIONS
First, we consider the natural exponential function
the Euler number, and has value 2.718281….
Example 1: Evaluate the
, where
is called
.
Solution: We will construct the table of values for
. We start by
approaching the number 0 from the left or through the values less than but
close to 0.
SMART TIP:
Using your scientific calculator to be familiar with the natural number ,
locate the symbol of natural number , and input the proper power of in
your calculator.
From
-1
- 0.5
-0.1
-0.01
-0.001
-0.0001
-0.00001
0.36787944117
0.60653065971
0.90483741803
0.99004983374
0.99900049983
0.999900049983
0.99999000005
the
.
table
on
the
left,
Now, we consider approaching 0
from its right or through values
greater than but close to 0.
From the given two tables, as the
values of get close and closer to 0,
the values of
get closer and closer
to 1. So,
. Combining the
two one-sided limits allows us to
conclude that
.
13
1
0.5
0.1
0.01
0.001
0.0001
0.00001
2.71828182846
1.6487212707
1.10517091808
1.01005016708
1.00100050017
1.000100005
1.00001000005
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
We can use the graph of
to determine its limit as
approaches 0. The figure provided is the graph of
We also have the following:
(a)
(b)
(c)
EVALUATING LIMITS OF LOGARITHMIC FUNCTIONS
Now, consider the natural logarithmic function
. Recall that
.
Moreover, it is the inverse of the natural exponential function
.
Example 2: Evaluate
.
Solution: We will construct the table of values for
. We first
approach the number 1 from the left or through values less than
but close to 1.
0.1
0.5
0.9
0.99
0.999
0.9999
0.99999
-2.30258509299
-0.69314718056
-0.10536051565
-0.01005033585
-0.00100050033
-0.000100005
-0.00001000005
From the table below,
as the
values of get close and closer to 1, the
values of
get closer and closer to 0.
In symbols,
.
From the two tables, we can infer that
the
. Now, we consider
approaching 1 from its right or through
values greater than but close to 1.
We now consider the common logarithmic function
.
14
2
1.5
1.1
1.01
1.001
1.0001
1.00001
2.99573227355
0.4054651081
0.09531017989
0.00995933085
0.00099950033
0.000099995
0.00000999995
. Recall the
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
TRIGONOMETRIC FUNCTIONS
Trigonometric functions can be a component of an expression and therefore
subject to a limit process. The question is that “Do you think that the
periodic nature of these functions, and the limited or infinity range of
individual trigonometric functions would make evaluating limits involving
these functions difficult?”
Limits with Trigonometric Functions
The limit rules discussed in week 1 of this module offer some, but not all, of
the tools for evaluating limits involving trigonometric functions.
Example 4: Evaluate
.
Solution: We will construct the table of values for
. We first
approach 0 from the left or through the values less than but close to 0.
-1
-0.5
-0.1
-0.01
-0.001
-0.0001
-0.00001
From the table on the left,
-0.8414709848
-0.4794255386
-0.09983341664
-0.00999983333
-0.00099999983
-0.00009999999
-0.00000999999
we consider approaching 0 from its right or
through values greater that but close to 0.
From the two tables, as the values of get
closer and closer to 1, the values of
get closer and closer to 0. In symbols,
.
We can also find the
Consider the graph of
. Now,
1
0.5
0.1
0.01
0.001
0.0001
0.00001
0.8414709848
0.4794255386
0.09983341664
0.00999983333
0.00099999983
0.00009999999
0.00000999999
by using the graph of the sine function.
.
15
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
The graph validates our observation in Example 4 that
. Also,
using the graph, we have the following:
(a)
(c)
(b)
(d)
Directions: Using the concepts of finding the limit using a table, find the
limit of the following.
(a)
(b)
In this lesson, I have learned that__________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
Directions: Evaluate the following limits by constructing the table of values.
1.
2.
3.
16
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
Directions: Given the graph, evaluate the
following limits:
1.
3.
2.
Directions: Evaluate the following limits by constructing their respective
tables of values.
1.
2.
17
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
18
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
WHAT I CAN DO:
1. 1
2. 2.2
3. 0.5
ASSESSMENT:
WHAT IS IT:
1. 3
2. -1
3. 0
1. 0
2. 0
Answer Key
References:
•
•
•
•
•
•
•
•
•
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines:
Anvil Publishing Inc.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of
America: Pearson Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second
Edition. United States of America: Alpha Books
Teaching Guide for Senior High School Basic Calculus (With permission
to use the concepts and examples)
https://tutorial.math.lamar.edu/Classes /Calc/ComputingLImits.aspx
https://www.analyzemath.com/calculus/limits/find_limits_functions.ht
ml
https://www.onlinemathlearning.com/limits-calculus.html
https://archives.math.utk.edu/visual.casual/1/limits.15/index.html
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondi
rectory/LimitConstant.html
19
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
SENIOR HIGH SCHOOL
(LEARNING
AREA)
BASIC
CALCULUS
(QUARTER NUMBER)
QUARTER 3 - Module 1
(MODULE NUMBER)
Lesson 3:
The Continuity of a
Function
20
(DO_Q3_BASICCALCULUS_MODULE1_LESSON2)
Targets:
1. Illustrate continuity of a function at a number (STEM_BC11LC-IIIe-1);
2. Illustrate whether a function is continuous at a number or not
(STEM_BC11LC-IIIe-2);
3. Illustrate continuity of a function at a function (STEM_BC11LC-IIIe-3);
and
4. Solve problems involving continuity of a function
(STEM_BC11LC-IIIe- 4).
What I Know
Directions: TRUE or FALSE: State whether the given statement is true or
false base on the idea about limits of trigonometric functions.
1. If a function 𝑓 is not defined at 𝑥 = 𝑎 then it is not continuous at 𝑥 =
𝑎.
2. If 𝑓 is a function such that lim 𝑓(𝑥) does not exist then 𝑓 is not
𝑥→0
continuous.
3. All polynomial functions are continuous.
4. If 𝑓(𝑥) is continuous everywhere, then |𝑓(𝑥)| is continuous
everywhere.
5. If the composition f o g is not continuous at x = a, then either g is
not continuous at x = a or f is not continuous at g(a).
Lesson The Continuity of a Function
3
As we have observed in our discussion of limits in Week 1, there are
functions whose limits are not equal to the function value at
, meaning,
.
is NOT NECESSARILY the same as
21
.
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
This leads us to the study of continuity of functions. In this section, we will
be focusing on the continuity of a function at a specific point.
Directions: Use the different laws of limits to perform the following.
1. Evaluate:
2. Given the function:
3. Evaluate the following limits, if they exist.
(a)
(b)
LIMIT AND CONTINUITY AT A POINT
What does “continuity at a point” mean? Intuitively, this means that in
drawing the graph of a function, the point in question will be traversed. We
start by graphically illustrating what it means to be continuity at a point.
Example 1: Consider the graph below.
Solution:
To check if the function is continuous at
, use the given graph. Note that one is
able to trace the graph from the left side of
the number
going to the right side of
, without lifting one’s pen. This is the
case here. Hence, we can say that the
function is continuous at
.
THREE CONDITIONS OF CONTINUITY
A function f(x) is said to be continuous at x = c if the following three
conditions are satisfied:
(i)
exists,
(ii)
exists, and
(iii)
.
If at least one of these conditions is not met, f is said to be
discontinuous at
22
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
Example 1: Determine if 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 2 is continuous or not at 𝑥 = 1 .
Solution: We have to check the three conditions for continuity of a function.
(a) If
(b)
, then
(c)
Therefore,
.
=
is continuous at
Example 2: Determine if 𝑓(𝑥) =
.
.
𝑥 2 −𝑥−2
𝑥−2
is continuous or not at 𝑥 = 0 .
Solution: We have to check the three conditions for continuity of a function.
(a) If
, then
.
(b)
.
(c)
therefore,
is continuous at
.
Directions: Supply the correct answer.
Let
Which of the following statements, I, II, III are true?
I.
exists
II.
exists
(a) only I
(b) only II
(c) I and II
III.
is continuous at
(d) none of them
(NOTE) Justify your answer by sustaining your complete solution)
In this lesson, I have learned that _________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
_________________________________________________________________________________.
23
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
Problem Solving
Let
Find the value of
so that
) is continuous at
Directions: Evaluate the following limits:
3𝑥(𝑥−1)
𝑥 2 −3𝑥+2
1. Suppose 𝑓(𝑥) = { −3
4
then
(a) at
(d) at
for 𝑥 ≠ 1, 2
𝑖𝑓 𝑥 = 1
𝑖𝑓 𝑥 = 4
is continuous except…
(b) at
(c) at
(e) at each real number
or 2
24
or 2
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
WHAT IS IT:
WHAT I CAN DO:
ASSESSMENT:
25
( b) only II
c = - 4/9
(B) except at x = 2.
Answer Key
References:
•
•
•
•
•
•
•
•
•
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines:
Anvil Publishing Inc.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of
America: Pearson Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second
Edition. United States of America: Alpha Books
Teaching Guide for Senior High School Basic Calculus (With permission
to use the concepts and examples)
https://tutorial.math.lamar.edu/Classes /Calc/ComputingLImits.aspx
https://www.analyzemath.com/calculus/limits/find_limits_functions.ht
ml
https://www.onlinemathlearning.com/limits-calculus.html
https://archives.math.utk.edu/visual.casual/1/limits.15/index.html
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondi
rectory/LimitConstant.html
26
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
SENIOR HIGH SCHOOL
BASIC
CALCULUS
(LEARNING
AREA)
QUARTER
- Module 1
(QUARTER3 NUMBER)
(MODULE
NUMBER)
Lesson
4:
Formulating and Solving
Accurately Situational
Problems Involving Extreme
Values
27
(DO_Q3_BASICCALCULUS_MODULE1_LESSON3)
Targets:
1. Illustrate the tangent line to the graph of a function at a given point
(STEM_BC11D-IIIe-1);
2. Apply the definition of the derivative of a function at a given number.
(STEM_BC11D-IIIe-2); and
3. Relate the derivative of a function to the slope of the tangent line.
(STEM_BC11D-IIIe-3).
What I Know
Directions: Choose the best answer in the following question base on the
idea about limits of trigonometric functions.
1. What is the slope of the tangent line to the graph of 𝑦 = 𝑥 3 at the
point (2, 8)?
a. 192
b. 12
c. 0
d. 4
2. Consider the parabola given by the equation 𝑦 = 𝑥 2 + 4𝑥 − 5. At
which point on the graph of this parabola is the slope of the tangent
line equal to 10?
a. (2, 7)
b. (10, 135)
c. (1, 0)
d. (3, 16)
3. The line 𝑦 = 2𝑥 − 9 is tangent to the parabola 𝑦 = 𝑥 2 + 𝑎𝑥 + 𝑏 at the
point (4, -1). What are the values of a and b?
a. a=-6, b=7 b. a=-4, b=-2
c. a=3, b=-30
d. a=-3, b=-6
4. What is the slope of the tangent line to the graph of 𝑓 at 𝑥 = 4, given
that 𝑓(𝑥) = −𝑥 2 + 4√𝑥?
a. -8
b. -10
c. -9
d. -5
5. What is the slope of the line tangent to the graph of 𝑦 = 𝑥 2 − 2 at the
point x=-8?
a. 2
b. -4
c. -8
d. -16
28
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
Lesson
4
Formulating and Solving Accurately
Situational Problems Involving Extreme
Values
TANGENT LINES TO CIRCLES
Recall from geometry class that a tangent line to a circle centered at O is a
line
intersecting the circle at exactly one point. It is found by constructing the
line, through a point A on the circle, that is perpendicular to the segment
(radius)
.
A secant line to a circle is a line intersecting the circle at two points.
The difficulty in defining the concept of the tangent line is due to an axiom
in Euclidean geometry that states that a line is uniquely determined by
two distinct points.
Thus, the definition of a tangent line is more delicate because it is
determined by only one point, and infinitely many lines pass through a
point.
HOW TO DRAW TANGENT LINES TO CURVES AT A POINT
The definition of a tangent line is not
very easy to explain without involving
limits. Students can imagine that locally,
the curve looks like an arc of a circle.
Hence, they can draw the tangent line to
the curve as they would to a circle.
29
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
One more way to see this is to choose
the line through a point that locally
looks most like the curve. Among all
the lines through a point
, the
one which best approximates the
curve
near the point
is the tangent line to the curve at that
point.
Note: Among all lines passing through
to the curve locally.
Another way of qualitatively
understanding the tangent line is to
visualize the curve as a roller coaster.
The tangent line to the curve at a point
is parallel to the line of sight of the
passengers looking straight ahead and
sitting erect in one of the wagons of the
roller coaster.
the tangent line is the closest
THE TANGENT LINE DEFINED MORE FORMALLY
The precise definition of a tangent line relies on the
notion of a secant line. Let be the graph of a
continuous function
and let be a point on
. A secant line to
through is any line
connecting and another point on . In the figure
on the right, the line
is a secant line of
through .
We now construct the tangent line to
at
Definition
Let C be the graph of a continuous function
and let be a point on .
1. A secant line to
through is any line connecting and another
point on .
2. The tangent line to
at is the limiting position of all secant lines
as
.
30
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
The Equation of the Tangent Line
In the previous lesson, we defined the tangent line at a point P as the
limiting position of the secant lines PQ, where Q is another point on the
curve, as Q approaches P. There is a slight problem with this definition
because we have no means of computing the limit of lines. Hence, we need
to work on the numbers that characterize the lines.
Recall the slope of a line passing through two points
Recall: Slope of a line
A line passing through distinct points
and
and
.
has
slope
Example 1: Given𝐴(1, −3), 𝐵(3, −2), and 𝐶(−1,0) , what are the slopes of the
lines 𝐴𝐵, 𝐴𝐶 and 𝐵𝐶 ?
Solution:
The Slope of
is
The Slope of
is
The Slope of
is
Therefore, the slopes of the
lines
1
3
1
̅̅̅̅, 𝐴𝐶
̅̅̅̅ , 𝑎𝑛𝑑 𝐶𝐵
̅̅̅̅ are , − , and − respectively.
𝐴𝐵
2
2
2
Recall the point-slope form of the equation of the line with slope
passing through the point
Recall: Point-Slope Form
A line passing through
with slope
and
has the equation
Question: What if we had chosen
as our point instead of , what
would be the point-slope form?
Answer: We would get an equivalent equation.
31
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
THE EQUATION OF THE TANGENT LINE
Given a function
a point
?
how do we find the equation of the tangent line at
Consider the graph of a function
whose graph is given below. Let
be a point on the graph of
. Our objective is to find the
equation of the tangent line
to the graph at the point
•
•
Observe that letting
approach
Find any point
on the
curve.
Get the slope of this secant
line
is equivalent to letting x approach
.
We use the formal definition of the tangent line:
Since the tangent line is the limiting position of the secant lines as
approaches ,
it follows that the slope of the tangent line
at the point is the limit of
the
slopes of the secant lines
as x approaches . In symbols,
Finally, since the tangent line passes through
given by
32
, then its equation is
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
Directions: Verify that the tangent line to the line
itself.
at
is the line
In this lesson, I have learned that _________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
Directions: Find the standard (slope-intercept form) equation of the tangent
line to the following functions at the specified points:
1.
at the point
2.
at the point
Directions: Find the standard (slope-intercept form) equation of the tangent
line to the following functions at the specified points:
at the point where
Directions: Find the standard (slope-intercept form) equation of the tangent
line to the following functions at the specified points:
1.
2.
at the point where
at the point where
33
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
34
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
WHAT IS IT:
1. 𝒚 = −𝟏𝟐𝒙 + 𝟏
2. 𝒚 = −𝟖𝒙 − 𝟗
ASSESSMENT:
1. 𝑦 = −
4𝑥
3
+
25
WHAT I CAN DO:
𝑥
+3
6
3
𝑦=
Answer Key
References:
•
•
•
•
•
•
•
•
•
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines:
Anvil Publishing Inc.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of
America: Pearson Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second
Edition. United States of America: Alpha Books
Teaching Guide for Senior High School Basic Calculus (With permission
to use the concepts and examples)
https://tutorial.math.lamar.edu/Classes /Calc/ComputingLImits.aspx
https://www.analyzemath.com/calculus/limits/find_limits_functions.ht
ml
https://www.onlinemathlearning.com/limits-calculus.html
https://archives.math.utk.edu/visual.casual/1/limits.15/index.html
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondi
rectory/LimitConstant.html
35
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
SENIOR HIGH SCHOOL
(LEARNING AREA)
(QUARTER
NUMBER)
BASIC
CALCULUS
(MODULE3NUMBER)
QUARTER
- Module 1
Lesson 5:
Differentiation Rules
36
(DO_Q3_BASICCALCULUS_MODULE1_LESSON4)
Targets:
1. Determine the relationship between differentiability and continuity of a
function (M11/STEM_BC11D - IIIf-1); and
2. Apply the differentiation rules in computing the derivative of an algebraic,
exponential, logarithmic, trigonometric functions and inverse trigonometric
functions (M11/STEM_BC11D-IIIf-3).
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Which of the following relationships are true?
I. All differentiable functions are continuous functions.
II. All continuous functions are differentiable functions.
III. Not all differentiable functions are continuous functions.
IV. Not all continuous functions are differentiable functions.
a. I & IV
b. I & III
c. I only
d. IV only
2. What is the derivative of the function 𝑓(𝑥) = 3 + 4𝑥 − 𝑥 2 at 𝑥 = 1?
a.
3. What
a.
b.
4. What
-2
b. 1
c. 0
3
is the derivative of 𝑓(𝑥) = 4 sec (5𝑥)?
60 sec 2 5𝑥
c. 60 sec 2 5𝑥 tan 5𝑥
60 tan 5𝑥
d. 60 sec 3 5𝑥 tan 5𝑥
is the derivative of the function 𝑓(𝑥) = (𝑦 − 2)2?
a. 2(𝑦 − 2)
𝑏. 4𝑦(𝑦 − 2)
𝑐.
𝑦−2
3
d. 2
𝑑.
(𝑦−2)3
3
5. Which of the following function will have a derivative of 8x-5?
a. 𝑔(𝑥) = 4𝑥 2 − 5𝑥 + 1
c. ℎ(𝑥) = 4𝑥 2 − 5𝑥 + 𝜋
2
b. 𝑓(𝑥) = 4𝑥 − 5𝑥 + 100
d. All of the above
37
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Lesson
5
Differentiation Rules
In the previous lessons, we discuss the basic concepts of differentiation. On
that lesson we have encountered different steps in finding the derivative of a
function. However, the process of finding the derivative of a function 𝑓(𝑥) is too
complex and complicated. So, to simplify the process, there are differentiation rules
that can be applied. The rules are derived so that we can find the derivative of
elementary functions easily.
What’s In
In the previous lessons about finding the derivative of the function, we have a series
of steps needed. For example. Let us find the derivative of 𝑓(𝑥) = 𝑥 3.
Solution:
𝑓(𝑥) = 𝑥 3
𝑓 ′ (𝑥) = lim
∆𝑥→0
= lim
∆𝑥→0
𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
(𝑥+∆𝑥)3 −𝑥 3
∆𝑥
𝑥 3 +3𝑥 2 (∆𝑥)+3𝑥(∆𝑥)2 +(∆𝑥)3−𝑥 3
∆𝑥
∆𝑥→0
= lim
3𝑥 2 (∆𝑥)+3𝑥(∆𝑥)2 +(∆𝑥)3
∆𝑥
∆𝑥→0
= lim
∆𝑥(3𝑥 2 +3𝑥∆𝑥+(∆𝑥)2
∆𝑥
∆𝑥→0
= lim
= lim 3𝑥 2 + 3𝑥∆𝑥 + (∆𝑥)2
∆𝑥→0
= 3𝑥 2 + 3𝑥(0) + (0)2
𝑓 ′ (𝑥) = 3𝑥 2
In our example, we find 𝑓′(𝑥) using the long method of finding the derivative. But if
you will look closely, what can you observe from the derivative 𝑓′(𝑥) = 3𝑥 2 in
reference with the actual function 𝑓(𝑥) = 𝑥 3? In this lesson we are going to find it
out.
38
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Let’s talk about the last example, the derivative of the function 𝑓(𝑥) = 𝑥 3 is
3𝑥 2 . If we are going to investigate, we need other examples. So, lets have some
monomial function with increasing degrees, and let’s compare it with their derivative.
The table below show the said functions with their corresponding derivative.
Table 1.1: Functions with corresponding derivative
𝒇(𝒙)
𝒇′(𝒙)
𝑥0
0
𝑥1
1
𝑥2
2𝑥
𝑥3
3𝑥 2
𝑥4
4𝑥 3
𝑥5
5𝑥 4
What can you observe from the table? Do you have an idea on how to find the
derivative of the functions?
Based from the table, we can find the derivative of the functions by multiplying the
exponent the expression and decreasing that exponent by one. But how about 𝑓(𝑥) =
𝑥 0 and 𝑓(𝑥) = 𝑥 1 ?? If we are going to use our observation, we will have:
𝑓(𝑥) = 𝑥 0
𝑓(𝑥) = 𝑥 1
𝑓 ′ (𝑥) = 0(𝑥 0−1 )
𝑓 ′ (𝑥) = 1(𝑥 1−1 )
= 0(𝑥 −1 )
= 1(𝑥 0 )
1
= 0 (𝑥)
= 1(1)
=0
=1
That goes for the pattern we have concluded in finding the derivative of a function.
We will later discuss it thoroughly on the next pages.
39
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Differentiability and Continuity
Before we proceed in finding the derivative of elementary function including
algebraic, exponential, logarithmic, trigonometric, and inverse trigonometric
functions. It will be safe for us to know whether we can find a derivative in the given
function.
There are three instances in which the derivative does not exist. In spite of the fact
that you can solve for the derivative of the function, it is not always valid because
the derivative does not actually exist. If the graph of the function contains (1) cusp,
(2) discontinuity, or (3) vertical tangent line, your function is not differentiable.
If your function has a cusp or sharp point in its graph, then the function has no
derivative at that certain point. Though function with these kinds of characteristics
are specific to absolute functions and piece-wise functions (continuous but not
smooth), it is necessary to know that these kinds of function must be inspected
thoroughly. In the figure below, you can see both nondifferentiable functions at 𝑥 =
2
𝑓(𝑥) = {
𝑥2, 𝑥 ≤ 2
6 − 𝑥, 𝑥 > 2
𝑓(𝑥) = |𝑥 − 2| − 2
If the function contains a vertical tangent line, then the function is not also
differentiable at that certain point. Recalling the slope of a line, a vertical line’s slope
is undefined. Moreover, derivative is defined as the slope of the line. For example,
2
𝑓(𝑥) = 𝑥 3 and we are going to find the derivative at 𝑥 = 0
Solution: We can also apply the pattern we discussed previously. So:
2
2
2
𝑓 ′ (0) = 3 (𝑥 3−1 )
= 3𝑥 1/3
1
2
3
= (𝑥 −3 )
2
=
1
2
3(0)1/3
2
= 3 ( 1)
=0
𝑥3
40
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
The tangent line’s slope is undefined, thus making the function nondifferentiable.
Lastly, the function is not differentiable if a function contains a point of discontinuity.
So, if a function is discontinuous at a certain x-value, it is automatically
nondifferentiable there. Differentiability implies continuity. If a function has a
derivative at a specific x-value, then the function must also be continuous at that x
value. Contra positively, we can also say that if the function is not continuous at a
certain x-value, then that function is not differentiable there either.
(𝑥−3)(𝑥+6)
For example: 𝑓(𝑥) = (𝑥+6)(𝑥+2), we know that the function is discontinuous at 𝑥 = −6
and 𝑥 = −2. Thus, the function is not differentiable at those values of x.
Differentiation Rules
There are different rules in order to find the derivatives of a certain function. Also,
there are different notations on derivatives. Leibniz’s notation is (but not limited to)
𝑑𝑦
𝑑𝑥
read as dee y dee x or derivative of y with respect to x. Also, Lagrange’s notation
is (but not limited to) a prime mark such as 𝑓′(𝑥) read as f prime of x. Lastly,
Newton’s notation is a dot notation where the first derivative of 𝑓(𝑥) is denoted by
𝑦̇ .
Rule 1: Constant Rule
Automatically, the derivative of any constant is zero
𝑑
[𝑐] = 0
𝑑𝑦
Rule 2: Power Rule
For all values of 𝑛, we can find the derivative of any expression in the form
𝑥 𝑛 as:
𝑑 𝑛
[𝑥 ] = 𝑛 ∙ 𝑥 𝑛−1
𝑑𝑥
Example 1: Find the derivative of 𝑓(𝑥) = 𝑥 3
Solution:
𝑓 ′ (𝑥) =
𝑑
[𝑥 3 ]
𝑑𝑥
= 3 ∙ 𝑥 3−1
by Rule 2
= 3𝑥 2
by simplifying
41
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Rule 3: Constant Multiple Rule
𝑑
[𝑐𝑓(𝑥)] = 𝑐[𝑓 ′ (𝑥)]
𝑑𝑥
Example 2: Find the derivative of 𝑓(𝑥) = 12𝑥 8
Solution: By applying power rule and constant multiple rule, we can have:
𝑓 ′ (𝑥) = 12 [
𝑑
[𝑥 8 ]]
𝑑𝑥
by Rule 3
= 12(8 ∙ 𝑥 8−1 )
by Rule 2
= 96𝑥 7
by simplifying
Rule 4: Sum and Difference Rule:
For all functions 𝑓(𝑥) and 𝑔(𝑥), we can the derivative of their sum/difference
as:
𝑑
[𝑓(𝑥) ± 𝑔(𝑥)] = 𝑓 ′ (𝑥) ± 𝑔′(𝑥)
𝑑𝑥
Example 3: Find the derivative of 𝑓(𝑥) = 𝑥 4 + 4𝑥 − 5
Solution: By applying several rules, starting by sum and difference
rule, we have,
𝑑
𝑓 ′(𝑥) = 𝑑𝑥 [𝑥 4 + 4𝑥 − 5]
𝑑
𝑑
𝑑
= 𝑑𝑥 [𝑥 4 ] + 𝑑𝑥 [4(𝑥)] − 𝑑𝑥 [5]
by Rule 4
= (4 ∙ 𝑥 4−1 ) + 4(1 ∙ 𝑥 1−1 ) + 0
by Rule 2 and Rule 1
= 4𝑥 3 + 4
by simplifying
Rule 5: Product Rule:
For all functions 𝑓(𝑥) and 𝑔(𝑥), we can have the derivative of their product as:
𝑑
[𝑓(𝑥) ∙ 𝑔(𝑥)] = 𝑓(𝑥) ∙ 𝑔′ (𝑥) + 𝑓 ′ (𝑥) ∙ 𝑔(𝑥)
𝑑𝑥
Example 4: Find the derivative of 𝑓(𝑥) = 𝑥 2 (𝑥 + 1)
𝑑
𝑓 ′ (𝑥) = 𝑑𝑥 [𝑥 2 (𝑥 + 1)]
𝑑
[𝑥
𝑑𝑥
= 𝑥2 (
𝑑
+ 1]) +
𝑑
(𝑥 2 )(𝑥
𝑑𝑥
𝑑
+ 1)
𝑑
by Rule 5
= 𝑥 2 (𝑑𝑥 [𝑥] + 𝑑𝑥 [1]) + 𝑑𝑥 [𝑥 2 ] ∙ (𝑥 + 1)
by Rule 4
= 𝑥 2 (1𝑥 1−1 + 0) + (2𝑥 2−1 )(𝑥 + 1)
by Rule 2
= 𝑥 2 (1) + (2𝑥)(𝑥 + 1)
by simplifying
= 𝑥 2 + 2𝑥 2 + 2𝑥
by simplifying
= 3𝑥 2 + 2𝑥
42
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Rule 6: Quotient Rule:
For all functions 𝑓(𝑥) and 𝑔(𝑥), we can have the derivative of their quotient
as:
𝑑 𝑓(𝑥)
𝑔(𝑥)𝑓 ′ (𝑥) − 𝑓(𝑥)𝑔′(𝑥)
[
]=
𝑑𝑥 𝑔(𝑥)
𝑔(𝑥)2
Example 5: Find the derivative 𝑓(𝑥) =
𝑓 ′ (𝑥) =
=
=
𝑥2
𝑥+1
𝑥2
𝑑
𝑑
[𝑥+1]−(𝑥+1) [𝑥 2 ]
𝑑𝑥
𝑑𝑥
(𝑥 2 )2
𝑥2(
𝑑
𝑑
𝑑
[𝑥]+ [1])−(𝑥+1) [𝑥 2 ]
𝑑𝑥
𝑑𝑥
𝑑𝑥
(𝑥 2 )2
𝑥 2 (1∙𝑥 1−1 +
by Rule 4
𝑑
[1])−(𝑥+1)(2∙𝑥 2−1 )
𝑑𝑥
(𝑥 2 )2
by Rule 2
=
𝑥 2 (1∙𝑥 1−1 +0)−(𝑥+1)(2∙𝑥 2−1 )
(𝑥 2 )2
by Rule 1
=
𝑥 2 (1+0)−(𝑥+1)(2𝑥)
𝑥4
by simplifying
=
𝑥 2 −2𝑥 2 −2𝑥
𝑥4
by simplifying
=
−𝑥 2 −2𝑥
𝑥4
Before we move further and discuss different rules in differentiation, it is important
to know that given 𝑦 as a function of 𝑥, it is convenient to use 𝑢 as an auxiliary
variable. We can use 𝑢 in different situations where in we cannot apply previous rules
in finding the derivative of a function. For example, if we are going to find the
derivative of 𝑓(𝑥) = √𝑥 3 − 8, we cannot directly apply Power Rule since the function
is not in the form 𝑥 𝑛 .
To solve the problem, we are going to assign 𝑥 3 − 8 into another variable. Let us use
𝑢, making 𝑢 = 𝑥 3 − 8. Thus, we can now denote 𝑓 ′(𝑥) =
Rule.
𝑑
𝑑𝑢
[√𝑢] ∙ 𝑑𝑥
𝑑𝑢
and apply Power
𝑓(𝑥) = √𝑢
𝑓(𝑥) = 𝑢1/2
𝑓 ′ (𝑥) =
1
𝑑
𝑑𝑢
2] ∙
[𝑢
𝑑𝑢
𝑑𝑥
1
1
𝑑𝑢
= 2 ∙ 𝑢2−1 ∙ 𝑑𝑥
1
1
by Rule 2
𝑑𝑢
= 2 ∙ 𝑢−2 ∙ 𝑑𝑥
1
=2∙
=2
1
1
𝑢2
by simplifying
𝑑𝑢
by simplifying
∙ 𝑑𝑥
1
𝑑𝑢
∙ 𝑑𝑥
𝑢
√
43
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Since 𝑢 = 𝑥 3 − 8
=
=
=
=
=
1
𝑑
∙ [𝑥 3
2√𝑥 3 −8 𝑑𝑥
1
2√𝑥 3 −8
1
2√𝑥 3 −8
1
2√𝑥 3 −8
𝑑
by substitution
− 8]
𝑑
∙ (𝑑𝑥 [𝑥 3 ] − 𝑑𝑥 [8])
by Rule 4
∙ (3𝑥 3−1 − 0)
by Rule 2 and 1
∙ 3𝑥 2
by simplifying
3𝑥 2
2√𝑥 3 −8
The previous process is what we call the Chain Rule. Finding the derivative of
exponential, logarithmic, and trigonometric functions rely heavily on Chain Rule so
you must bear it in mind.
Rule 7: Chain Rule
For all functions 𝑓(𝑥) and 𝑔(𝑥), we can have the derivative of the composite
function as:
𝑑
[𝑓(𝑔(𝑥)] = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔(𝑥)
𝑑𝑥
Derivative of Exponential, Logarithmic, and Trigonometric
Function
In general, an exponential function is in the form 𝑓(𝑥) = 𝑎 𝑥 , where 𝑎 is a positive
constant.
Rule 8: Derivative of Exponential Function
8.1. If 𝑎 > 0, 𝑎 ≠ 1 and 𝑓(𝑥) = 𝑎𝑢 , where 𝑢 is a differentiable function of 𝑥,
then:
𝑑 𝑢
𝑑𝑢
[𝑎 ] = (𝑎𝑢 ln 𝑎)
𝑑𝑥
𝑑𝑥
8.2. If 𝑎 = 𝑒, then ln 𝑒 = 1 and 𝑓(𝑥) = 𝑒 𝑢 , where 𝑢 is a differentiable function
of 𝑥, then:
𝑑 𝑢
𝑑𝑢
[𝑒 ] = 𝑒 𝑢
𝑑𝑥
𝑑𝑥
Rule 9: Derivative of Logarithmic Function
9.1. The derivative of the logarithmic function with base 𝑎, where 𝑎 is a
positive constant and 𝑎 ≠ 1 defined by 𝑓(𝑥) = log 𝑎 𝑥 is given by:
𝑑
1
[log 𝑎 𝑥] =
𝑑𝑥
𝑥 ln 𝑎
44
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
9.2. If 𝑎 = 𝑒, you will obtain a natural logarithmic function defined by
𝑓(𝑥) = ln 𝑥. Since 𝑒 = 1 then,
𝑑
1
[ln 𝑥] =
𝑑𝑥
𝑥
9.3. If 𝑢 is a differentiable function of 𝑥, then;
𝑑
1
𝑑𝑢
[log 𝑢 𝑎] =
∙
𝑑𝑥
𝑢 ln 𝑎 𝑑𝑥
Example: Find the derivative of 𝑓(𝑥) = 52𝑥+1 + log 5 𝑥
Solution:
𝑓 ′ (𝑥) =
𝑑
𝑑
[52𝑥+1 ] + [log 5 𝑥]
𝑑𝑥
𝑑𝑥
𝑑
by Rule 4
𝑑
= 𝑑𝑥 [5𝑢 ] + 𝑑𝑥 [log 5 𝑥]
𝑑𝑢
letting 𝑢 = 2𝑥 + 1
1
= 5𝑢 ln 5 𝑑𝑥 + 𝑥 ln 𝑎
by Rule 8.1 & 9.1
1
= 52𝑥+1 ln 5 ∙ 2 + 𝑥 ln 5
= 2 ln(5) 52𝑥+1 +
by substituting 𝑢
1
ln(5)𝑥
by rearranging
Rule 10: Derivative of Trigonometric Functions
Listed below are the derivatives of the 6 trigonometric functions:
i.
ii.
iii.
iv.
v.
vi.
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
[sin 𝑥] = cos 𝑥
[cos 𝑥] = −sin 𝑥
[tan 𝑥] = sec 2 𝑥
[csc 𝑥] = − csc 𝑥 cot 𝑥
[sec 𝑥] = sec 𝑥 tan 𝑥
[cot 𝑥] = − csc 2 𝑥
Example: Find the derivative of 𝑓(𝑥) = 𝑥 3 cos 𝑥
Solution: Since 𝑓(𝑥) is a product of two function. So, we are going to use
Product Rule first:
𝑑
𝑓 ′ (𝑥) = 𝑥 3 𝑑𝑥 [cos 𝑥] + cos 𝑥
𝑑
[𝑥 3 ]
𝑑𝑥
by Rule 5
= 𝑥 3 (− sin 𝑥) + cos 𝑥 (3𝑥 2 )
by Rule 10 & 2
= −𝑥 3 sin 𝑥 + 3𝑥 2 cos 𝑥
by simplifying
45
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
What’s More
Activity 1: Finding Derivatives
Apply different rules of differentiation to find the derivatives of the following
functions:
1. 𝑚(𝑥) = 5𝑥 2
2. 𝑛(𝑥) = 2𝑥 3 − 7𝑥 + 1
1
3. 𝑝(𝑥) = √𝑥 − 𝑥
√
4. 𝑞(𝑥) = (−7𝑥 2 )(3𝑥 2 − 6)
5𝑥+4
5. 𝑟(𝑥) = 4𝑥 3 −4𝑥
6. 𝑠(𝑥) = sin 𝑥 + csc 𝑥
7. 𝑡(𝑥) = sin 𝑥 cos 𝑥
8. 𝑢(𝑥) = ln 𝑥 + log10 2𝑥
9. 𝑣(𝑥) = 𝑒 𝑥 sin 𝑥
10. 𝑤(𝑥) = 10𝑥
What I Have Learned
Complete the following sentences.
1. The function is not differentiable if there are cusp, vertical tangent line, and
____________________.
2. If the function is not continuous at a certain x-value, then that function is not
________________ there either.
3. There are several differentiation rules that can be used to find the
______________ of a function.
4. There are 6 basic differentiation rules: (1) _______________, (2) ____________,
(3) _____________________, (4) ___________________, (5) __________________, and
(6) __________________________.
5. It is essential to know the Chain Rule since it will be used heavily on the
following ______________________________.
6. The derivative of the _____________________ can be obtained using the rules:
(1)
𝑑
[𝑎𝑢 ]
𝑑𝑥
𝑑𝑢
𝑑
𝑑𝑢
= (𝑎𝑢 ln 𝑎) 𝑑𝑥 and (2)𝑑𝑥 [𝑒 𝑢 ] = 𝑒 𝑢 𝑑𝑥 .
7. The derivative of the _____________________ can be obtained using the rules:
(1)
𝑑
[log 𝑎
𝑑𝑥
𝑥] =
1
,
𝑥 ln 𝑎
(2)
𝑑
[ln 𝑥]
𝑑𝑥
1
𝑥
= , and (3)
46
𝑑
[log 𝑢 𝑎]
𝑑𝑥
=
1
𝑑𝑢
∙
𝑢 ln 𝑎 𝑑𝑥
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
8.
The derivative of the ______________________ can be obtained using the rules:
(1)
(2)
(3)
(4)
(5)
(6)
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
[sin 𝑥] = cos 𝑥
[cos 𝑥] = −sin 𝑥
[tan 𝑥] = sec 2 𝑥,
[csc 𝑥] = − csc 𝑥 cot 𝑥
[sec 𝑥] = sec 𝑥 tan 𝑥, and
[cot 𝑥] = − csc 2 𝑥
Rate of Change
Layla has a balloon making business. Her balloon-blowing machine can inflate
spherical balloons whose surface area increases at the rate of 25 𝑐𝑚2 /𝑠𝑒𝑐. What is
the rate of increase of the radius of the balloon when its radius is 3 𝑐𝑚?
Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
𝑑𝑦
𝑑𝑥
2𝑥+7
(5−2𝑥)2
1. Determine
a.
2𝑥+7
of 𝑦 = 5−2𝑥 ?
b.
−8𝑥−4
(5−2𝑥)2
c.
2. If 𝑓(𝑥) = 3𝑥 2 , then 𝑓 ′ (5) is
a. 0
b. 15
24
(5−2𝑥)2
d.
c. 30
8𝑥+4
(5−2𝑥)2
d. 75
𝑥 3 −9𝑥+10
3. Which of the following is true about 𝑓(𝑥) = 𝑥−2 ?
a. 𝑓(𝑥) is continuous and differentiable
b. 𝑓(𝑥) is discontinuous but differentiable
c. 𝑓(𝑥) is continuous but not differentiable
d. 𝑓(𝑥) is discontinuous and not differentiable
4. What is the derivative of the function 𝑓(𝑥) = csc(𝑥 3 )
a. 3𝑥 2 sin 𝑥 3
c. −3𝑥 2 cot 𝑥 3 csc 𝑥 3
b. 3𝑥 2 cos 𝑥 3
d. 3𝑥 2 − cot 𝑥 3 csc 𝑥 3
𝑑𝑉
4
5. Find 𝑑𝑟 of 𝑉 = 3 𝜋𝑟 3 .
a. 𝑉 ′ = 4𝑟 2
4
b. 𝑉 ′ = 3 𝜋𝑟
c. 𝑉 ′ = 4𝜋𝑟 2
d. 𝑉 ′ = 4𝜋𝑟 3
47
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Additional Activities
The Hidden Code
Alicia opens an email send to her. Inside the email, there is a link that the sender
describes as the link to protect your account from professional hackers. After
opening the link, her computer monitor suddenly blacks out. Two minutes later,
her computer monitor opens and shows an unclosabe window. The window shows
this message:
“To delete the virus, crack the code first!”
_12_ _6_ _1_ _3_ _11_
_7_ _8_ _14_
_9_ _10_ _4_ _8_ _13_ _10_
_2_ _5_ _1_ _2_ _11_
Hint: Solve the derivative of the functions below to access the code.
CODE
1. 𝑓(𝑥) = 3𝑥
b. 250𝑥 9
c. 1
2. 𝑓(𝑥) = 𝑥 + 2
e. − sin 𝑥
3. 𝑓(𝑥) = 𝑥 3
4. 𝑓(𝑥) = 4𝑥 7 + 1
f. 28𝑥 6
5. 𝑓(𝑥) = 2𝑥 3 + 𝑥 2
g. 2𝑥 3
6. 𝑓(𝑥) = √𝑥
h.
1
2√𝑥
i. 3
4𝑥
7. 𝑓(𝑥) = 𝑥 2
8. 𝑓(𝑥) = 7𝑥(𝑥 3 − 1)
k. cos 𝑥 + sec 2 𝑥
9. 𝑓(𝑥) = 25𝑥10
l. 6𝑥 2 + 2𝑥
10.𝑓(𝑥) = cos 𝑥
n. 3𝑥 2
11. 𝑓(𝑥) = sin 𝑥 + tan 𝑥
o. 28𝑥 3 − 7
12. 𝑓(𝑥) = 4𝑥 3 sin 𝑥
r.
13. 𝑓(𝑥) = ln 𝑥 4
4
𝑥
t. 4𝑥 3 cos 𝑥 + 12𝑥 2 sin 𝑥
14. 𝑓(𝑥) = 5 ln 𝑥 + csc 𝑥
5
u. 𝑥 − csc 𝑥 cot 𝑥
4
y. − 𝑥 2
48
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
49
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
Key Concept: 𝑆𝐴 = 4𝜋𝑟 2
1. i
2. c
3. n
4. f
5. l
6. h
7. y
8. o
9. b
10. e
11. k
12. t
13. r
14. u
Hidden Message:
What I Can Do
Additional Activities
“Think before you click”
What I Know
1.
2.
3.
4.
5.
A
D
D
A
D
What's More
1. 5𝑥
2. 6𝑥 2 − 7
𝑥+1
3. 2𝑥 𝑥
√
4. −84𝑥 3 + 84𝑥
5.
−40𝑥 3 −48𝑥 2 +16
(4𝑥 3 −4𝑥)2
Given:
𝑑𝑆𝐴
𝑑𝑡
𝑑𝑟
𝑑𝑡
= 25𝑐𝑚2 /𝑠𝑒𝑐
=? 𝑤ℎ𝑒𝑛 𝑟 = 3
Working Eq:
𝑑𝑆𝐴 𝑑𝑟
=
(4𝜋𝑟 2 )
𝑑𝑡
𝑑𝑡
Answer:
𝑑𝑟
≈ 0.33 𝑐𝑚/𝑠𝑒𝑐
𝑑𝑡
Assessment
1.
2.
3.
4.
5.
C
C
D
C
C
6. cos 𝑥 − csc 𝑥 cot 𝑥
7. cos(2𝑥)
1
1
8. 𝑥 + ln(10)𝑥
9. 𝑒 𝑥 sin 𝑥 + 𝑒 𝑥 cos 𝑥
10.ln(10)10𝑥
Answer Key
References
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines: Anvil
Publishing Inc. pp. 67-82.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of America: Pearson
Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second Edition. United
States of America: Alpha Books.
50
(DO_Q3_BASICCALCULUS_MODULE1_LESSON5)
SENIOR HIGH SCHOOL
(LEARNING AREA)
(QUARTER
NUMBER)
BASIC
CALCULUS
(MODULE3NUMBER)
QUARTER
- Module 1
Lesson 6:
Extreme Value Theorem
51
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Targets:
1. Illustrate the Extreme Value Theorem;
2. Solve optimization problems that yield polynomial functions; and
3. Illustrate the Chain Rule of differentiation. (M11/STEM_BC11D-IIIh-2)
What I Know
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. At what point does the function 𝑓(𝑥) = 𝑥 2 − 3 has an absolute minimum at
the interval [−2, 2]?
a. (0, 0)
b. (0, −3)
c. (1, -2)
d. (-1, -2)
2. The function 𝑓(𝑥) = 3𝑥 + 5 is continuous on [0, 3]. Where can we find the
absolute maximum of the function?
a. origin
b. at x = 1
c. at x = 2
d. at x = 3
3. What is the derivative of 𝑓(𝑥) = (𝑥 + 1)5
a. 5(𝑥 + 1)4
b. 5(𝑥 + 1)4 + 𝑥
c. 5(𝑥 + 1)4 − 𝑥
4. What is the derivative of the function 𝑓(𝑥) = cos 𝑥 3 ?
a. sin 𝑥 3
b. 3 sin 𝑥 2
c. −3𝑥 2 sin 𝑥 2
d. 5𝑥(𝑥 + 1)4
d. − sin 𝑥 3 + 3𝑥 2
5. Find two positive numbers whose sum is 30 such that the maximum
product is possible.
Lesson
6
Extreme Value Theorem
Being extreme means a lot of different things. It can mean an outermost or
furthermost from a given point. Also, it can mean reaching high or highest degrees.
But in Mathematics, extreme means the maximum or the minimum point on a
graph of a function. In this lesson, we are going to find the extremum (extrema in
plural form), the minimum or maximum value of a function.
52
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
What’s In
In the previous lessons, we discuss how to find whether the function is differentiable
or not. As we recall, if the function is continuous at 𝑥 = 𝑐, then it is also differentiable
at 𝑥 = 𝑐 . For example, let determine if the function 𝑓(𝑥) = 𝑥 5 + 4𝑥 3 − 11𝑥 2 + 1 is
continuous at 𝑥 = 2.
Solution: Recalling the steps to determine the continuity of a function, we the
function needs to pass the 3 steps—𝑓(𝑐) exists, lim 𝑓(𝑥) exists, and lim 𝑓(𝑥) =
𝑥→𝑐
𝑥→𝑐
𝑓(𝑐).
i.
ii.
𝑓(𝑥) = 𝑥 5 + 4𝑥 3 − 11𝑥 2 + 1; 𝑥 = 2
𝑓(2) = (2)5 + 4(2)3 − 11(2)2 + 1
= 32 + 4(8) − 11(4) + 1
= 21
𝑓(2) 𝑒𝑥𝑖𝑠𝑡𝑠
lim 𝑥 5 + 4𝑥 3 − 11𝑥 2 + 1
𝑥→2
= (2)5 + 4(2)3 − 11(2)2 + 1
= 32 + 4(8) − 11(4) + 1
= 21
lim 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑠
𝑥→2
iii.
?
lim 𝑓(𝑥) = 𝑓(𝑐)
𝑥→2
✓
21 = 21
∴ the function 𝑓(𝑥) is continuous.
In finding the extrema of a function, it is necessary that the function is continuous.
Also, in the process of finding the extrema of a function, we need to find the
derivative, so continuity is really necessary.
53
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Let’s take a look at the graph of the function 𝑓(𝑥) = −𝑥 2 + 3𝑥 − 2. What can you
observe at its graph?
As we can see, if we are going to find the lowest point of the graph we are going to
find noting because the graph extends to negative and positive infinity. But, we can
3 1
2 4
have the highest point which is located at the vertex of a parabola ( , ). Thus, we
3 1
can locate the extremum of the function at (2 , 4).
Let’s have another example. The graph below is the graph of the active cases of
Coronavirus infection in the Philippines.
The function above is continuous from Feb 15 to Sep 22. As we can see, the
function have extrema, where the number of cases is at the maximum.
54
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Extreme Value Theorem
The Extreme Value Theorem states that If the function 𝑓(𝑥) is continuous (no holes
or jumps) on a closed interval [𝑎, 𝑏], then the function 𝑓(𝑥) has an absolute maximum
and absolute maximum on [𝑎, 𝑏].
We can call the maximums and minimums as the extrema (singular extremum) of the
functions. The function 𝑓(𝑥) has a maximum at 𝑥 = 𝑐 if the 𝑓(𝑐) ≥ 𝑓(𝑥) for all values
of x. On the other hand, the function 𝑓(𝑥) has a minimum at 𝑥 = 𝑐 if the 𝑓(𝑐) ≤ 𝑓(𝑥)
for all values of x. Meaning, the function’s maximum and minimum are the highest
and lowest point respectively.
It is important to note that functions must be continuous to guarantee extrema. But
still, discontinuous function can have extrema, the only thing is that is not always
guaranteed.
For example. The function below is continuous on [−5, 5]
The function has an absolute minimum at 𝑥 = −2, and absolute maximum at 𝑥 = 3.
Since the interval is closed, we can also say that the relative minimum at 𝑥 = 5 and
relative maximum at 𝑥 = −5. Thus, the extreme values of the function are 𝑓(−2) = −2
and 𝑓(3) = 4.
The figure above also shows relative extrema. Relative extrema occurs if there is more
than one extrema point. Absolute extrema is the lowest or highest point, and relative
extrema as the next highest or lowest point. So, if we are going to find the function’s
absolute extrema, we can just look for two places—at the relative extrema point or at
an endpoint.
Algebraically, we can locate the extreme values in an interval [𝑎, 𝑏] by finding the
derivative of the function, say 𝑓(𝑥), and substitute the roots of 𝑓′(𝑥) to 𝑓(𝑥). After
that, we will compare it to 𝑓(𝑎) and 𝑓(𝑏). The absolute extrema are the highest and
the lowest upon comparison.
55
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
For example. Find the absolute extrema of the function 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 1 on [−3, 3].
Solution: First thing to do is identify if the function is continuous at the given
interval. Since the function is a polynomial function, it is continuous at [−3, 3] (it is
a polynomial function which is continuous everywhere). We can now proceed to
finding 𝑓′(𝑥).
𝑓(𝑥) = 𝑥 3 − 3𝑥 + 1
𝑑
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑𝑥 (𝑥 3 ) − 𝑑𝑥 (3𝑥) + 𝑑𝑥 (1)
𝑓 ′ (𝑥) = 3𝑥 2 − 3𝑥
Make the derivative equal to zero then get the value of x:
𝑓 ′ (𝑥) = 0
3𝑥 2 − 3 = 0
𝑥2 − 1 = 0
(𝑥 + 1)(𝑥 − 1) = 0
𝑥+1=0
𝑥 = −1
𝑥−1=0
𝑥=1
Thus, the critical values are 𝑥 = ±1
Substitute the critical values and interval values to 𝑓(𝑥)
𝑓(1) = 𝑥 3 − 3𝑥 + 1
𝑓(−3) = 𝑥 3 − 3𝑥 + 1
= (1)3 − 3(1) + 1
= (−3)3 − 3(−3) + 1
= 1 − 3(1) + 1
= −27 + 9 + 1
= −1
= −17
𝑓(−1) = 𝑥 3 − 3𝑥 + 1
𝑓(3) = 𝑥 3 − 3𝑥 + 1
= (−1)3 − 3(−1) + 1
= (3)3 − 3(3) + 1
= −1 + 3 + 1
= 27 − 9 + 1
=3
=19
By comparing the values of the function, since the
lowest value is at
𝑓(−3) = −17 and the
highest value is at 𝑓(3) = 19, we can conclude that
the absolute minimum is at 𝑥 = −3 and the
absolute maximum is at 𝑥 = 3. The values of 𝑓(−1)
and 𝑓(1) are called the relative extrema. We can
confirm the result by looking at the graph at the
right.
56
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Optimization Problems
One of the applications of Calculus that deals with the absolute maximum and
absolute minimum of a function is what we called optimization problems. Commonly,
you are the one to analyze and formulate function based on the problem that is why
these kinds of problems are hard to solve.
To help you solve optimization problems, Barnett (2011) suggests a step-by-step
strategy:
1. Determine the variables, look for relationships among them, and establish a
mathematical model of the form:
Maximize (or minimize) 𝑓(𝑥) on the interval [𝑎, 𝑏]
2. Find the critical values of 𝑓(𝑥)
3. Find the absolute maximum and minimum value of 𝑓(𝑥) on the interval
[𝑎, 𝑏]and the values of 𝑥 where it occurs
4. Use the solution to the mathematical model to answer all the questions
asked in the word problem.
Example Problem 1:
A box is to be made of a piece of cardboard with a dimension 11 by 11
inches. What is the largest possible volume that the box can make?
Solution: Let 𝑥 be the height of the box (or the length of the side of the
square to be cut on all corners)
𝑥
𝑥
𝑥
11 − 2𝑥
11 − 2𝑥
11 − 2𝑥
Since you are going to cut 2 squares from each side cardboard piece, the
dimensions will become 11 − 2𝑥 and 11 − 2𝑥 inches.
Step 1: Find/formulate mathematical model
𝑉 = 𝑙𝑤ℎ
= (11 − 2𝑥)(11 − 2𝑥)𝑥
𝑉 = 𝑥(11 − 2𝑥)2
= 4𝑥 3 − 44𝑥 2 + 121𝑥
57
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Step 2: Find the critical values of 𝑓(𝑥)
To find the critical value, find the derivative of 𝑉:
𝑑
𝑑
𝑑
𝑉 ′ = 𝑑𝑥 [4𝑥 3 ] + 𝑑𝑥 [44𝑥 2 ] + 𝑑𝑥 [121𝑥]
= (4 ∙ 3𝑥 3−1 ) − (44 ∙ 2𝑥 2−1 ) + (121)
= 12𝑥 2 − 88𝑥 + 121
Equate to zero to get the critical values
12𝑥 2 − 88𝑥 + 121 = 0
We can get the root by using factoring:
(6𝑥 − 11)(2𝑥 − 11 = 0
6𝑥 − 11 = 0
2𝑥 − 11 = 0
6𝑥 = 11
2𝑥 = 11
𝑥=
11
6
𝑜𝑟 1.83
𝑥=
11
2
𝑜𝑟 5.5
Step 3. Since we are going to find the maximum volume. We are going to find
the absolute maximum of 𝑉.
However, we cannot use 𝑥 =
the dimension will become:
11
𝑜𝑟
2
5.5. Because, if we substitute it to 𝑉
11 − 2𝑥
11 − 2(5.5)
11 − 11 = 0
We cannot have 0 as dimension so we will exclude that. Thus, 𝑥 =
11
6
Step 4: Since we are task to find the maximum volume of the box, we will
substitute the value of x to the equation:
𝑉 = 𝑥(11 − 2𝑥)2
=
11
(11
6
=
2662
𝑜𝑟
27
11
2
− 2 ( 6 ))
98.52
The box dimension is
𝟏𝟏
𝒊𝒏
𝟔
by
𝟐𝟐
𝒊𝒏
𝟑
maximum volume is 𝟗𝟖. 𝟓𝟐 𝒊𝒏 .
58
𝟑
by
𝟐𝟐
𝒊𝒏.
𝟑
Therefore, the
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Chain Rule
Chain rule was introduced previously to solve for the derivative of logarithmic and
trigonometric function. This time we are going to illustrate the chain rule thoroughly.
If we have 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = 11𝑥 + 7 , we can find the derivative of the both
functions. Using also the rules of derivatives we can also get the derivative of their
sum and difference, their product, and also their quotient. But what if we are task
to find the derivative of two function plugged into one another?
Mathematically, we can denote that function as a composite function 𝑓(𝑔(𝑥)). This
kind of function is too complicated to get the derivative since it is not in the form of
𝑓(𝑥) = 𝑥 𝑛 . So, if a function contains something other than the variable, then we
should use the chain rule.
Chain Rule: To find the derivative of ℎ(𝑥) = 𝑓(𝑔(𝑥)) , where 𝑓(𝑥) and 𝑔(𝑥) are
differentiable functions,
ℎ′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′ (𝑥)
Example: Find the derivative of 𝑦 = (𝑥 2 + 1)5
Solution: To illustrate chain rule, it is helpful to denote the inner function as
other variable (auxiliary variable), say 𝑢, to make the function more
workable. Let 𝑢 = 𝑥 2 + 1, hence 𝑦 = 𝑢5 .
𝑦 = 𝑢5
𝑦′ =
𝑑
𝑑
[𝑢5 ] ∙ [𝑢]
𝑑𝑢
𝑑𝑥
𝑑
= (5 ∙ 𝑢5−1 ) (𝑑𝑥 [𝑢])
𝑑
= (5𝑢4 ) (𝑑𝑥 [𝑢])
𝑑
= 5(𝑥 2 + 1)4 (𝑑𝑥 [𝑥 2 + 1])
𝑑
𝑑
= 5(𝑥 2 + 1)4 (𝑑𝑥 [𝑥 2 ] + 𝑑𝑥 [1])
= 5(𝑥 2 + 1)4 ((2 ∙ 𝑥 2−1 ) + 0)
= 5(𝑥 2 + 1)4 (2𝑥)
𝑦′ = 10𝑥(𝑥 2 + 1)4
59
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
What’s More
Activity 1: Number Problems
Use the steps in solving optimization problem to solve the problem:
Find the two negative number whose sum is -50 such that their product is
at the maximum.
Activity 2. Chain Rule
Apply Chain Rule to solve for the derivative of the following functions:
1. 𝑓(𝑥) = √𝑥 2 + 3𝑥 − 1
2. 𝑓(𝑥) = (4𝑥 3 + 3𝑥)11
3. 𝑓(𝑥) = sin 3𝑥 2
What I Have Learned
1. The ____________________ states that if the function 𝑓(𝑥) is continuous on a
closed interval [𝑎, 𝑏], then the function has an absolute minimum and absolute
maximum at the interval [𝑎, 𝑏]
2. There are strategic steps in solving optimization problems.
a. Determine the variables, look for relationships among them, and
establish a mathematical model of the form:
Maximize (or minimize) 𝑓(𝑥) on the interval [𝑎, 𝑏]
b. Find the ____________ values of 𝑓(𝑥)
c. Find the absolute ________________________________ of 𝑓(𝑥) on the
interval [𝑎, 𝑏]and the values of 𝑥 where it occurs
d. Use the solution to the mathematical model to answer all the questions
asked in the word problem.
3. Chain rule is used in finding the derivative of a composite function
____________________. Where:
ℎ′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥)
60
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
More on Optimization Problem
Find the dimensions of the isosceles triangle (height and length of sides (for it to
have a maximum area given that the perimeter of the triangle is 20 cm.
𝑥
𝑥
ℎ
10 − 𝑥
10 − 𝑥
Hint: Let 𝑥 be the side of isosceles triangle. Use Pythagorean Theorem to find
equation for 𝑥.
Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1. The function 𝑓(𝑥) = 𝑥 4 + 4𝑥 3 − 5 is continuous on [−5, 1] . What is the
absolute minimum of the function?
a. f(-5)
b. f(-3)
c. f(0)
d. f(1)
2. In the interval [−1, 1], where can we find the absolute maximum of the
function 𝑓(𝑥) = 5𝑥 7 + 𝑥 3 + 1?
a. (-1, -5)
b. (0, 1)
c. (1, 7)
d. none of the above
3. The function 𝑓(𝑥) = sin(𝑥 2 + 1) is continuous at the interval [−2, 2]. What is
the absolute maximum of the function?
a. -1.92 and 1.92 b. -1.46 and 1.46 c. 1
d. 0.84
4. What is the derivative of the function 𝑓(𝑥) = tan(8𝑥) ?
a. 8 sec 2(8𝑥)
b. 8 sec 2 𝑥
c. sec 2 (8𝑥) + 8
d.
sec2(8𝑥)
8
5. You have 50 𝑓𝑡 of railings for the enclosure of your pet. What will be the
dimension of the enclosure for your pet to have a maximum area to live?
a. 8𝑓𝑡 𝑏𝑦 17 𝑓𝑡
c. 12.5 𝑓𝑡 𝑏𝑦 12.5 𝑓𝑡
b. 15 𝑓𝑡 𝑏𝑦 10 𝑓𝑡
d. none of the above
61
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Additional Activities
Manny the Farmer
Manny is a local farmer from Cebu. The local government give him enough
budget to have to make 2000 meters of fencing on a free lot for agriculture. The
local government tell him to maximize his lot to produce more crops. However, his
rectangular lot is beside a straight river, so he plans to fence only the three sides
(river side is not included). What dimension of the rectangular lot can you suggest
to Manny for him to have a maximum area to grow crops.
Hint: Let x be the length of the lot.
62
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
63
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Mathematical Model:
Mathematical Model:
What I Can Do
Additional Activities
𝐴 = (𝑥)(2000 − 2𝑥)
Critical Values:
𝐴′ = 2000 − 4𝑥
𝑥 = 500
Maximum Area:
𝐴 = 500,000 𝑚2
1
𝐴 = (20 − 2𝑥)(√20𝑥 − 100)
2
Critical Values:
𝐴′ =
What's More
10√3
3
−30𝑥 + 200
100√3
𝑐𝑚2
9
𝑐𝑚 𝑏𝑦
10
3
𝑐𝑚
Assessment
6. B
7. C
8. C
9. A
10.C
Activity 1. 25 and 25
Activity 2.
1.
2.
√20𝑥 − 100
𝑥 = 20/3
Maximum Area:
At 500 𝑚 by1000 𝑚2
𝐴=
At
What I Know
6. B
7. D
8. D
9. C
10.C
2𝑥+4
2√𝑥 2 +3𝑥−1
(132𝑥 2 + 33)(4𝑥 3 + 3𝑥)10
3. 6𝑥 cos(3𝑥 2 )
Answer Key
References:
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines: Anvil
Publishing Inc. pp. 67-82.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of America: Pearson
Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second Edition. United
States of America: Alpha Books.
64
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
SENIOR HIGH SCHOOL
(LEARNING AREA)
(QUARTER
NUMBER)
BASIC
CALCULUS
(MODULE3NUMBER)
QUARTER
- Module 1
Lesson 7:
Chain Rule
65
(DO_Q3_BASICCALCULUS_MODULE1_LESSON6)
Targets:
1. Illustrate Chain Rule of differentiation. (M11/STEM_BC11D-IIIh-2)
2. Solve problems using Chain Rule. (M11/STEM_BC11D-IIIh-i-1)
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. When do we use chain rule?
a. If there are sum/difference of two functions
b. If there are product of two functions
c. If there are quotient of two functions
d. If there are composite functions
𝑑𝑦
2. Find 𝑑𝑥 of 𝑦 = (5𝑥 2 + 3𝑥)5.
a. 𝑦 ′ = 5(5𝑥 2 + 3𝑥)4 (10𝑥 + 3)
b. 𝑦 ′ = 5(5𝑥 2 + 3𝑥)(10𝑥 + 3)
c. 𝑦 ′ = 5(5𝑥 2 + 3𝑥)4
d. 𝑦 ′ = 5(10𝑥 + 3)
3. What is the derivative of 𝑓(𝑥) = (3𝑥 + 1)5
a. 5(3𝑥 + 1)4
c. 15(3𝑥 + 1)4 + 3
b. 5(3𝑥 + 1)4 + 3
d. 15(3𝑥 + 1)4
4. What is the derivative of the function 𝑓(𝑥) = cos ln 𝑥?
sin ln 𝑥
a. sin ln x
b. -sin ln x
c.−
𝑥
d.
sin ln 𝑥 cos ln 𝑥
𝑥
5. What is 𝑓′(𝑥) of 𝑓(𝑥) = (𝑥 6 + 1)11 ?
a. 11(𝑥 6 + 1)
c. 66𝑥 5 (𝑥 6 + 1)10
b. 11(𝑥 6 + 1)10
𝑑. 66𝑥 6 (𝑥 6 + 1)10 + 1
66
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Lesson
7
Chain Rule
Composite functions are function that are too complex to find the derivative with.
But there is a rule where we can find it easily. In our last lessons we have already
illustrated on how to use Chain Rule in different scenarios. In this lesson, we are
going to apply Chain Rule to solve different problems.
In the previous lesson we have discuss the different differentiation rules. We
can apply those rules in finding the derivative of any continuous function. Let’s say
we have functions 𝑓(𝑥) and 𝑔(𝑥), we can find the derivative of the both functions.
And we can also find the derivative of the functions sum, difference, product and
quotient.
For example, let’𝑓(𝑥) = 𝑥 2 and 𝑔(𝑥) = 𝑥 + 1.
Derivative of the sum or difference of 𝑓(𝑥) and 𝑔(𝑥)
(𝑓 ± 𝑔)(𝑥) = 𝑓(𝑥) ± 𝑔(𝑥)
(𝑓 ± 𝑔)′ (𝑥) = 𝑓 ′ (𝑥) ± 𝑔′ (𝑥)
=
𝑑
𝑑
[𝑥 2 ] ± [𝑥
𝑑𝑥
𝑑𝑥
+ 1]
= 2𝑥 ± 1
Derivative of the product of 𝑓(𝑥) and 𝑔(𝑥)
(𝑓 ∙ 𝑔)(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥)
(𝑓 ∙ 𝑔)′ (𝑥) = 𝑓(𝑥)𝑔′ (𝑥) + 𝑓 ′ (𝑥)𝑔(𝑥)
𝑑
𝑑
= 𝑥 2 (𝑑𝑥 [𝑥 + 1]) + (𝑑𝑥 [𝑥 2 ]) (𝑥 + 1)
= 𝑥 2 (1) + 2𝑥(𝑥 + 1)
= 𝑥 2 + 2𝑥 2 + 2𝑥
= 3𝑥 2 + 2𝑥
67
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Derivation of the quotient of 𝑓(𝑥) and 𝑔(𝑥)
𝑓
𝑓(𝑥)
(𝑔) (𝑥) = 𝑔(𝑥)
𝑓 ′
𝑔
( ) (𝑥) =
=
=
=
𝑔(𝑥)𝑓′ (𝑥)−𝑓(𝑥)𝑔′ (𝑥)
2
(𝑔(𝑥))
𝑑
𝑑
[𝑥 2 ])−(𝑥 2 )(𝑑𝑥[𝑥+1])
𝑑𝑥
(𝑥+1)2
(𝑥+1)(
(𝑥+1)(2𝑥)−(𝑥 2 )(1)
(𝑥+1)2
2𝑥 2 +2𝑥−𝑥 2
(𝑥+1)2
𝑥 2 +2𝑥
= (𝑥+1)2
It is possible to find the derivative of all the basic operations regarding the two
functions 𝑓(𝑥) and 𝑔(𝑥). However, is it possible to use derivation rules to find the
derivative of a composite functions? For our example above, can we use it to find
(𝑓°𝑔)(𝑥) = 𝑓(𝑔(𝑥)) ? To illustrate more, how can we find the derivative of
(𝑓°𝑔)(𝑥) = (𝑥 + 1)2 ?
In the previous lesson we can define the derivative of a composite function as:
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥)
where 𝑓(𝑥) and 𝑔(𝑥) are continuous functions. But there are other processes like
making the inner function as another variable. Let say we have 𝑓(𝑥) and 𝑔(𝑥), if we
are going to find the derivative of 𝑓(𝑔(𝑥)), we can use another variable to denote 𝑔(𝑥),
let say 𝑢, making 𝑔(𝑥) = 𝑢. Hence, making the composite function 𝑓(𝑢).
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑢) ∙ 𝑢′
So, let’s have two functions: 𝑓(𝑥) = 𝑥 5 and 𝑔(𝑥) = 9𝑥 + 3 . Find the derivative of
𝑓(𝑔(𝑥)).
Solution: If we are going to use the first process, we are going to substitute
𝑔(𝑥) to the x-value of 𝑓(𝑥).
𝑓(𝑔(𝑥)) = (9𝑥 + 3)5
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥)
= 5(9𝑥 + 3)5−1 ∙ 9
= 45(9𝑥 + 3)4
68
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Therefore, the derivative of 𝑓(𝑔(𝑥)) is 45(9𝑥 + 3)4 .
Solution: If we are going to use the second process, we are going to denote
𝑔(𝑥) = 𝑢
𝑔(𝑥) = 𝑢
𝑢 = 9𝑥 + 3
𝑓(𝑔(𝑥)) = 𝑓(𝑢) ,
where 𝑢 = 9𝑥 + 3
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑢) ∙ 𝑢′
𝑑
𝑑
= 𝑑𝑢 [𝑢5 ] ∙ 𝑑𝑥 [𝑢]
𝑑
= 5𝑢4 ∙ 𝑑𝑥 [𝑢]
𝑑
= 5(9𝑥 + 3)4 ∙ 𝑑𝑥 [9𝑥 + 3]
= 5(9𝑥 + 3)4 ∙ (9)
= 45(9𝑥 + 3)4
Chain Rule
Chain Rule is the rule we can use to find the derivative of a composite function. If we
have 𝑓(𝑥) and 𝑔(𝑥), we can have the composite function (𝑓°𝑔)(𝑥) = 𝑓(𝑔(𝑥)). The rule
to find the derivative is denoted by:
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥).
There are many process or strategies that will help us find the derivative of a
composite function. It can be the normal process, or we can use an auxiliary
variable to denote another function then proceed to finding the derivative of the
function in respect to that variable. The latter process can be denoted by:
(𝑓°𝑔)′ (𝑥) = 𝑓 ′ (𝑢) ∙ 𝑢′,
where 𝑢 is another function of x.
So, let’s have examples on how to apply chain rule.
69
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Example 1. 𝑓(𝑥) = (𝑥 2 + 1)8
Solution: To find the derivative, we will apply the chain rule. We can
denote 𝑓(𝑥) = 𝑥 8 and 𝑔(𝑥) = 𝑥 2 + 1
𝑑
𝑓 ′ (𝑥) = 𝑑𝑥 [(𝑥 2 + 1)8 ]
𝑑
𝑑
= 𝑑(𝑥 2 +1) [(𝑥 2 + 1)8 ∙ 𝑑𝑥 [𝑥 2 + 1]
= 8(𝑥 2 + 1)8−1 ∙ (2𝑥 2−1 + 0)
= 8(𝑥 2 + 1)7 (2𝑥)
= 16𝑥(𝑥 2 + 1)7
𝑑
[(𝑥 2
𝑑(𝑥 2 +1)
As you can see, we denote
+ 1)8 ] which is read as
“derivative of (𝑥 2 + 1)8 with respect to (𝑥 2 + 1).” This is done because
we just wanted to find 𝑓′(𝑔(𝑥)). Thus, the derivative is 16𝑥(𝑥 2 + 1)7 .
Example. 𝑓(𝑥) = √4𝑥 + 7
Solution: We can denote 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = 4𝑥 + 7.
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑(4𝑥+7) [√4𝑥 + 7] ∙ 𝑑𝑥 [4𝑥 + 7]
𝑓 ′ (𝑥) =
𝑑
[(4𝑥
𝑑(4𝑥+7)
1
+ 7)2 ] ∙
1
1
1
1
𝑑
[4𝑥
𝑑𝑥
+ 7]
= 2 ∙ (4𝑥 + 7)2−1 ∙ (4 ∙ 1𝑥 1−1 + 0)
= 2 (4𝑥 + 7)−2 (4𝑥)
1
= 2𝑥(4𝑥 + 7)−2
=
2𝑥
1
(4𝑥+7)2
=
2𝑥
√4𝑥+7
Therefore, the derivative is
2𝑥
√4𝑥+7
Example 3: 𝑓(𝑥) = csc(4𝑥 2 )
Solution: We can denote 𝑓(𝑥) = csc 𝑥 and 𝑔(𝑥) = 4𝑥 2 .
𝑓 ′ (𝑥) =
𝑑
𝑑
[cos(4𝑥 2 )] ∙ [4𝑥 2 ]
𝑑(4𝑥 2 )
𝑑𝑥
= − sin(4𝑥 2 ) ∙ (4 ∙ 2𝑥 2−1 )
= − sin(4𝑥 2 ) (8𝑥)
= −8𝑥 sin(4𝑥 2 )
70
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Example 4: 𝑓(𝑥) = tan 𝑒 𝑥
Solution: We can denote 𝑓(𝑥) = tan 𝑥 and 𝑔(𝑥) = 𝑒 𝑥
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑(𝑒 𝑥 ) [tan(𝑒 𝑥 )] ∙ 𝑑𝑥 [𝑒 𝑥 ]
= sec 2 𝑒 𝑥 ∙ 𝑒 𝑥
= 𝑒 𝑥 sec 2(𝑒 𝑥 )
Example 5: 𝑓(𝑥) = ln(4𝑥 3 + 3𝑥 + 1)
Solution: We can denote 𝑓(𝑥) = ln 𝑥 and 𝑔(𝑥) = 4𝑥 3 + 3𝑥 + 1
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑(4𝑥 3 +3𝑥+1) [ln(4𝑥 3 + 3𝑥 + 1) ∙ 𝑑𝑥 [4𝑥 3 + 3𝑥 + 1]
1
= 4𝑥 3 +3𝑥+1 ∙ (4 ∙ 3𝑥 3−1 + 3 ∙ 𝑥 1−1 + 0)
=
1
4𝑥 3 +3𝑥+1
∙ (12𝑥 2 + 3)
12𝑥 2 +3
= 4𝑥 3 +3𝑥+1
Example 6: (𝑒 𝑥 + 11)5
Solution: We can denote 𝑓(𝑥) = 𝑥 5 and 𝑔(𝑥) = 𝑒 𝑥 + 11
𝑓 ′ (𝑥) =
𝑑
[(𝑒 𝑥
𝑑(𝑒 𝑥 +11)
+ 11)5 ] ∙
𝑑
[𝑒 𝑥
𝑑𝑥
+ 11]
= 5(𝑒 𝑥 + 11)5−1 ∙ (𝑒 𝑥 + 0)
= 5𝑒 𝑥 (𝑒 𝑥 + 11)4
Example 7: 𝑓(𝑥) = log 5 𝑥 2 + 1
Solution: We can denote 𝑓(𝑥) = log 5 𝑥 and 𝑔(𝑥) = 𝑥 2 + 1
𝑓 ′ (𝑥) =
𝑑
[log 5(𝑥 2
𝑑(𝑥 2 +1)
+ 1) ∙
𝑑
[𝑥 2
𝑑𝑥
+ 1]
1
= (𝑥 2 +1) ln 5 (2𝑥)
2𝑥
= (𝑥 2 +1) ln 5
Example 8: 𝑓(𝑥) = 10cot 𝑥
Solution: We can denote 𝑓(𝑥) = 10𝑥 and 𝑔(𝑥) = cot 𝑥
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑(cot 𝑥) [10cot 𝑥 ] ∙ 𝑑𝑥 [cot 𝑥]
= 10cot 𝑥 ln 10 ∙ − csc 2 𝑥
= −10cot 𝑥 ln 10 csc 2 𝑥
71
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Example 9: 𝑓(𝑥) = 𝑒 csc 𝑥
Solution: We can denote 𝑓(𝑥) = 𝑒 𝑥 and 𝑔(𝑥) = csc 𝑥
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑑(csc 𝑥) [𝑒 csc 𝑥 ] ∙ 𝑑𝑥 [csc 𝑥]
= 𝑒 csc 𝑥 ∙ − csc 𝑥 cot 𝑥
= −𝑒 csc 𝑥 csc 𝑥 cot 𝑥
Example 10: 𝑓(𝑥) = 𝑥 2 cos(𝑥 3 − 1)
Solution: We have cos(𝑥 3 − 1) as the composite function on 𝑓(𝑥). We
will apply product rule to find the derivative of 𝑓(𝑥)
𝑑
𝑑
𝑓 ′ (𝑥) = 𝑥 2 𝑑𝑥 [cos(𝑥 3 − 1) + 𝑑𝑥 (𝑥 2 ) cos(𝑥 3 − 1)
𝑑
𝑑
= 𝑥 2 [𝑑(cos(𝑥 3 −1)) [cos(𝑥 3 − 1)] ∙ 𝑑𝑥 [𝑥 3 − 1]] + (2𝑥)(cos(𝑥 3 − 1))
= 𝑥 2 (− sin(𝑥 3 − 1) ∙ 3𝑥 2 ) + 2𝑥 cos(𝑥 3 − 1)
= −3𝑥 4 sin(𝑥 3 − 1) + 2𝑥 cos(𝑥 3 − 1)
It is really helpful to understand chain rule because finding the derivative of
logarithmic, exponential, and trigonometric functions lies heavily on chain rule. The
examples above are some of the functions that needed to be solved using chain rule.
Apply Chain Rule
Find the derivative of the following function
1. 𝑓(𝑥) = (𝑒 𝑥 + 4𝑥 2 + 1)5
2. 𝑓(𝑥) = sec(𝑒 𝑥 )
3. 𝑓(𝑥) = ln(ln 𝑥)
4. 𝑓(𝑥) = cos(4𝑥 3 + 𝑒 𝑥 )
5. 𝑓(𝑥) = 𝑥 3 sec(𝑥 5 )
72
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
1. The derivative of a composite function (𝑓 ∘ 𝑔)(𝑥) can be obtain using
_______________________.
2. If we have two continuous functions 𝑓(𝑥) and 𝑔(𝑥), then the derivative of
𝑓(𝑔(𝑥)) is denoted by: _______________________________.
Too complex!
The function below is a combination of composite functions in one expression. Find
the derivative of the function using Chain rule.
5
𝑓(𝑥) = [tan (𝑠𝑖𝑛 (√𝑥 2 + 8𝑥))]
Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1. Let 𝑓(𝑥) = (𝑥 5 + 2)20. 𝑓 ′ (𝑥) is?
a. 20(𝑥 5 + 2)19
b. 100𝑥 4 (𝑥 5 + 2)19
c. 100𝑥 5 (𝑥 5 + 2)20
d. 100𝑥 4
2. Differentiate 𝑦 = √𝑥 2 − 1
𝑎. √𝑥 2 − 1
b.
𝑥
c.
√𝑥 2 −1
𝑥
2√𝑥 2 −1
3. What is the derivative of the function 𝑓(𝑥) = sin2 𝑥?
a. 2 sin 𝑥
b. 2 cos 𝑥
c. 2 sin 𝑥 cos 𝑥
1
d. 𝑥(𝑥 2 − 1)2
d. 2x cos x
4. Let 𝑓(𝑥) = cos 5𝑥 4 . What is the 𝑓′(𝑥)?
a. sin 20𝑥 3
b. − sin 20𝑥 3
c. 20𝑥 3 sin 5𝑥 4
d. −20𝑥 3 sin 5𝑥 4
5. Find the derivative of 𝑓(𝑥) = 𝑒 tan 𝑥 .
a. tan 𝑥 𝑒 tan 𝑥 b. tan 𝑒 𝑥
d. 𝑒 tan 𝑥 sec 2 𝑥
c. tan 𝑒 sec
73
2𝑥
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
The Chain of Chain Rules
Exit the maze by finding the correct path from the starting position to the finish
line. Use crayon to color the path
START
(2𝑥 2 − 3)5
20𝑥(2𝑥 2 − 3)4
45𝑥 2 (3𝑥 3 + 5𝑥)4
35(5𝑥 − 7)6
(𝑥 5 + 7)2
74
(5𝑥 − 7)7
7(5𝑥 − 7)6
45𝑥 2 (3𝑥 3 − 4)4
(−2𝑥 2 − 2)4
−16𝑥(−2𝑥 2 − 2)3
(6𝑥 2 − 1)5
12𝑥(6𝑥 2 − 1)4
49(7𝑥 + 1)6
(3𝑥 3 − 4)5
(−3𝑥 3 − 5𝑥)5
5(6𝑥 2 − 1)5
60𝑥(6𝑥 2 − 1)4
(4𝑥 − 1)6
24(4𝑥 − 1)5
24𝑥 2 (2𝑥 3 + 𝑥)3
8𝑥(𝑥 3 + 𝑥)3
(7𝑥 + 1)7
3(3𝑥 3 + 2𝑥)2 (9𝑥 2 + 2)
3(3𝑥 3 + 2𝑥)
2𝑥 2 (𝑥 2 − 3)4
(2𝑥 3 + 𝑥)4
(3𝑥 3 + 2𝑥)3
10𝑥 4 (𝑥 5 + 7)
FINISH
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
75
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Additional Activities
What I Can Do
𝑓 ′ (𝑥) =
4 tan4 (sin √𝑥 2 + 8𝑥) sec 2(sin √𝑥 2 + 8𝑥) cos(√𝑥 2 + 8𝑥 )(𝑥 + 4)
√𝑥 2 + 8𝑥
What I Know
11.D
12.A
13.D
14.C
15.C
What's More
1. (5𝑒 𝑥 + 40𝑥)(𝑒 𝑥 +
4𝑥 2 + 1)4
2. 𝑒 𝑥 tan 𝑒 𝑥 sec 𝑒 𝑥
1
3. 𝑥 ln 𝑥
4. − sin(4𝑥 3 +
𝑒 𝑥 )(12𝑥 2 + 𝑒 𝑥 )
Assessment
11.B
12.B
13.B
14.C
15.D
5. 3𝑥 2 sec 𝑥 5 +
5𝑥 7 tan 𝑥 5 sec 𝑥 5
References:
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines: Anvil
Publishing Inc. pp. 67-82.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of America: Pearson
Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second Edition. United
States of America: Alpha Books
76
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
SENIOR HIGH SCHOOL
(LEARNING AREA)
(QUARTER
NUMBER)
BASIC
CALCULUS
(MODULE3NUMBER)
QUARTER
- Module 1
Lesson 8:
Implicit Differentiation
77
(DO_Q3_BASICCALCULUS_MODULE1_LESSON7)
Target:
1. Illustrate implicit differentiation (M11/STEM_BC11D-IIIi-2).
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Which of the following is an example of implicit functions?
12
a. 𝑦 = 4𝑥 + 2
b. 𝑥 2 + 𝑦 2 = 4
𝑐. 𝑓(𝑥) = 𝑥
2. Find
𝑑𝑦
𝑑𝑥
of 𝑥 + 4𝑦 = 1.
1
a. − 4
b. 4
3. What is the
𝑑𝑦
𝑑𝑥
𝑑𝑦
of
𝑑𝑥
1
a.
9
4. Find
𝑑𝑦
𝑑𝑥
𝑥
−𝑦
a.
c. -1
d. 1
c. 3𝑥 2 + 3𝑦 2
d. 0
of 𝑥 3 + 𝑦 3 = 36 ?
𝑥2
a. 6 − 𝑥
5. What is
d. 𝑉 = 𝑙𝑤ℎ
b. − 𝑦2
9𝑦 + 𝑥 = 4𝑦 by implicit differentiation.
b. −
1
5
c.
1
5
6𝑦
𝑥
c. − 𝑥
d. −
1
9
of 𝑥𝑦 = 6?
b. −
𝑦
78
d.
𝑦
𝑥
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
Lesson
8
Implicit Differentiation
A function can be explicit or implicit. Explicit functions are functions which
the dependent variable 𝑦 is isolated on one side of the equation; and the other side
of the equations are for the other variable 𝑥 only. Examples of explicit functions are
𝑦 = 4𝑥 2 ; 𝑦 = √𝑥 2 + 1, and many more. Thus, taking the form 𝑦 = 𝑓(𝑥). In our past
lessons in derivatives, we usually get the derivative of these kinds of functions.
However, there are 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 functions, which are the equations that are usually
express in terms of both dependent 𝑦 and independent 𝑥 variable. These function
takes the form 𝐹(𝑥, 𝑦) = 0. In this lesson, we are going to find the derivative of the
implicit functions.
The functions that we already discussed belong to explicit functions, wherein the
function is in the for 𝑦 = 𝑓(𝑥). In finding the derivative of a function, it is necessary
to know different rules to apply in a given situation. Listed below are the rules that
we need to remember.
Basic differentiation rule
𝑑
[𝑐]
𝑑𝑥
=0
𝑑
[𝑥 𝑛 ]
𝑑𝑥
= 𝑛 ∙ 𝑥 𝑛−1
𝑑
[𝑐(𝑓(𝑥)]
𝑑𝑥
𝑑
[𝑓(𝑥) ±
𝑑𝑥
= 𝑐𝑓′(𝑥)
𝑔(𝑥)] = 𝑓 ′ (𝑥) ± 𝑔′(𝑥)
𝑑
[𝑓(𝑥)𝑔(𝑥)]
𝑑𝑥
𝑑 𝑓(𝑥)
[
]
𝑑𝑥 𝑔(𝑋)
=
= 𝑓(𝑥)𝑔′ (𝑥) + 𝑓 ′ (𝑥)𝑔(𝑥)
𝑓 ′ (𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)
𝑔(𝑥)2
Derivative of Composite Function/ Chain Rule
𝑑
[𝑓(𝑔(𝑥))]
𝑑𝑥
= 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′(𝑥)
In solving the derivative of the implicit functions, it is useful to have remember or to
have a list of rules we can apply to find the derivative of any functions.
79
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
Let us recall the equation of a circle when its center is at the origin. The
equation is written in the form 𝑥 2 + 𝑦 2 = 𝑟 2 . As we are talking earlier, there are
implicit and explicit functions. If we take the equation of the circle for example, we
can have two forms—explicit form and implicit form.
Explicit Form
Implicit Form
𝑦 = √𝑟 2 − 𝑥 2
𝑥2 + 𝑦2 = 𝑟2
In the explicit form, the variable 𝑦 is expressed as a function of 𝑥. However, in the
implicit form, the function is expressed in terms of both 𝑥 and 𝑦.
Now, if we are going to find the derivative of the explicit function 𝑦. We can simply
use the derivative rules to find the derivative of 𝑦 = √𝑟 2 − 𝑥 2
𝑦′ =
𝑑
[√𝑟 2
𝑑(𝑟 2 −𝑥 2 )
1
− 𝑥2] ∙
1
= 2(
√𝑟 2 −𝑥 2
𝑦′ =
𝑑
[𝑟 2
𝑑𝑥
− 𝑥2]
) (2𝑥)
𝑥
√𝑟 2 −𝑥 2
However, if we are given an implicit function, we cannot use the differentiation rules
we have discuss to find the derivative of the said given implicit function. Thus,
implicit differentiation is needed to find the derivative of the whole equation.
Implicit Differentiation
We have discussed earlier the difference between explicit and implicit
functions. Explicit functions are functions that are expressed in the form 𝑦 = 𝑓(𝑥),
while implicit functions are functions that are expressed in the form 𝐹(𝑥, 𝑦) = 0. Since
we have differentiation rules, we can simply find the derivative of an explicit
functions. But, if we have implicit functions, we can’t just simply apply the rules
since those are designed for explicit functions.
Implicit differentiation allows us to find the derivative of an equation when the
equation is cannot be expressed in terms of y. We can find the derivative of the
implicit function by differentiating each term of the equation. But there will be terms
with variables 𝑦. We can differentiate these terms by finding the derivative of the
terms in ‘respect to 𝑥’ while treating 𝑦 as a function of 𝑥 then solving
80
𝑑𝑦
𝑑𝑥
(derivative of
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
𝑦 with respect to 𝑥). To understand the concept of implicit differentiation, let have
some examples.
Example 1: Find the derivative of 2𝑥 3 = 2𝑦 2 + 5
Solution: Find the derivative of each term of the equation
𝑑
[2𝑥 3 ]
𝑑𝑥
𝑑
𝑑
= 𝑑𝑥 [2𝑦 2 ] + 𝑑𝑥 [5]
find the derivative of each term
𝑑𝑦
(6𝑥 2 ) = (4𝑦 𝑑𝑥 ) + (0)
apply differentiation rules
Note that we will treat the 𝑦 as another function of 𝑥. So, for us to find the
derivative of 2𝑦 2 with respect to 𝑥 (which the variable in the expression doesn’t
match the variable we are respecting with), we will use chain rule. To
differentiate 2𝑦 2 with respect to x, we will get the derivative of it with respect
to 𝑦 then multiply it by
denoted by
𝑑𝑦
.
𝑑𝑥
𝑑𝑦
.
𝑑𝑥
Wherein the derivative in implicit functions is
To continue the solution above:
6𝑥 2 = 4𝑦
𝑑𝑦
𝑑𝑥
by simplifying
𝑑𝑦
4𝑦 𝑑𝑥 = 6𝑥 2
4𝑦
𝑑𝑦
𝑑𝑥
4𝑦
𝑑𝑦
𝑑𝑥
=
=
isolating terms with
6𝑥 2
4𝑦
solving for
3𝑥 2
2𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
to left side
by dividing both sides by 4y
simplifying the fraction
Therefore, the derivative of the implicit function is
3𝑥 2
2𝑦
Example 2: Find the derivative of 4𝑥 2 = 2𝑦 3 + 4𝑦
Solution: Find the derivative of each term
𝑑
[4𝑥 2 ]
𝑑𝑥
=
𝑑
𝑑
[2𝑦 3 ] + [4𝑦]
𝑑𝑥
𝑑𝑥
(8𝑥) = (6𝑦 2
𝑑𝑦
)+
𝑑𝑥
𝑑𝑦
find the derivative of each term
𝑑𝑦
apply differentiation rules
(4 𝑑𝑥 )
𝑑𝑦
by simplifying
8𝑥 = 6𝑦 𝑑𝑥 + 4 𝑑𝑥
𝑑𝑦
𝑑𝑦
6𝑦 𝑑𝑥 + 4 𝑑𝑥 = 8𝑥
isolating terms with
𝑑𝑦
(6𝑦
𝑑𝑥
factoring out
+ 4) = 8𝑥
𝑑𝑦
(6𝑦+4)
𝑑𝑥
(6𝑦+4)
𝑑𝑦
𝑑𝑥
8𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
dividing both sides by 6𝑦 + 4
= (6𝑦+4)
4𝑥
by simplifying fractions
= (3𝑦+2)
81
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
Example 3: Find the derivative of 5𝑥 3 + 𝑥𝑦 2 = 5𝑥 3 𝑦 3
Solution: Start with finding the derivative of each term with respect to 𝑥
𝑑
𝑑
[5𝑥 3 ] + [𝑥𝑦 2 ]
𝑑𝑥
𝑑𝑥
(15𝑥 2 ) + [𝑥 ∙ 2𝑦
𝑑𝑦
𝑑𝑥
=
𝑑
[5𝑥 3 𝑦 3 ]
𝑑𝑥
finding derivative of each term
𝑑𝑦
+ 1 ∙ 𝑦 2 ] = 5 [𝑥 3 ∙ 3𝑦 2 𝑑𝑥 + 3𝑥 2 ∙ 𝑦 3 ]
As we can see, to find the derivative of 𝑥𝑦 2 and 5𝑥 3 𝑦 3, we use product rule:
𝑑𝑦
𝑑𝑦
15𝑥 2 + 2𝑥𝑦 𝑑𝑥 + 𝑦 2 = 15𝑥 3 𝑦 2 𝑑𝑥 + 15𝑥 2 𝑦 2
𝑑𝑦
𝑑𝑦
by simplifying
𝑑𝑦
2𝑥𝑦 𝑑𝑥 − 15𝑥 3 𝑦 2 𝑑𝑥 = 15𝑥 2 𝑦 2 − 15𝑥 2 − 𝑦 2
by isolating
𝑑𝑦
(2𝑥𝑦
𝑑𝑥
by factoring out
𝑑𝑦
by isolating the
𝑑𝑦
− 15𝑥 3 𝑦 2 ) = 15𝑥 2 𝑦 2 − 15𝑥 2 − 𝑦 2
𝑑𝑦
(2𝑥𝑦−15𝑥 3 𝑦 2 )
𝑑𝑥
(2𝑥𝑦−15𝑥 3 𝑦 2 )
𝑑𝑦
𝑑𝑥
=
=
15𝑥 2 𝑦 2 −15𝑥 2 −𝑦 2
(2𝑥𝑦−15𝑥 3 𝑦 2 )
15𝑥 2 𝑦 2 −15𝑥 2 −𝑦 2
2𝑥𝑦−15𝑥 3 𝑦 2
𝑑𝑥
terms on one side
𝑑𝑥
𝑑𝑥
by simplifying
Example 4: Find the derivative of 5𝑦 2 = 2𝑥 3 − 5𝑦
Solution: start with finding the derivative of each term with respect to x
𝑑
[5𝑦 2 ]
𝑑𝑥
(10𝑦
𝑑
𝑑
= 𝑑𝑥 [2𝑥 3 ] − 𝑑𝑥 [5𝑦]
𝑑𝑦
)
𝑑𝑥
= (6𝑥 2 ) − (5
𝑑𝑦
finding derivative of each term
𝑑𝑦
)
𝑑𝑥
applying differentiation rules
𝑑𝑦
10𝑦 𝑑𝑥 = 6𝑥 2 − 5 𝑑𝑥
𝑑𝑦
by simplifying
𝑑𝑦
𝑑𝑦
10𝑦 𝑑𝑥 + 5 𝑑𝑥 = 6𝑥 2
isolate
𝑑𝑦
(10𝑦
𝑑𝑥
factoring out
+ 5) = 6𝑥 2
𝑑𝑦
(10𝑦+5)
𝑑𝑥
(10𝑦+5)
𝑑𝑦
𝑑𝑥
6𝑥 2
𝑑𝑥
terms in one side
𝑑𝑦
𝑑𝑥
dividing both sides to isolate
= 10𝑦+5
6𝑥 2
𝑑𝑦
𝑑𝑥
simplifying
= 10𝑦+5
Example 5. Find the derivative of 2𝑥 3 = (3𝑥𝑦 + 1)2
Solution. Use chain rule to find the derivative of the right side of equation
𝑑
[2𝑥 3 ]
𝑑𝑥
= 𝑑𝑥 [(3𝑥𝑦 + 1)2 ]
𝑑
𝑑
[2𝑥 3 ]
𝑑𝑥
=
𝑑
[2𝑥 3 ]
𝑑𝑥
= 𝑑(3𝑥𝑦+1) [(3𝑥𝑦 + 1)2 ] ∙ [3 𝑑𝑥 [𝑥𝑦] + 𝑑𝑥 [1]]
𝑑
[(3𝑥𝑦
𝑑(3𝑥𝑦+1)
finding derivative of each term
+ 1)2 ] ∙
𝑑
[3𝑥𝑦
𝑑𝑥
𝑑
𝑑
82
+ 1] use chain rule
𝑑
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
You will use product rule to find the derivative of the inner function 3𝑥𝑦 + 1
with respect to x.
6𝑥 2 = 2(3𝑥𝑦 + 1) ∙ [3 (𝑥
𝑑𝑦
𝑑𝑥
+ 𝑦) + 0]
𝑑𝑦
6𝑥 2 = (6𝑥𝑦 + 2) (3𝑥 𝑑𝑥 + 3𝑦 )
6𝑥 2 = 18𝑥 2 𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑦
+ 18𝑥𝑦 2 + 6𝑥
apply rules of derivative
by simplifying
𝑑𝑦
𝑑𝑥
+ 6𝑦
𝑑𝑦
by using FOIL method
𝑑𝑦
18𝑥 2 𝑦 𝑑𝑥 + 6𝑥 𝑑𝑥 = 6𝑥 2 − 18𝑥𝑦 2 − 6𝑦
isolating
𝑑𝑦
(18𝑥 2 𝑦
𝑑𝑥
factoring out
+ 6𝑥) = 6𝑥 2 − 18𝑥𝑦 2 − 6𝑦
𝑑𝑦
(18𝑥 2 𝑦+6𝑥)
𝑑𝑥
(18𝑥 2 𝑦+6𝑥)
=
6𝑥 2 −18𝑥𝑦 2 −6𝑦
(18𝑥 2 𝑦+6𝑥)
𝑑𝑦
𝑑𝑥
=
6(𝑥 2 −3𝑥𝑦 2 −𝑦)
6(3𝑥 2 𝑦+𝑥)
𝑑𝑦
𝑑𝑥
=
𝑥 2 −3𝑥𝑦 2 −𝑦
3𝑥 2 𝑦+𝑥
finding
𝑑𝑦
𝑑𝑥
𝑑𝑥
terms on left side
𝑑𝑦
𝑑𝑥
by dividing (18𝑥 2 𝑦 + 6𝑥)
by simplifying fractions
What do you imply?
Use implicit differentiation to find
1.
2.
3.
4.
5.
𝑑𝑦
𝑑𝑥
in terms of x and y.
1. 9𝑦 6 + 9𝑥 5 = 2𝑥 6 + 7𝑦 2
2𝑥 4 𝑦 5 = 5𝑥 4 − 5𝑥 2 𝑦 3
5𝑦 3 + 8𝑥 2 = 7𝑦 2
3𝑥 4 = 9𝑥 5 𝑦 6
(3𝑥 5 𝑦 5 + 6)6 = −3𝑥 4
1. The function can be classified into two forms: ___________________ and
__________________________.
2. Explicit functions are functions that can be written in the form
____________________.
83
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
3. Implicit functions are functions that can be written in the form ___________.
4. Implicit differentiation allows us to find the derivative when the equation is
___________________________.
5. In implicit differentiation, we treat 𝑦 as another function of 𝑥, thus, making
it differentiable by __________________.
Slope of Tangent Line
We can find the slope of the equation of the tangent line by using implicit
differentiation. After finding
of the tangent point.
𝑑𝑦
,
𝑑𝑥
to find the slope, we will substitute the 𝑥 and 𝑦 values
Problem: Find the slope of the tangent line to the graph of 4𝑥 + 𝑥𝑦 − 3𝑦 2 = 6 at the
point (3, 2).
Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.
1. Which of the following is an explicit function?
a. 𝑦 = 20(𝑥 5 + 2)19
b. 𝑥 2 𝑦 2 = 4𝑥𝑦 + 7
c. 𝑥 4 − 𝑦 4 = 1
d. √𝑥𝑦 + 1 = 9𝑥 2 − 2𝑦 2
84
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
𝑑𝑦
2. What is the 𝑑𝑥 of the equation: 2𝑥𝑦 3 − 𝑥 2 𝑦 = 2?
a. 2𝑥𝑦 − 2𝑦 3
2𝑦
b. − 𝑥
c.
2𝑦(𝑥−𝑦 2 )
𝑥(6𝑦 2 −𝑥)
d. 2
3. What is derivative of y with respect to x of the function 𝑥𝑦 + 𝑦 2 = 2
𝑦
a. − (𝑥+2𝑦)
3𝑦
𝑥
𝑦
𝑥+2𝑦
3𝑥
−𝑦
b. −
c.
d.
4. Find the slope of the tangent line to 𝑥 3 + 2𝑥𝑦 − 𝑦 2 = 11 at (2, 3)
4
a. − 7
b. −9
c. 12
d. 9
5. Find the slope of the tangent line to the curve 𝑥 3 + 𝑥 2 𝑦 + 2𝑦 4 = 8 at the point
(2,0).
a. -3
b. 0
c. 8
d. 9
The Valentine Equation
The equation (𝑥 2 + 𝑦 2 − 1)3 − 𝑥 2 𝑦 3 = 0 is a function containing the point (1,1). Find
the slope of the tangent line at that point.
85
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
86
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
What I Can Do
Additional Activities
Given:
Given:
(𝑥 2 + 𝑦 2 − 1)3 − 𝑥 2 𝑦 3 = 0 Derivative:
4𝑥 + 𝑥𝑦 − 3𝑦 2 = 6
𝑑𝑦 𝑥(−24𝑥 4 + 4𝑥 2 − 4𝑥 2 𝑦 2 + 𝑥𝑦 3 + 4𝑦 2 − 2𝑦 4 − 2)
=
𝑑𝑥
𝑦(2𝑥 4 − 𝑥 3 𝑦 + 4𝑥 2 𝑦 2 − 4𝑥 2 + 2𝑦 4 + 2 − 4𝑦 2 )
Slope of tangent at (3,2)
𝑑𝑦
= −1
𝑑𝑥
What I Know
=
𝑑𝑦
𝑑𝑥
7.
=
𝑑𝑦
𝑑𝑥
6.
12𝑥 5 −45𝑥 4
54𝑦 5 −14𝑦
20𝑥 2 −10𝑦3 −8𝑥 2 𝑦 5
10𝑥 3 𝑦 4 +15𝑥𝑦 2
=
𝑑𝑦
𝑑𝑥
9.
= 15𝑦2 −14𝑦
𝑑𝑦
𝑑𝑥
8.
𝑑𝑥
Derivative:
𝑑𝑦 −𝑦 − 4
=
𝑑𝑥 𝑥 − 6𝑦
Slope of tangent at (3,2)
𝑑𝑦
= 2/3
𝑑𝑥
What's More
16.B
17.A
18.B
19.B
20.D
𝑑𝑦
10.
Assessment
16.A
17.C
18.A
19.D
20.A
16𝑥
4−15𝑥𝑦 6
18𝑥 2 𝑦 5
5
=
−12𝑥 3 −90𝑥 4 𝑦 5 (𝑥 5 𝑦 5 +6)
90𝑥 5 𝑦 4 (𝑥 5 𝑦 5 +6)5
References:
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines: Anvil
Publishing Inc. pp. 67-82.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of America: Pearson
Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second Edition. United
States of America: Alpha Books
87
(DO_Q3_BASICCALCULUS_MODULE1_LESSON8)
SENIOR HIGH SCHOOL
(LEARNING AREA)
(QUARTER
NUMBER)
BASIC
CALCULUS
(MODULE3NUMBER)
QUARTER
- Module 1
Lesson 9:
Related Rates
88
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Targets:
1. Solve problems (including logarithmic, and inverse trigonometric functions)
using implicit differentiation (M11/STEM_BC11D-IIIi-j-1),and
2. Solve situational problems involving related rated (M11/STEM_BC11D-IIIi-j-2)
Choose the letter of the best answer. Write the chosen letter on a separate sheet of
paper.
1. Rachel is standing atop a 13 ft ladder. The ladder is leaning against a
vertical wall. The ladder starts sliding away from the wall at a rate of 3
ft/sec. How fast is the ladder sliding down the wall when the tip of the
ladder is 5 ft high?
a. 3 𝑓𝑡/𝑠
c. −7.2 𝑓𝑡/𝑠
b. 7.2 𝑓𝑡/𝑠
d. 12 𝑓𝑡/𝑠
2. Bry is standing atop a 13 ft ladder. The ladder is leaning against a vertical
wall. The ladder starts sliding away from the wall at a rate of 3 ft/sec. How
fast is the angle between the tip of the ladder and the house changing when
the ladder is 5 ft high? Hint: Use a trig function.
a. 1 𝑑𝑒𝑔/𝑠
c. −0.5 𝑑𝑒𝑔/𝑠
b. 0.6 𝑑𝑒𝑔/𝑠
𝑑. The angle is not changing
3. A spherical snowball melts so that its radius decreases at a rate of 4 𝑖𝑛/𝑠𝑒𝑐.
At what rate is the volume of the snowball changing when the radius is 4 𝑖𝑛?
a. −262𝜋 𝑖𝑛3 /𝑠
c. −247𝜋 𝑖𝑛3 /𝑠
b. −256𝜋 𝑖𝑛3 /𝑠
d. 262𝜋 𝑖𝑛3 /𝑠
4. Oil spilling from a ruptured tanker spreads in a circle on the surface of the
ocean. The radius of the spill increases at a rate of 5 𝑚/𝑚𝑖𝑛. How fast is the
area of the spill increasing when the radius is 5 𝑚?
a. 50𝜋 𝑚2 /𝑚𝑖𝑛
c. 47𝜋 𝑚2 /𝑚𝑖𝑛
b. 52𝜋 𝑚2 /𝑚𝑖𝑛
d. 40𝜋 𝑚2 /𝑚𝑖𝑛
5. A spherical snowball is rolled in fresh snow, causing it to grow so that its
radius increases at a rate of 4 in/sec. How fast is the volume of the snowball
increasing when the radius is 9 in?
a. 1296𝜋 𝑖𝑛3 /𝑠
c. −1296𝜋 𝑖𝑛3 /𝑠
b. 1303𝜋 𝑖𝑛3 /𝑠
d. −1303𝜋 𝑖𝑛3 /𝑠
89
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Lesson
9
Related Rates
One of the most popular Calculus applications are what we call related rates
problems. Related rates problems are problems that figures out how quickly the
variable changes if you know how the other variable changes. The problems regarding
related rates varies from one another, but the procedure is the same. In the previous
lesson, we discuss implicit differentiation. In this lesson, we are going apply implicit
differentiation to answer problems involving related rates.
In solving related rates problem, it necessary to know the process of implicit
differentiation. Most of the problems rely on using implicit differentiation. Let’s have
a recall on how to use implicit differentiation.
Find
𝑑𝑦
𝑑𝑥
of 4𝑥 2 = 2𝑦 3 + 4𝑦
Solution: Differentiate each term in the equation.
𝑑
[4𝑥 2 ]
𝑑𝑥
𝑑
𝑑
= 𝑑𝑥 [2𝑦 3 ] + 𝑑𝑥 [4𝑦]
𝑑𝑦
differentiate each term
𝑑𝑦
4(2𝑥) = 2(3𝑦 2 ) 𝑑𝑥 + 4 𝑑𝑥
𝑑𝑦
apply differentiation rules
𝑑𝑦
8𝑥 = 6𝑦 2 𝑑𝑥 + 4 𝑑𝑥
6𝑦 2
𝑑𝑦
𝑑𝑥
+4
(6𝑦 2 + 4)
𝑑𝑦
𝑑𝑥
𝑑𝑦
(6𝑦 2 +4)𝑑𝑥
6𝑦 2 +4
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
by simplifying
by isolating terms with
= 8𝑥
factor out
= 8𝑥
8𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
divide both sides by 6𝑦 2 + 4
= 6𝑦2 +4
4𝑥
simplify fractions.
= 3𝑦2 +2
The process above shows the process of implicit differentiation. We need to be used
to this process to answer related rates problems.
90
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
When the baby is born, then its height and weight will be increasing over time.
The increase in both height and weight depends on different factors. These factors
are (but not limited to) the nutrient intake of the baby, the baby’s gender, the length
of pregnancy, mother’s nutrition during pregnancy, mother’s lifestyle during
pregnancy, and birth order. For example, if we are going to find out the relationship
of the changes between these factors, then we may wish to know how that factors
affects the rate of change in the baby’s weight and height. These considerations give
rise to related rates problem.
Implicit Differentiation (Logarithmic & Trigonometric)
In the previous lessons, we illustrated the process on how to find the
𝑑𝑦
𝑑𝑥
of the implicit
functions by using basic differentiation rules. But, not all implicit equations are not
that simple. Some equations include logarithmic, exponential, and even
trigonometric expression. So, to have an ease about this problem, let us recall the
rules regarding these kinds of functions.
Derivative of Exponential Functions
𝑑
[𝑎 𝑥 ]
𝑑𝑥
= 𝑎 𝑥 ln 𝑎
𝑑
[𝑒 𝑥 ]
𝑑𝑥
= 𝑒𝑥
Derivative of Logarithmic Functions
𝑑
[log 𝑎
𝑑𝑥
𝑥] = 𝑥 ln 𝑎
1
𝑑
[ln 𝑥]
𝑑𝑥
=𝑥
1
Derivative of Trigonometric Functions
𝑑
[sin 𝑥]
𝑑𝑥
= cos 𝑥
𝑑
[csc 𝑥]
𝑑𝑥
= − csc 𝑥 cot 𝑥
𝑑
[cos 𝑥]
𝑑𝑥
= − sin 𝑥
𝑑
[sec 𝑥]
𝑑𝑥
= sec 𝑥 tan 𝑥
𝑑
[tan 𝑥]
𝑑𝑥
= sec 2 𝑥
𝑑
[cot 𝑥]
𝑑𝑥
= − csc 2 𝑥
Along with these rules, we must also consider remembering chain rule since most
of the problems regarding the functions above heavily use chain rule. So with that
in mind, let’s continue differentiating implicit functions.
91
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Example 1. Find
𝑑𝑦
𝑑𝑥
of 𝑒 𝑥𝑦 = 3𝑒 3𝑥 + 𝑒 5𝑦
Solution: Start with differentiating each term
𝑑
[𝑒 𝑥𝑦 ]
𝑑𝑥
𝑑
𝑑
= 3 𝑑𝑥 [𝑒 3𝑥 ] + 𝑑𝑥 [𝑒 5𝑦 ]
𝑑
𝑑
[𝑒 𝑥𝑦 ] ∙ [𝑥𝑦]
𝑑(𝑒 𝑥𝑦 )
𝑑𝑥
[
= 3[
differentiate each term
𝑑
𝑑
𝑑
𝑑
[𝑒 3𝑥 ] ∙ [3𝑥]] + [ 5𝑦 [𝑒 5𝑦 ] ∙ [5𝑦]]
𝑑𝑒 3𝑥
𝑑𝑥
𝑑𝑒
𝑑𝑥
As we can see, there more than 1 composite expressions. So, we will use
chain rule to find the derivative.
𝑒 𝑥𝑦 (𝑥
𝑑𝑦
𝑑𝑥
+ 𝑦) = 3(3𝑒 3𝑥 ) + 5𝑒 5𝑦
𝑑𝑦
𝑑𝑦
𝑑𝑥
by applying rules of derivatives
𝑑𝑦
𝑒 𝑥𝑦 𝑥 𝑑𝑥 + 𝑒 𝑥𝑦 𝑦 = 9𝑒 3𝑥 + 5𝑒 5𝑦 𝑑𝑥
𝑒 𝑥𝑦 𝑥
𝑑𝑦
𝑑𝑥
− 5𝑒 5𝑦
(𝑒 𝑥𝑦 𝑥 − 5𝑒 5𝑦 )
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
(𝑒 𝑥𝑦 𝑥−5𝑒 5𝑦 )𝑑𝑥
(𝑒 𝑥𝑦 𝑥−5𝑒 5𝑦 )
𝑑𝑦
𝑑𝑥
by simplifying
= 9𝑒 3𝑥 + 𝑒 𝑥𝑦 𝑦
isolating
= 9𝑒 3𝑥 + 𝑒 𝑥𝑦 𝑦
𝑑𝑦
𝑑𝑥
terms on the left side
by factoring out
9𝑒 3𝑥 +𝑒 𝑥𝑦 𝑦
𝑑𝑦
𝑑𝑥
dividing both sides to isolate
= (𝑒 𝑥𝑦 𝑥−5𝑒 5𝑦 )
𝑑𝑦
𝑑𝑥
9𝑒 3𝑥 +𝑒 𝑥𝑦 𝑦
= 𝑒 𝑥𝑦 𝑥−5𝑒 5𝑦
Example 2. Find
𝑑𝑦
𝑑𝑥
of sin 𝑥 2 = 4𝑥 3 𝑒 𝑦
Solution: As always, differentiate each term.
𝑑
[sin 𝑥 2 ]
𝑑𝑥
=
𝑑
[4𝑥 3
𝑑𝑥
𝑒𝑦]
𝑑
𝑑
[sin 𝑥 2 ] ∙ [𝑥 2 ]
𝑑(sin 𝑥 2 )
𝑑𝑥
differentiate each term
𝑑
𝑑
= 4𝑥 3 [𝑑𝑥 (𝑒 𝑦 )] + 𝑒 𝑦 [𝑑𝑥 (4𝑥 3 )]
For us to find the derivative of each term, we will use chain rule and product
rule.
𝑑𝑦
cos 𝑥 2 ∙ 2𝑥 = 4𝑥 3 𝑒 𝑦 𝑑𝑥 + 𝑒 𝑦 ∙ 12𝑥 2
𝑑𝑦
2𝑥 cos 𝑥 2 = 4𝑥 3 𝑒 𝑦 𝑑𝑥 + 12𝑥 2 𝑒 𝑦
by simplifying
𝑑𝑦
4𝑥 3 𝑒 𝑦 𝑑𝑥 = 2𝑥 cos 𝑥 2 − 12𝑥 2 𝑒 𝑦
𝑑𝑦
𝑑𝑥
4𝑥 3 𝑒 𝑦
4𝑥 3 𝑒 𝑦
𝑑𝑦
𝑑𝑥
=
=
differentiate in respect to x
by simplifying
2𝑥 cos 𝑥 2 −12𝑥 2 𝑒 𝑦
4𝑥 3 𝑒 𝑦
dividing both sides to isolate
cos 𝑥 2 −6𝑥𝑒 𝑦
2𝑥 2 𝑒 𝑦
𝑑𝑦
𝑑𝑥
by simplifying the fraction.
92
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Example 3. Find
𝑑𝑦
𝑑𝑥
of 4𝑥 3 𝑦 2 + ln 𝑥 2 = 7𝑥 + 𝑦
Solution: Start with differentiating each term
𝑑
𝑑
[4𝑥 3 𝑦 2 ] + [ln 𝑥 2 ]
𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑑
= 𝑑𝑥 [7𝑥] + 𝑑𝑥 [𝑦]
𝑑
𝑑
𝑑
differentiate each term
𝑑
4 𝑑𝑥 [𝑥 3 𝑦 2 ] + 𝑑(ln 𝑥 2 ) [ln 𝑥 2 ] ∙ 𝑑𝑥 [𝑥 2 ] = 7 𝑑𝑥 [𝑥] + 𝑑𝑥 [𝑦]
As we can see above, we use different differentiation rules to differentiate the
equation
4 (𝑥 3 ∙ 2𝑦
𝑑𝑦
𝑑𝑥
+ 3𝑥 2 𝑦 2 ) +
𝑑𝑦
2
1
𝑥2
∙ 2𝑥 = 7 +
𝑑𝑦
8𝑥 3 𝑦 𝑑𝑥 + 12𝑥 2 𝑦 2 + 𝑥 = 7 + 𝑑𝑥
𝑑𝑦
𝑑𝑦
2
𝑑𝑦
𝑑𝑥
𝑑𝑦
(8𝑥 3 𝑦−1)𝑑𝑥
(8𝑥 3 𝑦−1)
𝑑𝑦
𝑑𝑥
=
=
= 7 − 12𝑥 2 𝑦 2 −
differentiate in respect to x
by simplifying
8𝑥 3 𝑦 𝑑𝑥 − 𝑑𝑥 = 7 − 12𝑥 2 𝑦 2 − 𝑥
(8𝑥 3 𝑦 − 1)
𝑑𝑦
𝑑𝑥
isolating
2
𝑥
2
𝑥
7−12𝑥 2 𝑦 2 −
𝑑𝑦
𝑑𝑥
terms on the left side
by factoring out
𝑑𝑦
𝑑𝑥
dividing both sides to isolate
(8𝑥 3 𝑦−1)
𝑑𝑦
𝑑𝑥
7𝑥−12𝑥 3 𝑦 2 −2
8𝑥 4 𝑦−𝑥
To further simplify
𝑑𝑦
,
𝑑𝑥
we multiply 𝑥 to both numerator and denominator. We
do that in order to have a simple fraction and not complex fraction.
Related Rates
One of the applications of Calculus in real-life situations is problems regarding
related rates. In these problems, we can see two variables are changing with respect
to time. In line with that, we can understand the changes of one variable in relation
to another variable. Related rates problems are one of the most feared problems by
the students, but there are strategies to help you deal with it.
According to Birkett (2014), there are four steps in solving related rates problems.
1. Draw a picture of the physical situation;
2. Write an equation that relates the quantities of interest;
3. Take the derivative with respect to time of both sides of equation.
Remember chain rule; and
4. Solve for the quantity you are after.
As one of the steps in solving related rates problem, we are going to write an equation
relating the variables together. Some of the relationship we are probably going to use
are: (1) geometric relationships, (2) trigonometric relationships, (3) Pythagorean
theorem, and (4) similar triangle relationships.
93
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Example 1: A certain particle is moving from left to right following the path generated
by 𝑦 = 5𝑥 2 𝑒 𝑥 . If the particle has a vertical rate of change of
𝑑𝑦
𝑑𝑥
= 12 𝑓𝑡/𝑠𝑒𝑐 when 𝑥 =
−1. What is the horizontal rate of change if 𝑥 = −1
Solution:
i.
Draw a picture of the situation.
The situation is simple since the function is already given.
𝑑𝑦
𝑓𝑡
=7
𝑑𝑡
𝑠𝑒𝑐
𝑑𝑦
=?
𝑑𝑡
ii.
iii.
Write an equation for that relates the variable
Since the equation is given, we don’t need to find it anymore.
𝑦 = 5𝑥 2 𝑒 𝑥
Find the derivative with respect to time.
𝑑𝑦
𝑑𝑡
𝑑𝑦
∙ 𝑒𝑥]
differentiate with respect to 𝑡
𝑑𝑥
𝑑𝑥
= 5𝑥 2 𝑒 𝑥 𝑑𝑡 + 10𝑥𝑒 𝑥 𝑑𝑡
𝑑𝑡
𝑑𝑦
=
𝑑𝑡
iv.
𝑑𝑥
[5𝑥 2
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
by using product rule
(5𝑥 2 𝑒 𝑥 + 10𝑥𝑒 𝑥 )
by simplifying
Solve for the variable you are after.
𝑑𝑦
𝑑𝑡
is already given on the problem which is
find the horizontal rate of change
𝑑𝑥
,
𝑑𝑡
𝑑𝑦
𝑑𝑡
= 7 when 𝑥 = −1. So, to
will substitute the given to the
function.
𝑑𝑦
=
𝑑𝑡
𝑑𝑥
(7) =
7=
(5𝑥 2 𝑒 𝑥 + 10𝑥𝑒 𝑥 )
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝑡
(5(−1)2 𝑒 −1 + 10(−1)𝑒 −1 ) by substitution
(5𝑒 −1 − 10𝑒 −1 )
7=
𝑑𝑥 5
(
𝑑𝑡 𝑒
7=
𝑑𝑥
5
(− 𝑒)
𝑑𝑡
7
5
𝑒
(− )
𝑑𝑥
𝑑𝑡
=
−
10
)
𝑒
by simplifying
𝑑𝑥
5
(− )
𝑑𝑡
𝑒
5
(− )
𝑒
=−
7𝑒
5
by isolating
𝑑𝑥
𝑑𝑡
≈ −3.81 𝑓𝑡/𝑠𝑒𝑐
94
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Therefore, the horizontal rate of change is −
7𝑒 𝑓𝑡
.
5 𝑠𝑒𝑐
The negative sign
indicates decreasing speed of the particle along the function.
Example 2. An ice cube with 5 in as the length of its side is melting. Assuming that
it resembles a cube while melting, at approximately what rate is its volume
changing when its side length is 2.5 in and is decreasing at a rate of 0.2 in/sec?
i.
Illustrate the given situation
0.2
𝑖𝑛
𝑠𝑒𝑐
5 𝑖𝑛
ii.
iii.
Write an equation for that relates the variable
We will use the volume of the cube since we are going to know the
rate of change in the volume considering its length of sides.
𝑉 = 𝑠3
where 𝑠 is the length of the side
Find the derivative with respect to time.
𝑑𝑉 𝑑𝑠 3
= [𝑠 ]
𝑑𝑡 𝑑𝑡
𝑑𝑉
𝑑𝑠
= 3𝑠 2
𝑑𝑡
𝑑𝑡
iv.
Solve for the variable you are after.
The given values are the rate of change of 𝑠 are the ice-cube melts
𝑑𝑠
𝑑𝑡
𝑖𝑛
= −0.2 𝑠𝑒𝑐 , and 𝑠 = 2.5 𝑖𝑛. To solve for
𝑑𝑉
𝑑𝑡
, we will substitute these
values to the formula above. Take note that we use negative sign
to the rate of change because the rate is decreasing.
𝑑𝑉
𝑑𝑠
= 3𝑠 2
𝑑𝑡
𝑑𝑡
𝑑𝑉
𝑑𝑡
𝑑𝑉
𝑑𝑡
= 3(2.5)2 (−0.2 )
𝑖𝑛3
= −3.75 𝑠𝑒𝑐
Therefore, the volume of the ice cube is changing at the rate of
𝑖𝑛3
−3.75 𝑠𝑒𝑐 .
95
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Example 3. Two cars leave the same station and travels in perpendicular
direction. Car A travels north at a speed of 40 km per hour; Car B travels east
at the speed of 65 km per hour. At what rate is the distance between the two
cars changing 3 hours into the trip.
i.
Illustrate the given situation
As we can see, the two cars and the station created a right triangle
after travelling 3 hours.
ii.
Write an equation for that relates the variable
Since we have a right triangle, we can use Pythagorean Theorem.
𝑎2 + 𝑏 2 = 𝑐 2
Find the derivative with respect to time
iii.
𝑑 2
𝑑
𝑑
[𝑎 ] + [𝑏 2 ] = [𝑐 2 ]
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑎
𝑑𝑏
𝑑𝑐
2𝑎 𝑑𝑡 + 2𝑏 𝑑𝑡 = 2𝑐 𝑑𝑡
iv.
Solve for the variable you are after.
From the problem, we are given the values
𝑑𝑎
𝑑𝑡
= 40 and
𝑑𝑏
𝑑𝑡
= 65. The
values 𝑎 and 𝑏 are given on the illustration. And lastly, 𝑐 are the
distance of the two cars after 3 hours. So, to find 𝑐, we will use the
Pythagorean Theorem
𝑎2 + 𝑏 2 = 𝑐 2
1202 + 1952 = 𝑐 2
𝑐 = √1202 + 1952
𝑐 = 15√233
To solve for the rate of change of the distance of the two cars after 3
hours, we will substitute the given values to the equation.
2𝑎
𝑑𝑎
𝑑𝑡
+ 2𝑏
𝑑𝑏
𝑑𝑡
= 2𝑐
𝑑𝑐
𝑑𝑡
𝑑𝑐
2(120)(40) + 2(195)(65) = 2 ∙ 15√233 𝑑𝑡
𝑑𝑐
9600 + 25350 = 30√233 𝑑𝑡
34950 = 30√233
𝑑𝑐
𝑑𝑡
96
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
34950
30√233
𝑑𝑐
𝑑𝑡
30√233 𝑑𝑐
√233 𝑑𝑡
= 30
𝑘𝑚
= 5√233 ≈ 76.32 ℎ𝑟
Therefore, the rate of change of the distance between the two cars is
approximately 76.32
𝑘𝑚
.
ℎ𝑟
What do you imply?
Use implicit differentiation to find
𝑑𝑦
𝑑𝑥
in terms of x and y.
1. sin 𝑦 − cos 𝑥𝑦 = 𝑥 − 𝑦
2. 𝑒 𝑥 + cos 𝑦 = ln 𝑦 6
The Falling Ladder
A 36 feet long leans against a vertical wall. Unfortunately, it rains and the ladder
slips downward at a rate of 4 feet per second. How fast does the lower end of ladder
slips on the ground when it is 20 feet from the wall?
𝑑𝑦
𝑓𝑡
= −3
𝑑𝑡
𝑠
𝑑𝑥
=?
𝑑𝑡
97
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
1. It is necessary to remember __________________ rules regarding exponential,
logarithmic and trigonometric if we are solving implicit equations including
those functions.
2. Chain rule is heavily used in differentiating ____________ equations including
exponential, logarithmic, and trigonometric functions.
3. The derivative of the function is similar to the _________________ with respect
to time
There are strategies in solving related rates problem. These are: (1) drawing a
picture of the physical situation; (2) writing an equation that relates the
quantities of interest; (3) differentiating with respect to time of both sides of
equation; and (4) solving for the quantity you are after.
There’s a hole in a tank!
A cylindrical tank 2 m tall, and 1 m wide, and was initially full of water. It is now
being drained at the rate of 15
𝑐𝑚3
.
𝑠
At what rate is the water level falling when the
water is halfway down the tank?
2𝑚
𝑑ℎ
=?
𝑑𝑡
1𝑚
98
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Multiple Choice. Choose the letter of the best answer. Write the chosen letter
on a separate sheet of paper.
1. An observer stands 2400 ft away from a launch pad to observe a rocket launch.
The rocket blasts off and maintains a velocity of 200 ft/sec. Assume the scenario can
be modeled as a right triangle. How fast is the observer to rocket distance changing
when the rocket is 700 ft from the ground?
𝑎. 56 𝑓𝑡/𝑠
c. 57 𝑓𝑡/𝑠
𝑏. 52 𝑓𝑡/𝑠
d. 61 𝑓𝑡/𝑠
2. Emma is starting to clean up after a birthday party. She begins deflating each
spherical balloon by puncturing a hole in each. The air leaves the balloon at a
constant rate of 2 cm3/sec. How fast is the diameter changing when the diameter is
8 cm?
1
1
𝑎. − 16 𝜋 𝑐𝑚/𝑠
c. − 16 𝑐𝑚/𝑠
1
4
1
4
b. − 𝜋 𝑐𝑚/𝑠
d. − 𝑐𝑚/𝑠
3. Ethan is sitting on the edge of a dock tossing rocks into the water. As each rock
hits the water, small circles appear traveling outward from the point of impact. The
radius of the circle is changing at a rate of 5 in/sec. How fast is the area of the outer
circle changing when the diameter is 8 in?
𝑎. 80𝜋 𝑖𝑛/𝑠
c. 20𝜋 𝑖𝑛/𝑠
d. 40𝜋 𝑖𝑛/𝑠
𝑏. 60𝜋 𝑖𝑛/𝑠
4. Louisa and Karis were each dropped off at the same bus stop. Louisa’s bus drops
her of at 3:30 whereas Karis is dropped off ten minutes later. Louisa runs home at a
constant rate of 6 mph and Karis runs home at 3 mph. Louisa lives north of the bus
stop and Karis lives to the east. How fast is the area formed by Louisa, Karis and
the bus stop changing at 4:00?
𝑎. 7.5 𝑚𝑖 2 /ℎ𝑟
c. 9 𝑚𝑖 2 /ℎ𝑟
d. 6.25 𝑚𝑖 2 /ℎ𝑟
𝑏. 8 𝑚𝑖 2 /ℎ𝑟
5. Devin set up a toy rocket. For safety, he stands 6 meters from the rocket. He sets
off the rocket and it heads straight up at a constant rate of 4 m/s. How fast is the
distance between the rocket and Devin changing after 2s?
c. 2.5 𝑚/𝑠
𝑎. − 2.5 𝑚/𝑠
𝑏. 3.2 𝑚/𝑠
d. −3.2 𝑚/𝑠
99
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Easy Funneling
For his mini-online business, Mr. Roger wants to use electronic funneling to
fill his ice-candy packets easily. His funnel has a perfect cone shape with
dimensions of 6 feet across the top and 4 feet deep. Ice candy mixture is dripping
from the funnel at the rate of 2 cubic feet of mixture per minute assuming the
funnel is full. Mr. Roger wants to maximize production efficiency, so he decided to
make the mixture at the funnel 3 feet deep. So, how fast does the mixture drip
when the mixture is 3 feet deep?
6 𝑓𝑡
𝑟
4 𝑓𝑡
ℎ
100
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
101
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
Key Concept:
Key Concept:
What I Can Do
Additional Activities
1
𝑉 = 𝜋𝑟 2 ℎ
3
Working Equation:
𝑑𝑉
9𝜋 2 𝑑ℎ
=
ℎ
𝑑𝑡 400 𝑑𝑡
9𝜋
𝑑ℎ
(3)2
−2 =
400
𝑑𝑡
Answer:
𝑉 = 𝜋𝑟 2 ℎ
Working Equation:
𝑑𝑉
𝑑ℎ
= 𝜋𝑟 2
𝑑𝑡
𝑑𝑡
−
What's More
11.
12.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
15𝑐𝑚3
𝑑ℎ
= 𝜋(100𝑐𝑚)2
𝑠
𝑑𝑡
Answer:
𝑑ℎ
𝑐𝑚
= −0.00047
𝑑𝑡
𝑠
𝑑ℎ
= −3.1438 𝑓𝑡/𝑚𝑖𝑛
𝑑𝑡
What I Know
21.C
22.B
23.B
24.A
25.A
Assessment
21.A
22.A
23.C
24.A
25.C
1−𝑦 sin 𝑥𝑦
cos 𝑦+𝑥 sin 𝑥𝑦+1
𝑒 𝑥𝑦
𝑦 sin 𝑦+6
The lower end of the
ladder moves
approximately 4.49 ft/sec
•
•
•
•
•
•
•
•
•
Alegre, H. C.(2016). Basic Calculus. Mandaluyong City 1550 Philippines:
Anvil Publishing Inc.
Sullivan, M. (2012). Pre-Calculus Ninth Edition. United States of
America: Pearson Education, Inc.
Kelly, W. M. (2006). The Complete Idiot’s Guide to Calculus Second
Edition. United States of America: Alpha Books
Teaching Guide for Senior High School Basic Calculus (With permission
to use the concepts and examples)
https://tutorial.math.lamar.edu/Classes /Calc/ComputingLImits.aspx
https://www.analyzemath.com/calculus/limits/find_limits_functions.ht
ml
https://www.onlinemathlearning.com/limits-calculus.html
https://archives.math.utk.edu/visual.casual/1/limits.15/index.html
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondi
rectory/LimitConstant.html
102
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)
For inquiries or feedback, please write or call:
Department of Education – SDO Valenzuela
Office Address: Pio Valenzuela Street, Marulas, Valenzuela City
Telefax: (02) 8292-4340
Email Address: sdovalenzuela@deped.gov.ph
103
(DO_Q3_BASICCALCULUS_MODULE1_LESSON9)