MATH 401 – DIFFERENTIAL CALCULUS
Batangas State University
PB Main I Rizal Ave., Batangas City
MODULE
in
MATH 401
DIFFERENTIAL CALCULUS
EMIL C. ALCANTARA, Ph.D.
RONIE A. MENDOZA, M.Sc.
RENSON A. ROBLES, Ph.D.
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MATH 401 – DIFFERENTIAL CALCULUS
Table of Contents
CHAPTER I Functions, Limits and Continuity
Definition and Notation of Functions
Domain and Range
Graph of Functions
Types of Functions
Operations on Functions
Limit of a Function
One – sided Limits
Continuity
Chapter Test
1
4
7
12
24
27
32
41
45
CHAPTER II Derivatives of Algebraic Functions
Differentition
The Increment Method
Theories on Differentiation of Algebraic Functions
Higher order Derivatives
Implicit Differentiation
Chapter Test
46
49
52
63
65
68
CHAPTER III Derivatives of Transcendental Functions
Trigonometric Functions
Chain Rule of Derivatives of Trigonometric Functions
Implicit Differentiation of Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Chain Rule of Inverse Trigonometric Functions
Higher – order Derivative of Inverse Trigonometric Functions
Implicit Differentiation of Inverse Trigonometric Functions
Logarithmic Functions
Higher – order Derivatives of Logarithmic Functions
Implicit Differentiation of Lograithmic Functions
Logarithmic Differentiation
Exponential Functions
Chain Rule of the Derivatives of Exponential Functions
Higher – order Derivaives of Exponential Functions
Implicit Differentiation of Exponential Functions
Hyperbolic Functions
Chain Rule of the Derivatives of Hyperbolic Functions
Higher – order Derivatives of Hyperbolic Functions
Implicit Differentiation of Hyperbolic Functions
Chapter Test
69
76
85
88
91
92
93
96
98
99
101
104
105
107
107
112
115
118
122
123
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER IV Applications of Derivatives of Algebraic and
Transcendental Functions
The Differential
127
Application of the Differential
130
Approximation Formulas
132
Error Propagation
135
Tangent Line and Normal Line to a given Curve
138
Relative Exrema
147
Increasing and Decresing Functions and the
150
First Derivative Test
Concavity, Points of Inflection and the Second
159
Derivative Test
Curve Tracing
169
Optimization Problems
174
Optimization Problems Involving Algebraic Functions
174
Optimization Problems Involving Transcedental Functions 189
Number Problems
195
Related Rates
199
Related Rates ProblemsInvolving Algebraic Functions
200
Related Rates Problems Involving Transcendental
206
Functions
Motion Problems
211
Chapter Test
218
CHAPTER V Partial Differentiation
Definition of Parital Derivatives of a Functions
Partial Derivatives by Formulas of Differentiation
Higher – order Partial Derivatives
Total Derivatives
Chain Rule of Partial Differentiation
Implicit Partial Derivatives
Chapter Test
220
220
230
233
236
239
241
REFERENCES
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER I
FUNCTIONS, LIMITS AND CONTINUITY
One of the most useful tools in modeling real – life problems and situations
is the concept of function. For instance, if we want to determine the dimension of
a rectangular field enclosed by 400m of fence, then we can express the area A
as a function of its length l or width w . Similarly, the concept of function is
used to model the total cost of a product C x as a function of the amount of
product ordered x . This function is used as a rule that describes the relationship
between the dependent and independent variables.
In the study of Differential Calculus, functions are used to present the
concept of differentiation which is based on the notion of limits. In this chapter,
we shall learn functions and their graphs, operations on functions, definition of
limits and how limit theorems are used to evaluate limits of a function. Also, this
chapter presents one–sided limits, limits at infinity and infinite limits, and
continuity of functions on an open or closed intervals. Definitions and theorems
presented in this chapter are taken from [1] Larson, R. (2018), [2] Leithold, L.
(2002) and [3] Stewart, J. (2016)
At the end of this chapter, the learners might be able to:
1. Define and sketch the different types of algebraic and transcendental
functions;
2. Classify functions and recognize combinations and operations of
functions;
3. Evaluate limits using properties of limits;
4. Determine continuity at a point and continuity on an open interval; and
5. Apply different theorems to evaluate one - sided limits and continuity
on a closed interval.
1.1.
FUNCTIONS
Function assigns rule to describe how a certain quantity depends on the
values of other variables. The equation of a line, y mx b , for example sets the
relationship between the variables x and y . This relationship can be thought of as
a correspondence from a set X of real numbers x to a set Y of real numbers y .
Below is the formal definition of a function and is due to Leithold, L. (2002)
Definition1.1. A function is a set of ordered pairs x, y in which no two
distinct ordered pairs have the same first number.
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MATH 401 – DIFFERENTIAL CALCULUS
Symbols such as f , g , and h are used to denote functions unless stated
otherwise. If the function is expressed in terms of the variable x , f x , g x , and
hx are used to denote this function. For instance, in the finding the area of
circle, Ar r 2 is used to describe the relationship between the area and the
radius of the given circle. Here, it can be observed that the area, A is expressed
as a function of r and the value of A depends on the value of r . So, we say that
A is the dependent variable and r is the independent variable.
Example1.1. Let f be defined by f x 2 x 1 . It can be observed that f is function
since when x is replaced by any real numbers there is exactly one value of y
obtained.
x
f x
3
7
2
5
1
3
0
1
1
1
3
5
2
3
Example1.2. The circle x2 y 2 9 cannot be a function since when x is 0, y
assumes two values such as 3 and –3. Thus, we have the ordered pairs 0,3 and
0,3 .
Similarly, the ordered pairs 3,0 , 3,0 , 1,2 2 and 1,2 2 can also be
obtained from the given circle. Observe that 1 is assigned to two values of y ,
that is 2 2 .
Figure 1.1. The circle x 2 y 2 9
Example1.3. Let g be defined by g x x 2 1. Refer to the next table for the
values of x and its corresponding values of g x . Although, several values of g x
appear similar in the table, g can still be thought as a function since no two or
more values of x are repeated.
x
g x
3
10
2
5
1
2
0
1
1
2
2
5
3
10
The relations presented in Example 1.1, 1.2 and 1.3 can be illustrated by
mapping. Here, the set of real numbers x is called the domain while the set of
real numbers f x , y and g x are the range.
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MATH 401 – DIFFERENTIAL CALCULUS
Figure 1.2 shows the mapping of the domain to the range of the given
relations in Example 1.1, 1.2 and 1.3. Figure 1.2 (a) shows a one–to–one mapping
since there is exactly one value of x mapped to exactly one value of y . In (b), a
mapping of one–to–many is observed since there is exactly one value of x , say 0
and 1 mapped to two values of y , i.e. -3 and 3; and 2 2 and 2 2 , respectively.
In (c), a mapping of many–to–one can be seen as there are two values of x
mapped to one value of y . The ordered pairs 3,10 and 3,10 ; 2,5 and 2,5
; and 1,2 and 1,2 show this relation.
Domain
Range
f x
x
Domain
-3
-7
-2
-5
-3
-1
-3
0
0
-1
1
1
2
3
3
5
a
Range
x
Domain
x
y
-3
-3
-2
2 2
0
1
0
5
2
3
3
Figure 1.2. Mapping of a f x 2 x 1 , b x y 9
2
2
1
2
2 2
b
g x
-1
1
3
Range
10
c
and c g x x 2 1
Based on the examples presented in Figure 1.2 and Definition 1.1, a
function possesses a one–to–one correspondence (bijection) and many–to–one
correspondence (surjection) but not one–to–many or many–to–many.
Another way to determine whether a relation is function or not is by using
the vertical line test. Given the graph of a function, draw vertical lines overlaying
the graph. If the vertical lines pass through exactly one point on the graph, then it
is a function. If it passes through two or more points, then it is not a function.
To evaluate function, a straight forward substitution is used. For instance,
given a function, f x x 2 2 , if we wish to find f 3 we shall replace x by -3
and perform the operation leading us with f 3 11 .
1
Example1.4. Let f be a function defined by f x x 2 2 x 3 , find a. f 2 ; b. f ;
2
c. f 1 ; d. f 2 x and e. f x h .
Solution:
a.
f x x 2 2 x 3
d. f x x 2 2 x 3
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MATH 401 – DIFFERENTIAL CALCULUS
f 2 2 22 3
f 2 3
2
b.
f x x 2 2 x 3
2
1 1
1
f 2 3
2 2
2
1 1
f 1 3
2 4
1 9
f
2 4
c.
f 2 x 2 x 22 x 3
f 2 x 4 x 2 4 x 3
2
e. f x x 2 2 x 3
f x h x h 2x h 3
2
f x h x 2 2 xh h 2 2 x 2h 3
f x x 2 2 x 3
f 1 1 2 1 3
f 1 6
2
1.1.1. Doman and Range
Domain and range as described on the first section helps
characterized a function. Domain is also classified as the set of the
independent variables while the range is the set of the dependent
variables.
Definition1.2. [2] The set of all admissible values of x is called the
domain, and the set of all the resulting values of y is called the range
of the function.
Example1.5. Let f be a function defined by f x 2 x 3 . Observed that x
can be replaced by any real numbers that will give a defined value for f x .
Thus, the domain of f is the set of real numbers, D x x . In interval
notation, it is given by , . Similarly, x is replaced by any number on
this interval, the range is also a number on this interval. So the range is
also an element of real numbers , .
Example1.6. Let g be a function defined by g x x .Since g involves a
square root, then x cannot be replaced by any negative numbers. Thus, the
domain of g is the set of non–negative real numbers. In interval notation,
it is given by 0, . Also, the range of this function are the numbers on the
same interval, 0, .
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MATH 401 – DIFFERENTIAL CALCULUS
Example1.7. Let f be a function defined by f x x 3 . Here, x 3 cannot
be a negative number, so we can let x 3 0 . Solving this inequality, we
have x 3 . So, the domain of this function is the set of all numbers
greater than or equal to -3 or 3, . The range of this function is 0, .
Example1.8. [2] Let f be a function defined by f x x 2 9 . Since f
involves square root, then x 2 9 0 . Here, we shall think of any number
that when replaced to x will give x 2 9 0 . Solving this inequality, we
have x 3 or x 3 . So the domain of f is given by ,3 3, and
the range is 0, .
Example1.9. Find the domain and range of the function y 3 x . Recall that
the cube root of any negative number is defined on the set of negative real
number, the cube root of 0 is 0 and the cube root of a positive real number
is still defined on the same set of positive real numbers. Therefore, the
domain of y 3 x is , and the range is , .
From this example, we can conclude that for any function f defined
by f x n x where n is any odd positive integer, then the domain and
range of f is , .
Example1.10. Find the domain and range of the following functions:
a.
f x
b. g x
f x
2
x
c.
x
x2
d. f x
1
x2
1
x 16
2
Solution:
a. Since x can be found on the denominator, then x cannot be replaced by
1
0. So the domain of f x is the set of real numbers except 0 or
x
interval notation we have ,0 0, and the range is also the set
of real numbers except 0, ,0 0, since the numerator is
constant.
x
b. For g x
, the denominator x 2 should be equal to 0, i.e. x 2
x2
. The domain of g is the set of real numbers except -2 or
,2 2, . For the range of g , observed that the numerator is
no longer constant so g x assumes any number except 1. Thus, the
range of g is ,1 1,.
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MATH 401 – DIFFERENTIAL CALCULUS
c. Similar with a, x appears on the denominator. So we have x 2 0 giving
us with the domain ,0 0, . For the range, take note that the
numerator is constant, 1 and the denominator is x 2 . Given this facts,
f x do not assumes values such as 0 and any negative numbers, thus
the range of this function is 0, .
d. The domain of f x
1
is the set of real numbers except 4
x 16
since when x is 4 or -4, the denominator becomes 0. The range is
2
1
, 0, .
16
Example 1.11. Let y be function defined by y 2 x . The function y 2 x is
an exponential function whose domain is the set of real numbers, ,
and the range is is the set of all positive real numbers, 0, .
Example 1.12. The domain of the function f x sin 2 x is the set of real
numbers , while the range is any numbers on the interval 1,1 .
Exercise 1.1. Find the domain and range of the following functions.
2
x2
3
2. y x 2
3. y x 2 2 x 4
4. y x 3 2
1. y
5.
6.
7.
8.
9.
10.
f x 3 2 x 3
f x 9 x 2
x2 4
g x
x2
x
g x
2 x
2x 2
f x
25 x 2
x
1
y
2
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MATH 401 – DIFFERENTIAL CALCULUS
1.1.2. Graph of Functions
Function is described as a set of ordered pairs x, y . This set of
ordered pairs or points leads to the graphical representation of a function
in R 2 plane. Below is the formal definition of the graph of a function.
Definition 1.3. [2] If f is a function, then the graph of f is the set of all
points x, y in the plane R 2 for which x, y is an ordered pair in f.
The graph of a function f can also be thought as a graph of the
equation y f x . This determines the behavior of the graph as the
variable x and y change. For instance, in the graph of the parabola y x 2 ,
as the value of x increases, y also increases approaching . See Figure
1.3.
As mentioned in section 1.1, one
way to determine whether a relation is a
function or not when its graph is given is
through the vertical line test (VLT).
Referring to the graph of y x 2 on the left,
when you draw vertical lines, you will
notice that the lines pass through exactly
one point on the graph or curve of the
parabola. Also as observed on its graph,
y x2
follows
many–to–one
correspondence as illustrated in Section
Figure1.3. Graph of y x 2
1.2.
Another way to sketch the graph of a function is by determining its
properties. This applies for the graphs of some special curves like the conic
sections. Knowledge on the standard forms of these curves will help us
trace their graphs. As example, it would be very easy to sketch the circle
x 2 y 2 9 if the center–radius form of this equation is obtained.
Particularly, x 2 y 2 9 is a circle with center at the origin 0,0 and
radius equal to 3 by following the center-radius form of a circle. To sketch
this curve, one might start at the center 0,0 and locate points 3 units
away to the left and right and above and below the point 0,0 . Doing it so,
we obtained the points 3,0 , 3,0 , 0,3 and 0,3 . See Figure 1.1.
Functions other than these curves can be graphed by plotting points on the
plane and smoothly tracing those points.
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MATH 401 – DIFFERENTIAL CALCULUS
To facilitate the graphing of a function, the following steps are suggested:
1. Identify the domain and range and the properties of the function.
2. Choose suitable values of x from the domain of a function and solve
for its corresponding value of y .
3. Determine the behaviour of x and y .
4. Plot the points x, y on the plane.
5. Smoothly trace the curve.
Example 1.13. Sketch the graph of the following functions:
a.
f x 2 x 3
b.
f x x 3
c.
f x 25 x 2
d.
f x x 2 4
f x
2
x
1
f. f x 2
x
x
g. y 2
e.
Solution:
a.
b.
Figure1.4. Graph of f x 2 x 3
Figure1.5. Graph of f x x 3
From Figure 1.4 it can be observed that the graph of f x 2 x 3 is a
3
2
line that passes through the points 0,3 and ,0 . The point 0,3 is the
3
x intercept and the point ,0 is the y intercept of f x 2 x 3 . We
2
can also verify that the domain and range of this function is any number on
the interval , . From Example 1.7 the domain of f x x 3 is
3, and the range is 0, . Also, it can be seen from Figure1.5 that as
x increases, y also increases.
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MATH 401 – DIFFERENTIAL CALCULUS
c.
d.
Figure1.6. Graph of f x
Figure1.7. Graph of f x x 2 4
25 x 2
The graph of f x 25 x 2 (see Figure 1.6) shows a semi – circle with
0,3 as y intercept and 3,0 and 3,0 as x intercepts. The graph of
f x x 2 4 in Figure 1.7 confirms that the domain of this function is
,2 2, and the range is 0, .
e.
f.
Figure1.8. Graph of f x
2
x
Figure1.9. Graph of f x
1
x2
g.
Figure1.10. Graph of f x 2 x
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MATH 401 – DIFFERENTIAL CALCULUS
The graphs of the functions f x
2
1
(Figure 1.8), f x 2 (Figure
x
x
x
1.9) and y 2 (Figure 1.10) confirm the result of Example 1.10 (a) and (c)
and Example 1.11. In Figure 1.8 we can see that as x increases
approaching , y decreases approaching 0; as x approaches 0 from the
right, y increases approaching ; as x decreases approaching , y
approaches 0 and finally as x approaches 0 from the left, y approaches
.
Example1.14. Sketch the graph of the following functions:
2x 2
x2 9
1
a. f x
b.
c.
f
x
f
x
x 2 16
x3
25 x 2
Solution:
Figure1.11. Graph of f x
2x 2
25 x
2
c.
Figure1.12. Graph of f x
x2 9
x3
Figure1.12. Graph of f x
1
x 16
2
2x2
The graph of f x
in Figure
25 x 2
1.11 shows that the domain of this
function is the set of real numbers
except 5 and the range is the set of
real numbers except (0,–2]. Figure
1.12 shows that the domain of
1
is the set of real
f x 2
x 16
numbers except 4 and the range is
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MATH 401 – DIFFERENTIAL CALCULUS
1
, 0, . Finally, it can be seen from Figure 1.12 that domain of
16
x2 9
is ,3 3, and the range is ,6 6, .
f x
x3
The lines x 5 and x 5 in Figure 1.11 and the lines x 4 and
x 4 are called vertical asymptotes; while the lines y 2 in Figure 1.11
and y 0 in Figure 1.12 are called horizontal asymptotes. The asymptotes
of functions determine its discontinuity. For instance in the function
1
1
, when x is replaced by 4 or -4, you’ll have f 4
and
f x 2
x 16
0
1
f 4 . We can also say that when x approaches 4 from the right, f x
0
gets arbitrarily large or approaching . Similarly, as x approaches 4
from the left f x approaches . This idea gives the formal definition of
asymptotes.
px
is a rational
q x
function in lowest terms and a is some real number where qx 0
Definition 1.4. Rees, R., (2003). Suppose that f x
i. The vertical line x a is a vertical asymptote of the graph of f if
as x a , then f x .
ii. The horizontal line y a is a horizontal asymptote of the graph
of f if as x , then f x a .
From this definition, we set the rules to determine the asymptote of
a function.
Suppose that the rational function
px a n x n an 1 x n1 ... a1 x a0
f x
where qx 0
qx bm x m bm1 x m1 ... b1 x b0
is in lowest terms.
If qa 0 , then x a is a vertical asymptote .
If n m , then the x axis is a horizontal asymptote.
If n m , then the horizontal asymptote is the line y
an
.
bm
If n m , then the graph of f has no horizontal asymptote.
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MATH 401 – DIFFERENTIAL CALCULUS
x2 9
in Figure 1.12, observe that the f is
x3
undefined at x 3 , i.e., when x is replaced by -3, the denominator
becomes 0. However, based on Definition 1.4, the rational function should
be in lowest term leading us with f x x 3 . So now, f 3 6 . The
For the function f x
point 3,6 is called point discontinuity.
1.1.3. Types of Functions
In the study of Calculus we might encounter different forms of
functions. Generally speaking functions can be classified into two such as
algebraic functions and transcendental functions. Some forms of
algebraic functions are linear function, f x ax b ; quadratic function,
f x ax2 bx c , for a, b, c and a 0 ; and polynomial functions,
f x an x n an1 x n1 ... a1 x a0 . We may also include the rational
functions of the forms f x
(a)
(b)
px
where qx 0
qx
(c)
(d)
Figure 1.13. Some examples of (a) linear function; (b) quadratic function; (c)
polynomial function, and (d) rational function
Transcendental functions are functions that transcend algebraic functions.
x x 2 x3
x3 x5 x7
For instance, 1 ... transcends e x ; sin x x ...
3! 5! 7!
1! 2! 3!
2
4
6
x
x
x
and cos x 1 ... . Some commonly used transcendental
2! 4! 6!
functions are:
1. Trigonometric functions (Six Circular Functions)
2. Inverse trigonometric functions
3. Exponential functions
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MATH 401 – DIFFERENTIAL CALCULUS
4. Logarithmic functions
5. Hyperbolic and inverse hyperbolic functions
Trigonometric Functions and Inverse Trigonometric Functions
One of the most commonly used transcendental functions is the six
circular functions or the trigonometric functions. These functions were
taken from the unit circle in Figure 1.14. The cosine, since, tangent,
cotangent, cosecant and secant are the six circular functions.
Px, y
y
Definition 1.5. [Stewart (2012)]. Let be an
angle in standard position and let Px, y be a
r
0
1
Figure 1.14. Unit Circle
x
point on the terminal side. If r x 2 y 2 is the
distance from the origin to the point Px, y ,
then
sin
y
r
csc
r
r
, y 0 sec , x 0
y
x
cos
x
r
tan
cot
y
, x0
x
x
, y0
y
The graphs of the six circular functions are presented on Figure 1.15.
s a function defined by x n where n x and
nZ .
Figure 1.15. Graphs of the six unit circle
Reference: https://sites.google.com/site/reimerprecalculus/home/unit-4b/4-6-graphs-other-trig-functions
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MATH 401 – DIFFERENTIAL CALCULUS
The domain and range of the trigonometric functions are:
Table1
Domain and Range of Trigonometric Functions
Domain
Range
x x n where n Z
2
x x n where n Z
,
x x n where n Z
2
x x n where n Z
,1 1,
,
,
sin
cos
tan
cot
sec
csc
1,1
1,1
,
,1 1,
Some trigonometric identities are as follows:
cos x cos x
sin x sin x
sin x csc x 1
cos x sec x 1
tan x cot x 1
sin 2 x cos 2 x 1
1 tan 2 x sec 2 x
1 cot 2 x csc 2 x
cos 2 x sin 2 x
cos 2 x 1 2 sin 2 x
2 cos 2 x 1
sin 2 x 2 sin x cos x
2 tan x
tan 2 x
1 tan 2 x
1 cos 2 x
sin 2 x
2
1 cos 2 x
2
cos x
2
1 cos 2 x
2
tan x
1 cos 2 x
cos x y cos x cos y sin x sin y
cos x y cos x cos y sin x sin y
sin x y sin x cos y cos x sin y
sin x y sin x cos y cos x sin y
tan x tan y
tan x y
1 tan x tan y
tan x tan y
tan x y
1 tan x tan y
1
1
sin x sin y cos x y cos x y
2
2
1
1
sin x cos y sin x y sin x y
2
2
1
1
cos x cos y cos x y cos x y
2
2
1
1
sin x sin y 2 sin x y cos x y
2
2
1
1
sin x sin y 2 cos x y sin x y
2
2
1
1
cos x cos y 2 cos x y cos x y
2
2
1
1
cos x cos y 2 sin x y sin x y
2
2
These identities will help students to simplify both trigonometric expressions
and equations.
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MATH 401 – DIFFERENTIAL CALCULUS
y
. If , we see from
r
2
2
Figure 1.14 that the sine function attains the value on the interval 1,1
Let us consider the sine function sin
exactly once and so is one–to–one. For the cosine function, if we restrict
the value of inclusively between 0 and
for tangent
2
2
function, these gives cosine and tangent a one–t –one correspondence. On
these intervals, we obtain their inverse functions as follows:
sin 1 x y
cos 1 x y
tan 1 x y
sin y x
cos y x
tan y x
Table2
Domain and Range of the Inverse Trigonometric Functions
Domain
Range
sin 1 x
1,1
cos 1 x
1,1
tan 1 x
all real numbers
2 , 2
0,
,
2 2
(a) y sin 1 x
(b) y cos 1 x
(c) y tan 1 x
Figure 1.16. Graphs of (a) y sin 1 x , (b) y cos 1 x and (c) y tan 1 x
Reference: https://www.onlinemathlearning.com/inverse-sine-cosine-tangent.html
Exponential and Logarithmic Functions
A function y defined by the relation, y a x where a is a positive
number except 1 is called an exponential function of x.
1
Example 1.15. Sketch the graph of (a) f x 2 and (b) f x
2
x
x
1
2
x
Solution: The graphs of f x 2 x and f x are shown in Figure 1.18.
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MATH 401 – DIFFERENTIAL CALCULUS
x
-3
-2
-1
0
1
2
3
f x
1/8
1/4
1/2
1
2
4
8
f x 2 x
1
f x
2
x
x
-3
-2
-1
0
1
2
3
1
Figure 1.18. Graphs of (a) f x 2 and (b) f x
2
f x
8
4
2
1
1/2
1/4
1/8
x
x
Properties of Exponential Function
For f x a x where a 1
1. Domain : ,
Range
: 0,
2. As x , f x increases and gets steeper
3. As x , f x decreases and flattens
4. Asymptotic with respect to negative x – axis
Figure 1.19. Graph of f x a x
where a 1
For f x a x where 0 a 1
1. Domain : ,
Range: 0,
2. As x , f x decreases and flattens
3. As x , f x increases and gets steeper
4. Asymptotic with respect to positive x – axis
Figure 1.20. Graph of f x a x
where 0 a 1
Some laws on exponential functions are listed below.
a0 1
a1 a
a x a y a x y
ax
a x y
ay
a x y a xy
e0 1
e1 e
e x e y e x y
ex
e x y
ey
e x y e xy
16
MATH 401 – DIFFERENTIAL CALCULUS
Logarithmic Functions
By definition of logarithm, the relation y a x , can be written as
x log a y . Thus a x and log a x are inverse functions of one another, and it
follows that a log x x and log a a x x .
a
Example 1.16. Convert the following exponential to logarithmic functions
1. 23 8
Ans.
log 2 8 3
1
1
1
2.
5
5
log 1
3. 100 1
1
4. 3 3
27
3
1
1
5.
27
3
log10 1 0
1
log 3
3
27
1
log 1
3
27
27
5
1
1
5
The logarithmic function takes a special when a e , that is called
natural logarithm, given by log e x ln x and when a 10 , called common
logarithm and is written as log10 x log x .
Properties of Logarithmic Function
For f x log a x where a 1
1. Domain : 0,
Range : ,
2. As x , f x increases and flattens
3. As x 0 from the right, f x decreases
and gets steeper
4. Asymptotic with respect to negative y – axis Figure 1.21. Graph of f x log a x
where a 1
For f x log a x where 0 a 1
1. Domain : 0,
Range : ,
2. As x , f x decreases and flattens
3. As x 0 from the right, f x increases
and gets steeper
4. Asymptotic with respect to positive y – axis
Figure 1.22. Graph of f x log a x
where 0 a 1
17
MATH 401 – DIFFERENTIAL CALCULUS
Some laws on logarithmic functions are as follows:
ln 1 0
ln e 1
log a 1 0
ln xy ln x ln y
log a xy log a x log a y
ln
log a a 1
x
ln x ln y
y
log a
r ln x ln x r
x
log a x log a y
y
r log a x log a x r
Hyperbolic Functions
Hyperbolic functions are functions possessing similar
characteristics with that of the six trigonometric functions derived from its
relation to the equilateral hyperbola. The hyperbolic functions are defined
as follows:
e x ex
2
x
e ex
cosh x
2
2
x
e e x
2
sec hx x
e e x
csc hx
sinh x
ex
ex
ex
coth x x
e
tanh x
ex
e x
ex
e x
Some properties and identities of hyperbolic functions and their
domain and range are presented in Table 3 while their graphs are on
Figure 1.23.
cosh 2 x sinh 2 x 1
tanh 2 x sec h 2 x 1
coth 2 x csc h 2 x 1
cosh x sinh x e x
cosh x sinh x e x
sinh( x y ) sinh x cosh y cosh x sinh y
cosh( x y ) cosh x cosh y sinh x sinh y
sinh 2 x 2 sinh x cosh x
cosh 2 x cosh 2 x sinh 2 x
Table 3
Domain and Range of Hyperbolic Functions
cosh x
sinh x
tanh x
sec hx
csc hx
coth x
Domain
Range
,
,
,
,
,0 0,
,0 0,
1,
,
1,1
0,1
,0 0,
,1 1,
18
MATH 401 – DIFFERENTIAL CALCULUS
Figure 1.23. Graph of Hyperbolic Functions
Reference: https://www.jobilize.com/calculus/test/graphs-of-hyperbolic-functions-by-openstax
Absolute Function
The next example shows an absolute function.
Example 1.17. Let
f be a function defined by (a) f x x and (b)
f x x 1 . Find the domain and range of f and sketch its graph.
Solution:
Observed from Figure 1.24 and
Figure 1.25 that the domain of
f is the set of real numbers
, and the range is the
set of non negative
numbers, 0, .
real
Recall that
Figure 1.24. The absolute function f x x
x
x0
x
if
x
x0
19
MATH 401 – DIFFERENTIAL CALCULUS
It can also be observed from
Figure 1.25, that when the
graph of f x x is shifted one
unit to the left, the graph of
f x x 1 is obtained.
Figure 1.25. The absolute function f x x 1
This observations lead to the following generalization:
To sketch or trace the graph of
i. f x x a shift the graph of f x x , a units to the left;
ii.
f x x a shift the graph of f x x , a units to the right;
iii.
f x x a shift the graph of f x x , a units to downward; and
iv. f x x a shift the graph of f x x , a units to upward.
this is true for all a
Another functions that you might want to study is the greatest
integer functions. This function is an example of step functions. To
investigate the behavior of this function, you may use the concept of one –
sided limits which will be introduced on the next section of this chapter.
Definition 1.5. [1]. The greatest integer function (gif) is a function
defined by x n where n x and n Z .
For instance, 2.4 2 , 2.2 3 , 5 5 and
1
0
2
Example 1.18. Find the domain and range of the function f x x and
sketch its graph.
Solution:
The domain of this function is the set of real numbers , and
the range is the set of all integers, Z . The graph is shown in Figure 1.26.
20
MATH 401 – DIFFERENTIAL CALCULUS
Figure 1.26. The greatest integer function f x x
Odd and Even Function
If the function f satisfies f x f x for all number on its domain,
then the function is said to be an even function. The graph of f is
symmetric with respect to the y axis. If the graph of f is symmetric with
respect to the origin then f is an odd function. This function satisfies
f x f x .
Example 1.19. Determine whether f defined by (a) f x 4 x 4 3x 2 3 ;
(b) f x x 3x 3 x 5 and (c) f x 2 x 3 3x 2 x 3 is even or odd function.
Solution:
To do this, we replace x by x and solve for f x
b. f x x 3x 3 x 5
a. f x 4 x 4 3x 2 3
f x 4 x 3 x 3
f x x 3 x x
f x 4 x 4 3x 2 3
f x x 3x 3 x 5
4
2
f x 4 x 4 3x 2 3 is an even function
3
5
f x x 3x 3 x 5
f x x 3x 3 x 5 is an odd function
c. f x 2 x 3 3x 2 x 3
f x 2 x 3 x x 3
3
2
f x 2 x 3 3x 2 x 3
f x 2 x 3 3x 2 x 3
Observe that f x f x and f x f x therefore f is neither even nor
odd function.
21
MATH 401 – DIFFERENTIAL CALCULUS
Piecewise Functions
Piecewise – defined functions will be useful in the study of limits
and continuity which will be presented in Section 1.2 of this chapter. For
x2 9
instance the graph of f x
in Figure 1.12 has a break point at
x3
x 3 . This point is called point discontinuity. This means that f x do not
exist at this point of f.
Let us consider the piecewise –
defined function
x 1
2 x 1
if x 1
f x 2
x 2
1 x
Here if we replace x by any
number less than 1, we shall use
f x 2 x 1 and if x 1 , use
f x x 2 . Obviously when x 1 ,
f x 2 .
2 x 1
Figure 1.27. The graph of f x 2
x 2
x 1
if x 1
1 x
The
graph
of
this
piecewise – defined function is
shown on the next figure. The
domain of this function is the set
of real numbers , and the
range is ,1 2 3, .
1 x 2
x0
if
3
x
1
1
x
Example 1.20. Sketch the piecewise – defined function hx
.
Solution: Observe that the function
h1 x 1 x 2 is a parabola that
opens downward with vertex at
0,1 while h2 x 3x 1 is a line that
1
3
will pass through 0,1 and ,0 .
Taking this into consideration, the
graph of h is shown below in
Figure 1.28.
Figure 1.28. The graph of h
22
MATH 401 – DIFFERENTIAL CALCULUS
The
domain
of
h
is
,0 1, and the range is
,1 4, .
Example 1.21. Sketch the piecewise function g defined by
x 1
x 2
2
g x 4 x if 2 x 2 .
1 x
2 x
Solution:
The domain of g is , and the
range is ,1 0,2 .
Figure 1.29. The graph of g
Exercise 1.2. Sketch the graph of the following functions. Also, find its
domain and range.
1. f x x 2
2x 1
x
2
4 x
3. y
x2
2. g x
4. y x 2 25
5. y
x3 2x2
x2
6. f x x 2
7. f x 2 x
3x 2
x 1
8. hx 2
if
x
1
x
x3
x 1
9. f x 5
if x 3
2 x 1
3 x
x 2
3
10. g x 1 if 2 x 2
2
2 x
Reference: [2] Leithold, L (2002)
23
MATH 401 – DIFFERENTIAL CALCULUS
1.1.4. Operations on Functions
A new function can be obtained by combining two or more
functions. These functions may undergo operations such as addition,
subtraction, multiplication, division and composition.
Definition 1.6. [2] Given two functions f and g :
i.
their sum, denoted by f g is a function defined by
f g x f x g x
ii.
their difference, denoted by f g is a function defined by
f g x f x g x
iii. their product, denoted by f g is a function defined by
f g x f x g x
iv. their quotient, denoted by
f
f
f x
is a function defined by x
g
g x
g
where g x 0 .
The domain of these resulting functions consists of those values of x
common to the set of domains of f and g . We only need to exclude those
values of x in Case iv where g x 0 .
Example 1.22. Let
f and g be functions defined by f x 2 x 1 and
g x x 2 2 x 1 , find (a) f g ; (b) f g ; (c) f g and(d)
Solution:
a. f g x f x g x
f g x 2x 1 x 2 2x 1
f g x x 2 2
c. f g x f x g x
f g x 2 x 1x 2 2 x 1
f g x 2 x 3 3x 2 1
f
.
g
b. f g x f x g x
f g x 2x 1 x 2 2x 1
f g x x 2 4 x
f
f x
d. x
g x
g
f
2x 1
x 2
x 2x 1
g
Another operation involving two or more functions is called
composition. Below is the formal definition of the composite function.
24
MATH 401 – DIFFERENTIAL CALCULUS
Definition 1.7. [2] Given two functions f and g , the composite function of
f and g denoted by f g is defined by
f g x f g x .
The domain of f g is the set of all numbers x in the domain of g such
that g x is in the domain of f.
Example 1.23. Let f and g be functions defined by f x x 1 and
g x x 2 2 find (a) f g x ; (b) g f x
Solution:
a.
f g x x 2 2 1
f g x x 2 1
b. g f x
2
x 1 2
g f x x 1 2 x 1
Observed that the domain of f is 1, and the domain of g is , ,
therefore the domain of f g x is ,1 1, and this set is in the
domain of g .
Example 1.24. Let f and g be functions defined by f x
2
and
x 1
1
g x 2 x 1 find (a) f g 2 ; (b) g f 3 ; (c) f g .
2
Solution:
a. f g 2
g x 2 x 1
g 2 22 1
g 2 3
f g 2 f 3
2
f g 2
3 1
f g 2 1
b. g f 3
2
f x
x 1
2
1
f 3
3 1
2
1
g f 3 g
2
1
g f 3 g
2
1
g f 3 2 1 2
2
g f 3 2
1
c. f g
2
1
Since g 2 , then
2
1
f g f 2
2
1
2
f g
2 1 1
2
1
4
f g
3
2
1
g f 3 2 1 2
2
g f 3 2
25
MATH 401 – DIFFERENTIAL CALCULUS
Exercise 1.3. Apply the concepts of operations on functions to solve the
following.
Let f and g be functions defined by in each of the following numbers. Find
f g; f g; f g ;
f
; f g and g f . Also, find the domain and range of
g
the resulting functions.
1.
f x x 3 1 ; g x x 2
2.
f x 4 x 2 ; g x
3.
4.
5.
1
x
x2
; g x 2 x 1
x 1
1
1
f x 2 ; g x
x
x
2x
f x e ; g x ln x
f x
Let f, g and h be functions defined by f x 3x 1; g x
hx x 2 1 , find the following:
6.
f g 2
7.
f h 1
8.
9.
1
x2
and
2
f
1
g
f h 1
10. f g h2
References: [1] Larson, R. (2018) and [2] Leithold, L (2002)
26
MATH 401 – DIFFERENTIAL CALCULUS
1.2. LIMIT OF FUNTION
Limits of a function play a very important role in finding the derivatives of
the given functions. For instance, given a function f x , if the limit of this function
exists at a certain number a , then f x is said to be differentiable on that value of
a.
1.2.1. Definition of Limit and Theorems on Limits
The formal definition of the limit of a function is given on Definition 1.8.
Here, we consider a function f .
Definition 1.8. [2] Let f be a function at every number in some open
interval containing a , except possibly at the number a itself. The limit of
a f x as x approaches a is L , written as
lim f x L
x a
if the following statement is true:
Given any 0 , however small, there exists a 0 such that if
0 x a , then f x L .
Geometric interpretation of Definition 1.8 is presented in Figure 1.30.
L 1
L
L 1
a 1
a
a 1
Figure 1.30. Geometric representation of lim f x L
x a
It can be seen from the figure on the left that as the value of x approaches
a from the right, f x approaches L . This is also true when you approach
the value of a from the left.
Theorem 1.1 shows the basic theorems on limits.
27
MATH 401 – DIFFERENTIAL CALCULUS
Theorem 1.1. [2] Let f and g be functions and let a
1.1.1. If c is constant, then lim c c .
x a
1.1.2. If c is constant, then lim cx c lim x ca .
xa
xa
1.1.3. lim mx b ma b (limit of a linear function)
xa
1.1.4. If lim f x L and lim g x M , then lim f x g x L M .
x a
x a
x a
x a
x a
x a
1.1.5. If lim f x L and lim g x M , then lim f x g x L M .
1.1.6. If lim f1 x L1 , lim f 2 x L2 ,…, lim f n x Ln then
x a
x a
x a
lim f1 x f 2 x f n1 x L1 L2 Ln .
xa
1.1.7. If lim f x L and n is any positive integer, then lim f x n Ln .
x a
f x
x a
1.1.8. If lim f x L and lim g x M , then lim
.
x a
x a g x
x a
M
L
1.1.9. If lim f x L and n is any positive integer, then lim n f x n L
x a
x a
1.1.10.
1.1.11.
1 1
.
x a x
a
If a 0 and n is positive integer, or if a 0 and n is an odd
positive integer, then lim n x n a .
If a is any real number except 0, then lim
x a
1.1.12.
c
0 if n Z and c any constant
x x n
lim
One easy way to evaluate the limits of a function is by straight forward
substitution. Consider the following illustrations.
Illustration 1.1. Find lim 4 .
x 2
Solution:
By Theorem 1.1.1 it is easy to show that lim 4 4 .
x 2
1
2
Illustration 1.2. Evaluate lim .
x 2
Solution:
1
2
1
2
By Theorem 1.1.1, lim .
x 2
Illustration 1.3. Evaluate lim x 2 2 x 3 .
x 2
Solution:
lim x 2 2 x 3 lim x 2 2 lim x lim 3 by Theorem 1.1.4
x 2
x2
x2
x2
28
MATH 401 – DIFFERENTIAL CALCULUS
2
2 2 2 3 by Theorem 1.1.7, 1.1.2 and 1.1.1
lim x 2 2 x 3 11
x2
3
Illustration 1.4. Evaluate lim1 2 x 3 .
x
2
Solution:
3
1
3
3
lim 2 x 3 2 3 2 by Theorem 1.1.7, 1.1.4, 1.1.2 and
1
x
2
2
1.1.1.
3
lim 2 x 3 8
1
x
2
Illustration 1.5. Evaluate lim
x 0
x 2 2x 1
.
x2
Solution:
x 2 2 x 1 0 20 1
x 0
0 2
x2
2
x 2x 1
1
lim
x 0
x2
2
2
lim
Illustration 1.6. Evaluate lim
x 4
Solution:
lim
x 4
2x 1
24 1
x2
42
lim
x 4
2x 1
.
x2
9
3
3 2
2
2
2
2x 1 3 2
x2
2
3 x2
.
x
Illustration 1.7. Evaluate lim
x 1
Solution:
3 1
3 x2
4
lim
2
x 1
x
1
1
2
Illustration 1.8. Evaluate lim
x 4
Solution:
x 2 16
.
x4
x 2 16 4 16 0
x4
44
0
2
lim
x 4
29
MATH 401 – DIFFERENTIAL CALCULUS
0
. The
0
0
x 2 16
expression is called indeterminate form. To evaluate lim
we
x 4
x4
0
By direction substitution we noticed that the limit is
first simplify this algebraically.
lim
x 4
x 4x 4 lim x 4 4 4 8 . So, lim x 2 16 8
x 2 16
lim
x 4
x 4
x 4
x4
x4
x4
Illustration 1.9. Evaluate lim
x 1
x3 1
.
x 1
Solution:
By straight forward substitution, we have lim
x 1
x 3 1 1 1 0
.
x 1
11
0
3
x 1
we have
x 1
x3 1
x 1x 2 x 1
2
lim
lim
lim x 2 x 1 1 1 1 3 .
x 1 x 1
x 1
x
1
x 1
3
x 1
3.
Therefore, lim
x 1 x 1
Simplifying
3
Illustration 1.10. Evaluate lim
x 1
1 x
.
1 x
Solution:
It is also an obvious observation that when x is replaced by 1, we’ll
0
0
1 x
, we first rationalize the numerator.
x 1 1 x
have . To evaluate lim
So, we have
1 x
1 x 1 x
1 x
1
1
1
lim
lim
lim
.
x 1 1 x
x 1 1 x
x 1 1
1 x x1 1 x 1 x
x 1 1 2
lim
1 x 1
x 1 1 x
2
lim
Illustration 1.11. Evaluate lim
h0
3 h3
.
h
Solution:
Replacing h
by 0, we’ll have lim
Rationalizing the numerator,
h0
3 h3
3 03 0
.
h
0
0
3 2 h3
3 h3 3 h3
lim
lim
h0
h0
h
3
h
3
h 3 h 3
. Simplifying,
2
30
MATH 401 – DIFFERENTIAL CALCULUS
3 h 3
3 h 3
3 h3
lim
lim
h0
h0 h
h
3 h 3 h0 h 3 h 3
h
1
lim
lim
.
h0 h
3 h 3 h0 3 h 3
Substituting 0 to h ,
1
1
3 h3
lim
h 0
h
3 03 2 3
3 h3
3
lim
h0
h
6
lim
Indeterminate form
0
0
is introduced in Illustration 1.8–1.11. This
expression should be avoided when evaluating limits of function. One way
to do that is to simplify the expression algebraically. However, in a more
advanced study of limits, we can employ the rule called L’Hospitals Rule.
But this rule needs the concept of differentiation. Other indeterminate
forms are
, , 0 , 0 0 , 1 and 0 .
2
.
x x 3
Illustration 1.12. Evaluate lim
Solution:
By theorem 1.1.12 we have lim
x
2
2
0.
3
x
3x 3 2 x 4
.
x 2 3 x 2 2 x 3
Illustration 1.13. Evaluate lim
Solution:
If we substitute to x we obtain an indeterminate form. To
evaluate the limit, we first divide every terms of the numerator and
denominator by x raised to the highest power, in this case by x 3 . So
we’ll have
3x 3 2 x 4
x 2 3 x 2 2 x 3
lim
3x 3 2 x 4
x 2 3 x 2 2 x 3
lim
3x 3 2 x 4
3 3
3
x
x
x . Simplifying,
lim
2
x 2
3x
2x3
x3 x3
x3
2
4
3 2 3
x
x . Applying Theorem 1.1.12
lim
x 2
3
2
3
x
x
31
MATH 401 – DIFFERENTIAL CALCULUS
2
4
3 2 3
3x 3 2 x 4
30 0 3
lim
2
3
x 2 3 x 2 x
2
3
002
2
2
3
3x 3 2 x 4
3
lim
x 2 3 x 2 2 x 3
2
x
Illustration 1.14. Evaluate lim sin 2 x
2
.
cos x
Solution:
2
2
2
lim sin 2 x
sin 2
0
2
x
cos x
cos
1
Two important theorems in finding the limits of trigonometric
functions are lim
x 0
sin x
1 cos x
1 and lim
0 . The proof of these theorems
x
0
x
x
are found in [2] Leithold, L. (2002). These theorems proved that the sine
and cosine functions are continuous at 0. The tangent, secant, cosecant and
cotangent functions are continuous on their domains.
1.2.2. One – sided Limits
The concept of one – sided limits lies when we approach the value of
a either from the left or from the right, or a number less than r greater
than a , that is when we choose a number on the open interval containing a
but not a itself.
For instance, the function f defined
by f x x 3 does not exists for all
x 3 . This means that the lim x 3
x 3
Figure1.31. Graph of f x x 3
from the right. So we, write lim
x 3
has no meaning. However, if x 3 , it
can be observed that that value of
f x gets closer and closer to 0. In
this case, we let x approach 3 from
the right and this is called the right–
hand limit or the one–sided limit
x 3 0 (read as limit of x 3 as x
approaches 3 from the right). Below is the definition of right–hand limit.
32
MATH 401 – DIFFERENTIAL CALCULUS
Definition 1.9. [2] Let f be a function at every number in some open
interval a, c . Then the limit of a f x as x approaches a from the right
is L , written as lim f x L if any 0 , however small, there exists a
x a
0 such that if 0 x a , then f x L .
When a number less than a is taken into consideration, we say x
approaches a from the left. This limit is called left–hand limit or the one–
sided limit from the left.
Definition 1.10. [2] Let f be a function at every number in some open
interval d, a . Then the limit of a f x as x approaches a from the left is
L , written as lim f x L if any 0 , however small, there exists a
x a
0 such that if 0 a x , then f x L .
x 4 x 4
if
, find
x 4 4 x
Example 1.25. Let f be a function defined by f x
lim f x and lim f x
x 4
x 4
Solution:
Since the limit is approach from the right of 4 , then we might
choice x 4 . So we have lim x 4 4 4 0 , therefore
x 4
lim f x 0 .
x 4
When x is approach from the left of 4 , we shall use x 4 , then
lim f x lim x 4 4 4 8 .
x 4
x 4
From Example 1.25 we can say that the lim f x does not exists since
x 4
lim f x lim f x . This leads to the following theorem.
x 4
x 4
Theorem 1.2. [2] lim f x exists and is equal to L if and only if lim f x
x a
xa
and lim f x exist and both are equal to L.
x a
11 x 2
if x 2 , find
2
2 x
3
x
Illustration 1.15. Let g be a function defined by g x
lim g x and lim g x and show that lim g x exists.
x 2
x 2
x 2
Solution:
33
MATH 401 – DIFFERENTIAL CALCULUS
lim g x lim 3 x 2 3 2 7
i.
2
x 2
x 2
x 2
x 2
ii. lim g x lim 11 x 2 11 22 7
Since lim g x and lim g x exist and are both equal to 7, then
x 2
x 2
lim g x exists and lim g x 7
x 2
x 2
if x 3
x 5
2
Illustration 1.16. Let f be a function defined by f ( x) 9 x if 3 x 3
3 x
if 3 x
Determine whether the lim f x and lim f x exist.
x 3
x 3
Solution:
i. To show that lim f x exists we have
x 3
to show that lim f x and lim f x
x 3
x 3
both exist and are equal.
lim f x lim 3 x 3 3 0
x3
x 3
lim f x lim 9 x 2 9 32 0
x3
x 3
Since lim f x and lim f x both exists
x 3
x 3
and both are equal to 0, then lim f x
Figure1.32. Graph of f
exists and lim f x 0 .
x 3
x 3
ii. lim f x lim 9 x 2 9 3 0
2
x3
x3
lim f x lim x 5 3 5 2
x3
x 3
Since lim f x and lim f x both exists however lim f x lim f x ,
x 3
x 3
x 3
then lim f x does not exists.
x 3
x 3
1.2.3. Infinite Limits
Let us consider the function f defined by f x . Observed that as x
1
x
f x tends to increase
1
1
approaching , so we say that lim . Also, we see that lim .
x 0 x
x 0 x
approaches 0 from the right, the value of
34
MATH 401 – DIFFERENTIAL CALCULUS
This case of limits is called infinite limits. It
also confirms that the line x 0 is the
vertical asymptote of f x .
1
x
Figure1.33. Graph of f x 1
x
Definition 1.11. [2] The line x a is a vertical asymptote of the graph
of the function f if at least one of the following statements is true:
i.
lim f x
x a
ii. lim f x
x a
iii. lim f x
x a
iv. lim f x
x a
Below are the theorems for infinite limits.
Theorem 1.3. [2] If r is any positive integer, then
1
x 0 x r
1
ii. lim r
x 0 x
i.
lim
if r is even
if r is odd
Illustration 1.17. Use Theorem 1.13 to evaluate lim
x 0
1
.
x5
Solution:
By Theorem 1.3 (ii), it is an obvious observation that r 5 which is
an odd positive integer and the limit is approaching 0 from the left.
So we have lim
x 0
3
.
x5
Illustration 1.18. Evaluate lim
x 0
x3
.
x 3x 3
4
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MATH 401 – DIFFERENTIAL CALCULUS
Solution:
lim
x3
can be simplified into
x 3x 3
lim
x 3
x 3
1
lim
lim
x 4 3x 3 x0 x 3 x 3 x0 x 3
x 0
x 0
4
Since the limit is approaching 0 from the right, it follows that
lim
x 0
x 3
.
x 4 3x 3
Theorem 1.4. [2] If a is any real number and if lim f x 0 and
lim g x c , where c is any constant not equal to 0, then
x a
x a
i. If c 0 and if f x 0 through positive values of f x ,
g x
x a f x
ii. If c 0 and if f x 0 through negative values of f x ,
g x
lim
x a f x
iii. If c 0 and if f x 0 through positive values of f x ,
g x
lim
x a f x
iv. If c 0 and if f x 0 through negative values of f x ,
g x
lim
x a f x
The theorem is also valid if “ x a ” is replaced by “ x a ” or “ x a ”
lim
Illustration 1.19. Apply Theorem 1.4 to evaluate the following limits:
x2
x 1 x 1
x2 x 2
(c) lim 2
x 3 x 2 x 3
(a) lim
Solution:
(b) lim
3 x2
x
(d) lim
x2 4
x2
x 0
x 2
x2
x 1 x 1
Let g x x 2 and f x x 1 , then
lim g x lim x 2 1 2 3
(a) lim
x1
x1
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MATH 401 – DIFFERENTIAL CALCULUS
lim f x lim x 1 1 1 0 . By applying Theorem 1.4 (iii)
x1
x 1
lim
x 1
(b) lim
x 0
lim
x 0
x 2 3
x 1 0
3 x2
x
3 x2
3 02
3
. The numerator
x
0
0
3 0 and the limit is
approached through the negative values, by applying Theorem
1.4 (ii)
lim
x 0
3 x2
3
x
0
x2 x 2
(c) lim 2
x 3 x 2 x 3
32 3 2 14 By Theorem 1.4 (i)
x2 x 2
lim 2
2
x 3 x 2 x 3
3 23 3 0
x 2 x 2 14
lim 2
x 3 x 2 x 3
0
(d) lim
x 2
x2 4
x2
x2 4
22 4 0
. It is noted from the theorem that the
x 2
x2
22
0
x2 4
numerator should not be 0. So we need to simply
before
x2
lim
applying the theorem. That is
lim
x 2
x2 4
lim
x 2
x2
x 2
x2
lim
x 2
x2
x 2
x2
x2
22
22
4 2
0
0
By Theorem 1.4 (i)
lim
x 2
x2 4 2
x2
0
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MATH 401 – DIFFERENTIAL CALCULUS
Theorem 1.5. [2]
i. If lim f x , and lim g x c , where c is any constant not
x a
x a
equal to 0, then lim f x g x
x a
ii. If lim f x , and lim g x c , where c is any constant not
x a
x a
equal to 0, then lim f x g x
x a
The theorem holds if “ x a ” is replaced by “ x a ” or “ x a ”
Illustration 1.20. Apply Theorem 1.5 to evaluate lim
x 3
1
1
.
x 3 x 3
Solution:
Since lim
x 3
1
1
1
1
1
1
and lim
, then
x
3
x 3 33 6
x 3 33 0
1
1
1
lim
.
x3
6
x 3 x 3
1
2x
2
.
2
x 4x 2 x 4
Illustration 1.21. Evaluate lim
x 2
Solution:
4
2 2
2
2x
lim 2
2
x 2 x 4 x 2
3
2 4 2 2 6
1
1
1
lim 2
2
x 2 x 4
2 4 0
Therefore,
1
2
2
2x
lim 2
2
.
3
3
x 4x 2 x 4
x 2
Theorem 1.6. [2] If lim f x , and lim g x c , where c is any
x a
x a
constant not equal to 0, then
i. if c 0 , lim f x g x
xa
ii. if c 0 , lim f x g x
xa
The theorem holds if “ x a ” is replaced by “ x a ” or “ x a ”
38
MATH 401 – DIFFERENTIAL CALCULUS
Theorem 1.7. [2] If lim f x , and lim g x c , where c is any
x a
x a
constant not equal, then
i. If c 0 , lim f x g x
xa
ii. If c 0 , lim f x g x
xa
The theorem holds if “ x a ” is replaced by “ x a ” or “ x a ”
Illustration 1.22. Apply Theorem 1.6 and 1.7 to evaluate (a)
9 x2 x 2
2 2x 1
and
(b)
lim
lim
x 4
x 3 3 x
x
2
x4 x2
Solution:
2 2x 1
x 4
x4 x2
2
2
2
lim
By Theorem 1.4 (ii)
x 4
x4 44 0
2 x 1 24 1 9
lim
x 4
42
2
x2
2 2x 1
9
By Theorem 1.7 (i) lim
x 4
x4 x2
2
(a) lim
9 x2 x 2
(b) lim
x 3 3 x
x
2
9 x2
lim
x 3 3 x
lim
x3
3 x 3 x
3 x 3 x
lim
2
3 x 2 x3 3 x
3 x
33
6
lim
By Theorem 1.4 (i)
x 3
33
0
3 x
x 2 3 2
lim
5
x 3 x 2
3 2
9 x2 x 2
5
By Theorem 1.6 (i) lim
x 3 3 x
x
2
39
MATH 401 – DIFFERENTIAL CALCULUS
Exercise 1.4. Evaluate the following by applying the different theorems on
limits.
x2 x 2
x 3 x 2 2 x 3
x2 4
2. lim
x 2
x2
x 4
3. lim
x 4
x4
1
4. lim
x 5 x 5
x2
5. lim
x 1 x 1
x2
6. lim 2
x 2 x 4
3 x2
7. lim
x 0
x
2
x 9
8. lim
x 3
x3
4 x2
9. lim
x 2
x2
5 x
10. lim 2
x 2 x 25
x2
11. lim
x 4 4 x
1 1
12. lim 2
x 0 x
x
3
1
13. lim
2
x 2 x 2
x 4
x 1
14. lim
x 1
2x x 2 1
x2
15. lim
x 2
2 4x x 2
1. lim
References: [1] Larson, R. (2018) and [2] Leithold, L (2002)
40
MATH 401 – DIFFERENTIAL CALCULUS
1.3. CONTINUITY
Another concept of a function that needs to be
understood is its continuity. For example the
function f defined by f x 2 x 3 is continuous at
every number on the set of real numbers, i.e.
, . Same is also true for polynomial function
(see Figure 1.13 (c)) which is continuous at every
number.
One obvious observation to identify the
continuity of a function is the behavior of its
domain. That is if the limit of a function f exits at a
number a, then f is continuous at a number a.
Figure1.35. Graph of f x
x2 9
x3
Figure1.34. Graph of f x 2 x 3
2
3x 1
Figure1.36. Graph of f x 1 x
Observe from Figure 1.35 that when f x
if x 0
if 1 x
x2 9
is simplified we have
x3
f x x 3 . When x is replaced by -3, f x 6 . However, we cannot substitute -
3 to the function f x
x2 9
x3
since the function is discontinuous at x 3 . The
point 3,6 is an example of point discontinuity. Another example is the function
f on Figure 1.36. Noticed that f is discontinuous on the interval 0,1 . This type of
discontinuity is called jump discontinuity. For the functions
f x
f x
1
and
x
1
, the line x 0 is the vertical asymptote. This type of discontinuity is
x2
called infinite discontinuity.
41
MATH 401 – DIFFERENTIAL CALCULUS
Definition 1.11. [2] The function f is said to be continuous at the number a if
and only if the following conditions are satisfied:
i. f a exists;
ii. lim f ( x) exists, and
xa
iii.
lim f x f a
x a
If one or more of these conditions fails to hold at a, the function f is said to be
discontinuous at a.
Example 1.26. Let f be a function defined by f x
continuous at x 3 ?
1
. Is the function f
x3
Solution:
Graphing this function, we’ll observe that the graph of f x
1
x3
has a break at x 3 . This line is the asymptote or the infinite discontinuity
of the function. Following Definition 1.11, we have f 3 , therefore f 3
1
0
does not exist. This violates the first condition of the definition of
continuity. Therefore, f x
1
is discontinuous at x 3 .
x3
2 x
6 x
Example 1.27. Determine whether the function g x
if
x2
is
2 x
continuous at x 2
Solution:
(i)
Find g 2 .
g x 6 x g 2 6 2 4
Since g 2 4 , then g 2 exists.
(ii) Does lim g x exist?
x 2
By Theorem 1.2 lim g x if lim g x and lim g x exist and are
x 2
x 2
x 2
equal.
a. lim g x lim 6 x 6 2 4
x2
x 2
lim g x exist.
x 2
b. lim g x lim 2 x 2 2 4
x 2
x 2
42
MATH 401 – DIFFERENTIAL CALCULUS
lim g x exist.
x 2
Therefore, lim g x since lim g x lim g x
x 2
x 2
x 2
(iii) Does lim g x f 2 ?
x 2
In (i)and (ii) we see that lim g x f 2
x 2
Since the three conditions hold for g then the g is continuous
at x 2 .
if x 3
x 5
2
Example 1.28. Is the function f defined by f ( x) 9 x if 3 x 3
3 x
if 3 x
continuous at (a) x 3 and (b) x 3 ?
Solution:
a. at x 3
(i) f 3
f x 9 x 2 f 3 9 3 0
2
(ii) Does lim f x exist?
x 3
lim f x lim x 5 3 5 2 , lim f x exists.
x3
x3
x 3
lim f x lim 9 x 2 9 3 0 , lim f x exists.
2
x 3
x 3
x 3
Since lim f x lim f x , therefore lim f x does not exist.
x 3
x 3
x 3
Therefore f x is NOT continuous at x 3 .
b. at x 3
(i) f 3
f x 9 x 2 f 3 9 3 0
2
(ii) Does lim f x exist?
x 3
lim f x lim 9 x 2 9 3 0 , lim f x exists.
2
x3
x3
x3
x 3
x 3
lim f x lim 3 x 3 3 0 , lim f x exists.
x 3
43
MATH 401 – DIFFERENTIAL CALCULUS
Since lim f x lim f x , therefore lim f x exists and
x 3
x 3
lim f x 0
x 3
x 3
(iii) lim f x f 3 0
x 3
Therefore f x is continuous at x 3 .
Exercise 1.5. Determine whether the following functions are continuous
at the indicated number.
if x 2
at x 2
2.
x2
1
f ( x) x 2 if
3
x2
at x 2
3.
x 2 1,
2 x
f ( x)
2 x 4
0
1. f ( x) x
2
0
x2
if
1 x 0
0 x 1
at x 0 , x 1 and x 2
1 x 2
2 x3
2 x 3 x
if x 1
1 x
x x
4. g x
at x 1
1
x3
5. h x x
if
at x 3
2
3 x
9 x
Reference: [2] Leithold, L. (2002)
44
MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER TEST (Problem Set 1)
I. Sketch the graph of the following function and determine its domain and
range.
1. f ( x)
2.
x 2 81
x9
1
x
f ( x) 1 if
x
1
x 1
1 x 0
x0
0 x 1
x 1
II. Let f and g be functions defined by f ( x) x 2; g ( x) x 2 4 . Find the
following:
3. f g
4. f g
5. f g
f
g
7. f g
8. g f
6.
Also, find the domain and range of the each resulting function.
III. Find the limits of the following functions.
1 x
x 1 1 x
( x h) 2 x 2
10. lim
h0
h
4
x x3
11. lim
x 12 x 3 128
2x
16 x 2
x 1
13. lim 2
x 2 x 4
12. lim
9. lim
x 4
IV. Sketch the graph of the function. Determine whether the function is
continuous at the indicated number.
2
14. f ( x) x 1
x 5
x 4
15. f ( x) 16 x 2
4 x
x3
at x 3
3 x
if
if
x 4
4 x 4 at x 4 and x 4
4 x
References: [1] Larson, R. (2018) and [2] Leithold, L. (2002)
45
MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER II
THE DERIVATIVES AND DIFFERENTIATION OF
ALGEBRAIC FUNCTIONS
The concepts of limit of a function and its continuity play a significant role
in finding the derivative of a function. If the function f is continuous on an open
interval (a,b) and the limit also exists on this interval, then we say that the
function f is differentiable on that interval. In this chapter, we shall introduce the
concepts of derivatives first by geometric interpretation as the slope of a tangent
line to the graph of the given function. This gives the formal definition of the
derivatives of a function. Second, this chapter also includes how theorems and
formulas are used to find the derivative of a function and the high order
derivatives of a function.
At the end of this chapter, the student might be able to:
1. Define the process of differentiation and determine the derivative of a
function by increments.
2. Define derivative as slope of tangent line to the curve and solve
problems related to it.
3. Apply the different theorems on differentiation of algebraic functions
4. Apply the concept of chain rule and the general power rule on algebraic
functions.
5. Determine higher order derivatives and derivatives of implicit
functions.
2.1
Differentiation
The process of finding the derivative of a function is called differentiation
and the branch of calculus that deals with this process is called differential
calculus. Differentiation is an important mathematical tool in physics, mechanics,
economics and many other disciplines that involve change and motion.
Consider a continuous function y f ( x) . Let P and Q be any distinct
points of the curve which determine secant PQ . Representing the point P by
( x, f ( x)) and Q by ( x x, f ( x x)) we know that the slope of PQ will be
f ( x x) f ( x)
x
46
MATH 401 – DIFFERENTIAL CALCULUS
We can see that as x approaches
zero, the point Q becomes nearer and
nearer to point P . At the same time that
point Q approaches P , the secant PQ
rotates about the point P . Intuitively, the
limiting position of the secant PQ as Q
approaches P is that of the position of
the tangent line to the curve y f ( x) at
point P . In symbols,
y
f x x f x
lim
x0 x
x0
x
lim
is equal to the slope of the tangent line at P . This shows that the derivative of
y f ( x) at the point P is equal to the slope of the tangent line at the same point.
Using the functional notation y f ( x) , the following are the usual symbols
used to mean the first derivative:
dy d
d
;
[ f ( x)] ; Dx y ; Dx f ( x) ; etc.
( y) ;
y ' ; f '( x) ;
dx dx
dx
Considering a continuous function y f ( x) , we define
y' f ' x
dy
f x x f x
lim
the first derivative of the function f .
dx x0
x
Note:
The symbol x , read as “delta x ,” is a single entity which means increment
or change in x .
To find the slope of the tangent line to the curve at point P means that we
are to find the value of the derivative at that point P.
There are two ways of finding the derivative of a function:
1. By using the increment method
2. By using the differentiation formulas
Definition2.1. Suppose that x1 is in the domain of the function f, the tangent line
to the curve y=f(x) at the point P(x1,f(x1)) is the line with equation,
y f ( x1 ) m( x x1 )
where m lim
x 0
f ( x1 x) f ( x1 )
provided the limit exists, and P( x1 , f ( x1 ))
x
is the point of tangency.
47
MATH 401 – DIFFERENTIAL CALCULUS
Definition2.2. The derivative of y = f(x) at point P on the curve is equal to the
slope of the tangent line at P, thus the derivative of the function f given by y= f(x)
with respect to x at any x in its domain is defined as:
f ' x
dy
y
f x x f x
lim
lim
dx x0 x x0
x
provided the limit exists.
Example1. Differentiate the function y 3x
Solution:
y 3x
y y 3x x
y y 3x 3x
y 3x 3x y
Since y 3x then
y 3x 3x 3x
y 3x
y
3
x
dy
y
lim
lim 3 3
dx x0 x x0
dy
3
dx
2
Example2. Differentiate the function y x x
Solution:
y x2 x
y y x x x x
2
y y x 2 2 xx x x x
2
y x 2 2 xx x x x y
2
2
Since y x x then
y x 2 2 xx x x x x 2 x
2
y 2 xx x x
2
y x2 x x 1
y
2 x x 1
x
48
MATH 401 – DIFFERENTIAL CALCULUS
dy
y
lim
lim 2 x x 1 2 x 0 1 2 x 1
dy x0 x x0
dy
2x 1
dy
2.2
The Increment Method
Based on the definition of the derivative, y' f ' x lim
x 0
f x x f x
x
we can write the steps in solving for this derivative.
These are as follows:
Step 1. Write down the expression for f ( x x) f ( x) and simplify.
Step 2. Divide the result in Step 1 by x , again simplify.
Step 3. Find the limit of the result in Step 2 as x approaches zero. The
obtained limit is the derivative.
Example1. Given
find the first derivative f '( x) using the threestep rule.
Solution:
Step1.
Step2.
f x x f x
lim 2 x 5 x
x 0
x 0
x
f x x f x
lim
2x 5 0
x 0
x
f x x f x
lim
2x 5
x 0
x
f x x f x
f ' x lim
x 0
x
Step3. lim
f ' x 2 x 5
49
MATH 401 – DIFFERENTIAL CALCULUS
Example2. Given y
1
x 1
, find y ' .
Solution:
Step1. f x x f x
1
x x 1
1
x 1 x x 1
x 1 ( x x 1)( x 1)
x 1 x x 1
( x x 1)( x 1)
x 1 x x 1
x 1 x x 1
2
2
x 1 x x 1
( x x 1)( x 1)[ x 1 x x 1]
( x 1) ( x x 1)
( x x 1)( x 1)[ x 1 x x 1]
x 1 x x 1
( x x 1)( x 1)[ x 1
f x x f x
Step2.
x x 1]
x
( x x 1)( x 1)[ x 1 x x 1]
f x x f x
x
1
x
( x x 1)( x 1)[ x 1 x x 1] x
f x x f x
1
x
( x x 1)( x 1)[ x 1 x x 1]
Step3. lim
x 0
lim
x 0
f x x f x
1
lim
x 0
x
( x x 1)( x 1)[ x 1
x x 1]
1
( x 0 1)( x 1)[ x 1 x 0 1]
1
( x 1)( x 1)[ x 1 x 1]
1
1
1
12
2( x 1) 3 2
( x 1)[2 x 1] 2( x 1)( x 1)
f x x f x
1
x
2( x 1) 3 2
f x x f x
x
1
y'
2( x 1) 3 2
y' lim
x 0
50
MATH 401 – DIFFERENTIAL CALCULUS
Exercise1. Find the derivative of the following using the definition of derivative.
1. y 1 x 2
Answer: -2x
2. f x x 2 x
Answer: 2x +1
3. y x
Answer:
2 x
1
3x
Answer:
1
3x 2
x
x 1
Answer:
x2
32
2 x 1
4. f x
5. y
1
51
MATH 401 – DIFFERENTIAL CALCULUS
2.3
Theories on Differentiation of Algebraic Functions
The increment-method (three-step rule) of finding the derivative of
a function gives us the basic procedures of differentiation. However these
rules are laborious and tedious when the functions to be differentiated are
“complex”, that is, functions with large exponents, functions with fractional
exponents and other rational functions
Understanding of the theorems of differentiation is very
important. This is the heart of differential calculus. All of the succeeding
topics such as applications of derivatives, differentiation of transcendental
functions etc. will be dependent on these theorems. Understanding of
these theorems will enable us to calculate derivatives more efficiently and
will make calculus easy and enjoyable.
Here we will be using the “dy/dx” notation (also called Leibniz's
notation) instead of limits. The following are the rules of differentiation for
Algebraic Functions.
Theorem 2.1. The Constant Rule of Differentiation
If c is a constant and any real number, then
d
(c ) 0
dx
Illustration 2.1.
d
5 0
dx
d 3
0
b.)
dx 4
d
3 0
dx
d
7 0
d.)
dx
a.)
c.)
Theorem 2.2. The Power Rule (for positive integer powers) of
Differentiation
If n is a positive integer, then
d n
x nx n 1
dx
In words, to differentiate a power function, decrease the constant
exponent by one and multiply the resulting power function by the original
exponent.
Illustration 2.2.
a.)
d 2
x 2 x 21 2 x
dx
52
MATH 401 – DIFFERENTIAL CALCULUS
b.)
d 4
x 4 x 41 4 x 5
dx
6
6
6 7
1
d 7 6 7 1 6 7
6 7
x x
x
x
c.)
dx 7
7
7
d 2
x
d.)
dx
3
3 23 1 3 322 3 25
x
x
x
2
2
2
Theorem 2.3. The Constant Multiple Rule of Differentiation
If f is a differentiable function at x and c is any real number,
then cf is also differentiable at x and
d
d
cf ( x) c f ( x)
dx
dx
In words, the derivative of a constant times a function is the constant times
the derivative of the function, if this derivative exists.
Illustration 2.3.
d
5 x 8 5 8 x 7 40 x 7
dx
d
9 x 4 9 4 x 5 36 x 5
b)
dx
a)
2
3
2 53
d
5
5 x 5 x 2 x 5
c)
5
dx
d)
d 4 3 4
2
2
r 3r 4r
dx 3
3
Theorem 2.4. The Sum Rule of Differentiation
If f and g are both differentiable functions at x,
d
f g d f d g or
dx
dx
dx
d
f ( x) g ( x) d f ( x) d g ( x)
dx
dx
dx
then so are f + g and f – g, and
In words, the derivative of a sum or of a difference equals the sum or
difference of their derivatives, if these derivatives exist.
Illustration 2.4.
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MATH 401 – DIFFERENTIAL CALCULUS
d
3x 4 d 3x d 4 3 0 3
dx
dx
dx
3
3
dy
d
d
d
d
2
2
2 x 4 x 5x 2 9 2 x 4 x 5x 2 9
b)
dx
dx
dx
dx
dx
3 1
22 x 41 5 x 2 0
2
3
dy
15 1
2
2 x 4 x 5 x 2 9 4 x 4 x 2
dx
2
a)
c)
d 2 4 3 d
d 4
r 2 r 3
r r
dx
3
dx 3
dx
4
2r 3 3r 2
3
d 2 4 3
3
2
r r 2r 4r
dx
3
Theorem 2.5. The Product Rule of Differentiation
If f and g are both differentiable functions at x, then so is the product
f g , and d f g f dg g df
or
dx
dx
dx
d
f ( x) g ( x) f ( x) d [ g ( x)] g ( x) d f ( x)
dx
dx
dx
In words, the derivative of a product of two functions is the first function
times the derivative of the second plus the second function times the
derivative of the first, if these derivatives exist.
Illustration 2.5.
a)
d
3x 4 4 x 2 3 3x 4 d 4 x 2 3 4 x 2 3 d 3x 4
dx
dx
dx
3x 48x 0 4 x 2 3 3 0
3x 48x 4 x 2 3 3
24 x 2 32 x 12 x 2 9
d
3x 4 4 x 2 3 36 x 2 32 x 9
dx
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MATH 401 – DIFFERENTIAL CALCULUS
b)
d 3
d
d 3
x 1 5 - 2x x 3 1 5 - 2x 5 - 2x
x 1
dx
dx
dx
x 3 1 0 - 2 5 - 2x 3x 2 0
x 3 1 - 2 5 - 2x 3x 2
2 x 3 2 15x 2 6 x3
d 3
x 1 5 - 2x 8 x 3 15 x 2 2
dx
Theorem 2.6. The Quotient Rule of Differentiation
If f and g are both differentiable functions at x, and if g(x) ≠ 0 then
g
f
g
df
dg
f
dx
dx or
g2
f
g
d
d
g ( x)
g ( x)
f ( x ) f ( x )
d f ( x)
dx
dx
dx g ( x)
g ( x)2
is differentiable at x and
d
dx
In words, the derivative of a quotient of two functions is the fraction
whose numerator is the denominator times the derivative of the
numerator minus the numerator times the derivative of the denominator
and whose denominator is the square of the given denominator.
Illustration 2.6.
d 4x 3
dx 1 2 x
2
a)
4 x
dx
1 2 x d
2
dxd 1 2 x
3 4x2 3
1 2 x 2
1 2 x 8 x 0 4 x 2 30 2
1 2 x 2
1 2 x 8 x 4 x 2 3 2
1 2 x 2
8x 16 x 8x
2
2
1 2 x
6
2
8 x 16 x 2 8 x 2 6
1 2 x 2
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MATH 401 – DIFFERENTIAL CALCULUS
d 4 x 2 3 8x 2 8x 6
dx 1 2 x
1 2 x 2
d 3x 4 x
b)
dx 3x 2
2
3x
dx
3x 2 d
2
dxd 3x 2
4 x 3x 2 4 x
3x 22
3x 26 x 4 3x 2 4 x 3
3x 22
18x 12 x 12 x 8 9 x
2
3x 2
2
12 x
2
18 x 2 8 9 x 2 12 x
3x 22
d 3x 2 4 x 9 x 2 12 x 8
dx 3x 2
3x 22
Theorem 2.7. The General Power Rule of Differentiation
If n is a positive integer and f is a differentiable functions at x, then
d
f x n n f x n1 d f x
dx
dx
In words, the derivative of the power of a function is equal to the power,
times the function raised to the power minus one, times the derivative of
the function.
Illustration 2.7.
a)
2 x 56 x
5
51 d
d
2x3 4x 2 2x 5 2x3 4x 2 2x 2x3 4x 2 2x
dx
dx
4
5 2 x 3 4 x 2 2 x 6 x 2 8x 2
d
2x3 4x 2
dx
5
2
8x 2 2 x 3 4 x 2 2 x
4
1
1
d 3
d
1
1 d
3
3
5x 4
5
x
4
5
x
4
5
x
4
b)
dx
dx
3
dx
2
1
5x 4 3 5 0
3
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MATH 401 – DIFFERENTIAL CALCULUS
2
d 3
5
5 x 4 5x 4 3
dx
3
Theorem 2.8. The Derivative of a Composite Function
If the function g is differentiable at x and the function f is
differentiable at g(x), then the composition function
f g is
differentiable at x and
f
g ' x f ' g x g ' x
Illustration 2.8.
2
5
a) Let f x x and g x x 3x 1 find f g ' x
Solution:
Because f x x then f ' x 5x , thus
4
5
f ' g x 5g x
4
f ' g x 5 x 2 3x 1 , furthermore
4
Because g x x 3x 1 then g ' x 2 x 3
2
Therefore
f
f
g ' x f ' g x g ' x
2x 3
g ' x 5 x 2 3x 1
4
Illustration 2.9.
3
b) Let f x x and g x
2
find f g ' x
x 1
Solution:
Because f x x then f ' x 3x , thus
2
3
f ' g x 3g x
2
2
2
f ' g x 3
, furthermore
x 1
2
2
Because g x
then g ' x
x 1
x 12
Therefore
f
g ' x f ' g x g ' x
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MATH 401 – DIFFERENTIAL CALCULUS
f g ' x 3 2
x 1
2
2
2
x 1
12
g ' x
2
x 1
f g ' x 24 4
x 1
2
2
x 1
f
Theorem 2.9. The Chain Rule of Differentiation
If g is differentiable at x and if f is differentiable at g(x), then the
composition f g is differentiable at x. Moreover, if y=f(g(x)) and
dy dy du
dx du dx
u=g(x) then y=f(u) and
In words, the derivative of a composite function of f and g is equal to the
product of their respective derivatives.
Illustration 2.10.
a) Given y u
10
and, u 2 x 5x 4 find
3
2
dy
dx
Solution:
dy
10u 9
du
from y u , the derivative is
10
from u 2 x 5x 4 , the derivative is
3
2
du
6 x 2 10 x , then
dx
dy dy du
10u 9 6 x 2 10 x , since u 2 x 3 5x 2 4 , then
dx du dx
9
dy
10 2 x 3 5x 2 4 6 x 2 10 x
dx
dy
3
4
b) Given y 5u 2 and, u 3 x find
dx
Solution:
dy
15u 2
du
du
4
4x 3 , then
from u 3 x , the derivative is
dx
from y 5u 2 , the derivative is
3
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MATH 401 – DIFFERENTIAL CALCULUS
dy dy du
15u 2 4 x 3 60u 2 x 3 , since u 3 x 4 , then
dx du dx
2
dy
60 x 3 3 x 4
dx
Theorem 2.10. If x f t and y g t are parametric equations, the
derivative of y with respect to x is equal to the quotient of the derivative
of y with respect to t divided by the derivative of x with respect to t .
dy
dy dt dx
0
;
dx
dx dt
dt
Illustration 2.11.
2
a) x 2t 4t and y 4t 2 , find
dy
dx
Solution:
dx
4t 4
dt
dy
4 , then
from y 4t 2 , the derivative is
dt
dy dy dt
4
4
dx dx dt 4t 4 4t 1
2
from x 2t 4t , the derivative is
dy
1
dx t 1
dy
3
2t 1
and y
, find
dx
t
4
dx 3
3
from x , the derivative is
dt t 2
t
dy 1
2t 1
, then
from y
, the derivative is
dt 2
4
1
2
dy dy dt
1 t
2
3 2 3
dx dx dt
t2
b) If x
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MATH 401 – DIFFERENTIAL CALCULUS
dy t 2
dx
6
dy
dx
is equal to the reciprocal of the derivative of the given function with
Theorem 2.11. If the variable x is a function of y then the derivative
respect to y . If x g y , then
dy
1
dx
dx
dy
Illustration 2.12.
a) Given x 3y , find
3
dy
dx
Solution:
from x 3y , the derivative is
3
dx
9y2
dy
3
dy
1
1
9x
3
2 , from x 3y the value of y
dx
dx
9y
3
dy
dy
1
dx
9 y2
dy
dx
1
3
81x 2
1
3 9x
9
3
3
3
2
9x
9x
3
3
9
3
1
9x
9
2
1
3
81x 2
9x
729 x3
3
dy
9x
dx
9x
Theorem 2.12. If u is a differentiable function of x, then
d
dx
du
u dx
2 u
The derivative of a radical whose index is two, is a fraction whose
numerator is the derivative of the radicand, and whose denominator is
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MATH 401 – DIFFERENTIAL CALCULUS
twice the given radical, if the derivative exists
Illustration 2.13.
a) Given f x
2 x 2 4 x , find f ' x
Solution:
d
2x 2 4x
d
f ' x
2 x 2 4 x dx
dx
2 2x 2 4x
4x 4
2 2x2 4x
22 x 2
2 2x2 4x
2 x 2
f ' x
2x2 4x
b) If y 3 2 x
2
find y '
d
3 2x2
d
y'
3 2 x 2 dx
dx
2 3 2x2
0 4x
2 3 2x2
4x
2 3 2x2
2x
y'
3 2x2
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise2. Differentiate the following function by applying the theorems
1.
f x
2.
1 2
x 4x 9
2
Answer: x+4
f x 2 x 3 10 x 5
Answer: 40 x 3 15x 2 - 20
3.
x 2 x
y
x 1
Answer:
4.
5
y
x 1
3
2
3
Answer:
find f g ' x
2
7. If x
t 2
dy
t
and y
2 , find
t 4
dx
t 4
8. If x 2 y y 4 , find
2
dy
dx
9. If f x x 3 2 x , find f ' x
2
2
1
2
375
x 14
2
Answer: 48x 3x 1
2
2
6. If y u 9 and u 3x 5 find
x
5. If f x 4x and g x 3x 1
2
x 4 5x 2 2
dy
dx
2
Answer: 12 x 3x 5
Answer:
t
4t 4
Answer:
1
1 84 x
Answer:
4x 2 3
3 2x 2
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MATH 401 – DIFFERENTIAL CALCULUS
2.4
Higher-Order Derivatives
The derivative f ' of a function f is itself a function and hence may have a
derivative of its own. If f ' is differentiable, then its derivative is denoted by f ''
and is called the second derivative of f . As long as we have differentiability, we
can continue the process of differentiating to obtain the third, fourth, fifth, and
even higher derivatives of f .
These successive derivatives are denoted by
f ', f '' ( f ')', f ''' ( f '')', f 4 ( f ''')', f 5 ( f 4 )',...
Using the same functional notation y f ( x) , we will have the following
symbols for higher derivatives:
First Derivative:
y ' , f '( x) ,
dy d
d
,
f x , Dx f x
y ,
dx dx
dx
Second Derivative:
y '' , f ''( x) ,
d2y d2
d2
,
y
,
f x , D2 x f x
2
2
2
dx
dx
dx
Third Derivative:
y ''' , f '''( x) ,
d3y d3
d3
,
y
,
f x , D3 x f x
3
3
3
dx
dx
dx
nth Derivative:
dn y dn
dn
,
y
,
f x , Dn x f x
n
n
n
dx
dx
dx
2
n
dy d y d y
The symbols
are called Leibniz notations.
,
,
dx dx2 dxn
y n , f n ( x) ,
3
2
Example1. Given f x x 2 x 3x 5 , find f '( x) , f ''( x) and f '''( x)
Solution:
f ' x 3x 2 4 x 3
f ' ' x 6 x 4
f ' ' ' x 6
2
1
Example2. Find f '( x) , f ''( x) and f '''( x) given that f x x 2 x
Solution:
f ' x 2 x 2 x 2 2 x 2 x 2
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MATH 401 – DIFFERENTIAL CALCULUS
f ' ' ' x 0 4 3x 12 x
f ' ' x 2 2 2 x 3 2 4 x 3
4
4
2
Example3. Find the first and second derivatives of f x x x 5x
Solution:
f x x 2 x 5 x f x x 2 x 5 x f x x 5 2 5 x
12
5
5 1
5
f ' x x 2 5 x 3 2 5
2
2
5 3 1
f ' ' x x 2 0
22
3
f ' ' x
15 1 2 15 x
x or
4
4
Exercise3. Determine the first and second derivative given the following
functions.
1. f x x 3 9 x 2 27 x 27
Answer: 3x 2 18x 27 , 6 x 18
1
x
2.
y
Answer:
3.
f x 4 x 2 1
2
4.
y
1 2
x 3x
x
2
1
,
x2
2
x3
Answer: 16 x 3 16 x , 48x 2 16
Answer: 3x 2 12 x 9 , 6 x 12
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MATH 401 – DIFFERENTIAL CALCULUS
5.
f x 9 x 2
Answer:
x
9 x2
,
9
9 x
2 32
2.5
Implicit Differentiation
There are two ways to define functions, implicitly and explicitly. Most of the
equations we have dealt with have been explicit equations, such as y 2x 3 , so
that we can write y f ( x) where f ( x) 2x 3 . But the equation 2x y 3
describes the same function. This second equation is an implicit definition of y
as a function of x . As there is no real distinction between the appearance of x
or y in the second form, this equation is also an implicit definition of x as a
function of y .
An implicit function is a function in which the dependent variable has
not been given "explicitly" in terms of the independent variable. To give a
function f explicitly is to provide a prescription for determining
the output value of the function y in terms of the input value x : y f ( x) . By
contrast, the function is implicit if the value of y is obtained from x by solving an
equation of the form: f ( x, y) 0 .
An equation of the form y f ( x) is said to define explicitly as a function of
x because the variable y appears alone on one side of the equation and does not
appear at all on the other side. However, sometimes functions are defined by
equations in which y is not alone on one side; for example the equation
yx y 1 x is not of the form y f ( x) , but still defines y as a function of x
x 1
. Thus we say that yx y 1 x defines
x 1
x 1
implicitly as a function of x , the function being f ( x)
.
x 1
since it can be rewritten as y
y
Suppose we have an equation f ( x, y) 0 where neither variable could
be expressed as a function of the other. In other words, it wouldn’t be possible, by
rearranging f ( x, y) 0 , to separate out one of the variables and express it as a
function of the other. Often we can solve an equation f ( x, y) 0 for one of the
variables obtaining multiple Solutions constituting multiple branches. Consider
the equation x2 y 2 1 0 which defines y as an implicit function of x . If we
solve for y in terms of x , we obtain two Solutions y 1 x2 and y 1 x2
thus we have found two functions that are defined implicitly by x2 y 2 1 0 .
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MATH 401 – DIFFERENTIAL CALCULUS
In general, it is not necessary to solve an equation for y in terms of x in
order to differentiate the functions defined implicitly by the equation. To find the
derivative of functions defined implicitly we use implicit differentiation.
Steps in Implicit Differentiation:
1. Differentiate both sides of the equation with respect to x .
dy
2. Collect all the terms with
on one side of the equation.
dx
dy
3. Factor out
and solve for it.
dx
2
dy
Example1. Given x 4 y 5 , find
.
dx
Step1. 2 x 4
dy
0
dx
dy
2 x
dx
dy 2x
Step3.
dx
4
Step2. 4
dy
1
x
dx
2
3
3
dy
Example2. Given x y 5xy 0 , solve for
dx
Step1.
d 3 d 3 d
d
x
y 5xy 0
dx
dx
dx
dx
dy dy
3x 2 3 y 2 5 x y1 0
dx dx
dy
dy
5x 5 y 0
dx
dx
dy
2 dy
5x 3x 2 5 y
Step2. 3 y
dx
dx
dy 2
3 y 5 x 3x 2 5 y
Step3.
dx
3x 2 3 y 2
Example3. Given 2 x x y y
2
2
2
3
dy 3x 2 5 y
dx
3 y 2 5x
4 , find y '
4
Step1. 4 x [( x )(2 yy ' ) ( y )(2 x)] (3 y y' ) 0
2
2
4 x 2 x 2 yy '2 xy 2 3 y 4 y' 0
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MATH 401 – DIFFERENTIAL CALCULUS
4
Step2. 2 x yy '3 y y' 4 x 2 xy
2
Step3. 2 x y 3 y
Exercise4. Find
2
4
2
y' 4x 2xy
2
4 x 2 xy
2
y'
2 x
2
y 3 y 4
dy
or y ' by implicit differentiation
dx
4x
9y
1.
4 x 2 9 y 2 36
Answer:
2.
x 2x y y 5
3x 4 x
Answer:
4 y 1
3.
4 x 5x y 4 y 3 y
4.
x y 3xy
5.
xy 2 y 3xy 2 x y 3 y
2
3
2
3
4
2
2
3
12x
2
y x y
Answer:
x y x
2
3
3
2
3
3
2
10 xy 2
Answer:
10 x 2 y 12 y 2 3
3
3
2
Answer:
3
2
4x y
3y
2 xy 3 y 3x 6 y
2
2
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER TEST (Problem Set 2).
Solve the following completely and neatly as possible.
Find the derivative of the following using the three-step rule
1. y
3x 1
2x 5
2
2
2. y a x
Differentiate the following function by applying the theorems on differentiation
of algebraic functions
3. ht t 3t 5 3t t
4.
4
2
4
2
2
3x 2
y
2 x 62
5. x
t2 2
t
y
and
t4
t 42
3
6. y u 9 and u 4 x 9
2
Find the first, second and third derivative of the given function
7. f x
8. y
4
t 22
1
x
x
Find y’ using implicit differentiation given the following functions
9.
x y xy 21
10. x y x y
2
2
2
2
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER III
DERIVATIVES OF TRANSCENDENTAL FUNCTIONS
So far you have studied one of the elementary functions – the algebraic
functions. In this chapter, the learners will study the derivatives of the remaining
elementary functions – the transcendental functions. Theorems presented in this
chapter are taken from [1] Larson, R. (2018), [2] Leithold, L. (2002) and [3]
Stewart, J. (2016)
At the end of this chapter, the student should be able to:
1. Define the derivatives of transcendental functions such as trigonometric,
inverse trigonometric, logarithmic, exponential and hyperbolic functions.
2. Apply the theorems of differentiation in finding the derivatives of
transcendental functions.
3.1
TRIGONOMETRIC FUNCTIONS
With this section we are going to start looking at the derivatives of
functions other than polynomials or roots of polynomials. We will start this
process off by taking a look at the derivatives of the six trigonometric functions.
The basic six trigonometric functions include the following six functions:
sine (sin x), cosine (cos x), tangent (tan x), cotangent (cot x ), secant (sec x), and
cosecant (csc x). All these functions are continuous and differentiable in their
domains.
Before we actually get into the derivatives of the trigonometric functions
we need to recall a couple of limits that will show up in the derivation of two of
the derivatives.
lim
lim
3.1.1. Derivatives of Trigonometric Functions
The derivatives of the six trigonometric functions are summarized
in the following theorem.
Theorem 3.1. Derivatives of Trigonometric Functions
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MATH 401 – DIFFERENTIAL CALCULUS
Proof: Let f be the sine function, so that
From the definition of the derivative of a function, the derivative for
the sine function can be written as
To evaluate this limit, we will need to use the trigonometric identity
For the proof of the derivative of cosine function, we can follow a
similar process
Let g be the cosine function, so that
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MATH 401 – DIFFERENTIAL CALCULUS
From the definition of the derivative of a function, the derivative for
the cosine function can be written as
To evaluate this limit, we will need to use the trigonometric identity
The derivatives of tangent, cotangent, secant and cosecant functions
are obtained from the trigonometric identities involving sine and cosine
functions as well as the above derivatives of the sine and cosine functions,
and the previous theorems on differentiation of algebraic functions.
For the tangent function, we will use the trigonometric identity
Applying the Quotient Rule of differentiation, we have
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MATH 401 – DIFFERENTIAL CALCULUS
For the cotangent function, we will use the trigonometric identity
Applying the Quotient Rule of differentiation, we have
For the secant function, we will use the trigonometric identity
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MATH 401 – DIFFERENTIAL CALCULUS
Applying the Quotient Rule of differentiation, we have
For the cosecant function, we will use the trigonometric identity
Applying the Quotient Rule of differentiation, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Let us have the following examples:
Example 3.1.1.1 Differentiate
Solution: We will need to use the Product Rule of differentiation on the
first term. Also the constant 2 will be considered part of the first function
in the product of the first term. Here is the derivative of the function y
Example 3.1.1.2 Find the derivative of
Solution: We will need to use the Product Rule of differentiation on the
second term. Be careful with the minus sign in front of the second term
and make sure it gets dealt with properly. There are two ways to deal with
this. One way is to make sure that you use a set of parentheses as follows:
Another way to do this is to consider minus sigh as part of the first
function in the product of the second term. Doing this gives
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MATH 401 – DIFFERENTIAL CALCULUS
So regardless how you approach this problem, you will get the same
derivative.
Example 3.1.1.3 Find the derivative of
at
Solution: Just differentiate each term using the formula above.
At
.
. the value of the first derivative of h(x) is
Example 3.1.1.4 Differentiate
Solution: We will use the quotient rule of differentiation to evaluate the
derivative of this function.
We can still simplify this by factoring out 3 in the last two terms in
the numerator and use the Pythagorean identity
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MATH 401 – DIFFERENTIAL CALCULUS
Since the numerator is exactly a factor of the denominator, we can
canceled them and gives
Example 3.1.1.5 Differentiate
Solution: We need to use the product rule of differentiation for both terms.
Combining like terms, we have
3.1.2. Chain Rule of the Derivatives of Trigonometric Functions
In this section, we will determine the derivatives of trigonometric
functions using the Chain Rule. This rule deals with composite functions
involving trigonometric functions.
The Chain Rule versions of the derivatives of the six trigonometric
functions are as follows: Let u be a differentiable function of x
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.1.2.1 Apply the Chain Rule to find the derivative of the following
functions:
(a)
(b)
Solution:
Solution:
(c)
(d)
Solution:
Solution:
(e)
(f)
Solution:
Solution:
Example 3.1.2.2 To understand the mathematical conventions regarding
parentheses, apply the Chain Rule to find the derivative of the following
functions:
Function
Derivative
(a)
(b)
(c)
(d)
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MATH 401 – DIFFERENTIAL CALCULUS
(e)
Example 3.1.2.3 Find the derivative of the following:
(a)
Solution:
We must use the Product Rule and Chain Rule.
(b)
Solution:
We must use the Quotient Rule along with the Chain Rule.
(c)
Solution: Recognize here that we have
function that
is inside the
function; that is
. We will start
using the General Power Rule, then the Chain Rule by approaching this
step-by-step.
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MATH 401 – DIFFERENTIAL CALCULUS
To express this in terms of trigonometric function only, we can use the
Pythagorean identity
3.1.3. Higher-Order Derivatives of Trigonometric Functions
The higher-order derivatives are denoted as follows:
First Derivative :
Second Derivative:
Third Derivative:
Fourth Derivative:
nth Derivative:
Let’s have the following examples:
Example 3.1.3.1 Find the second derivative of the following:
(a)
Solution:
( )
(b)
Solution:
(d)
Solution:
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
(d)
Solution:
Using the Pythagorean identity
Example 3.1.3.2 Find the third derivative of the following:
at
Solution:
at
(b)
Solution:
Use the Power Rule and Chain Rule
Use the Product Rule, Power Rule and Chain Rule
Using the Pythagorean identity
Combining like terms, we have
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MATH 401 – DIFFERENTIAL CALCULUS
As an alternative, the final answer can be expressed in terms of
tangent function only.
(c)
Solution:
Use the Product Rule, Power Rule, and Chain Rule
Use the Product Rule, Power Rule, and Chain Rule in each term
(d)
Solution:
Using the Pythagorean identity
Its third derivative is
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MATH 401 – DIFFERENTIAL CALCULUS
Expressing this in terms of one function only, use the Pythagorean identity
As an alternative, the final answer can be expressed in terms of cotangent
functions only.
Example 3.1.3.3 Find the 4th derivative of the following:
(a)
Solution:
(b)
Solution:
Remarks: The higher-order derivatives of sin x and cos x follow a repeating
pattern. Once we recognize the pattern of derivatives, we can find any
higher-order derivative by determining the step in the pattern to which it
corresponds. For example, every fourth derivative of sin x equals sin x , so
while
The pattern of derivatives is that the first derivative of sin x equals
its
derivative. The 2nd derivative of sin x equals its 6th derivative. The
nth derivative of the function
. For n greater
than 4, we can perform division algorithm n/4 = q + r/4, where q is the
quotient and r is the remainder.
It is now evident that for the nth
derivative of sin x,
, it follows that
, where
.
5th
Example 3.1.3.4 Evaluate the indicated nth derivative of sin x.
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MATH 401 – DIFFERENTIAL CALCULUS
(a) Find the 50th derivative of sin x.
Solution:
Using the algorithm, 50 = 4(12) + 2
Since r = 2, then
(b)Find the 111th derivative of sin x.
Solution:
Using the algorithm, 111 = 4(27) + 3
Since r = 3, then
(c)Find the 1200th derivative of sin x.
Solution:
Using the algorithm, 1200 = 4(300) + 0
Since r = 0, then
The same pattern for the nth derivative of the function
which is
.
Example 3.1.3.5 Evaluate the indicated nth derivative of cos x.
(a)Find the 25th derivative of cos x.
Solution:
Using the algorithm, 25 = 4(6) + 1
Since r = 1, then
(b)Find the 335th derivative of cos x.
Solution:
Using the algorithm, 335 = 4(83) + 3
Since r = 3, then
Example 3.1.3.6 Find the 4th derivative of the following:
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MATH 401 – DIFFERENTIAL CALCULUS
(a)
(b)
Solution:
Solution:
x
Remarks: A similar pattern of higher-order derivatives can be observed for
the functions
and
, where is a constant. The only
difference is the numerical coefficient which is equal to .
For instance, the 4th derivative of
is
.
In (b), the 3rd derivative of
is
.
Example 3.1.3.7 Evaluate the indicated nth derivative:
(a) Find the 99th derivative of
Solution:
Using the algorithm, 99 = 4(24) + 3
Since r = 3 and
then
(b) Find the 77th derivative of
Solution:
Using the algorithm, 77 = 4(19) + 1
Since r = 1 and
then
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3.1.4. Implicit Differentiation of Trigonometric Functions
In this section, we will apply the process of implicit differentiation
for equations involving trigonometric functions. Implicit differentiation is
a technique based on a Chain Rule that is used to find the derivative when
the relationship between the variables is given implicitly rather than
explicitly (solved for one variable in terms of the other).
Example 3.1.4.1 Find the
(a)
given that
.
Solution: Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
(b)
Solution: Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
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Example 3.1.4.2 Find the
(a)
given that
Solution : Using the result of of the Example 3.1.4.1 (b), we can evaluate
the 2nd derivative
as follows
By implicit differentiation with respect to x, we have
Substitute the expression of
(b)
Solution: Start with the implicit differentiation with respect to x
To evaluate the , perform another implicit differentiation with respect to
x. Use Product Rule on the right side.
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MATH 401 – DIFFERENTIAL CALCULUS
Substitute the expression for y’
Exercise 3.1. Find the derivative of the following functions:
1.
2.
3.
4.
ans.
ans.
ans.
ans.
at
at
5.
6.
ans.
ans.
Find the indicated nth derivative of the following:
7. 250th derivative of
ans.
8. 85th derivative of
ans.
Use implicit differentiation to find the
9.
10.
of
ans.
of sin
ans.
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3.2.
Inverse Trigonometric Functions
In this section, we are to explore the process of finding the derivatives of
the six inverse trigonometric functions.
3.2.1. Derivatives of Inverse Trigonometric Functions
The following theorem lists the derivatives of the six inverse
trigonometric functions.
Theorem3.2 Derivatives of Inverse Trigonometric Functions
Let u be a differentiable function of x.
It can be noted that the derivative of Arccos u, Arccot u, and Arccsc u
are the negatives of the derivatives of Arcsin u, Arctan u, and Arcsec u,
respectively.
Proof for Arcsin u:
Let
. So,
implicit differentiation with respect to x as follows.
, and you can use
Then use the Pythagorean identity
So
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MATH 401 – DIFFERENTIAL CALCULUS
If u is a differentiable function of x, then you can use the Chain Rule to
write
Proof for Arctan u:
Let
. So,
implicit differentiation with respect to x as follows.
, and you can use
Then use the Pythagorean identity
If u is a differentiable function of x, then you can use the Chain Rule to
write
Proof for Arcsec u:
Let
. So,
implicit differentiation with respect to x as follows.
, and you can use
Then use the Pythagorean identity
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MATH 401 – DIFFERENTIAL CALCULUS
If u is a differentiable function of x, then you can use the Chain Rule to write
Let us analyze the following examples:
Example 3.2.1.1 Find the derivative of the following functions:
(a)
(d)
Solution:
Solution:
Let
, then
Let
, then
(b)
(e)
Solution:
Let
Solution:
Let
, then
(c)
Solution:
Let
, then
(f)
, then
Solution:
since the derivative of Arccsc u if
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MATH 401 – DIFFERENTIAL CALCULUS
the negative of that of Arcsecu, then
Example 3.2.1.2 Find the derivative of
Solution: Apply the Product Rule on the second term.
3.2.2 Chain Rule of the Derivatives of Inverse Trigonometric
Functions
In this section, we will determine the derivatives of inverse
trigonometric functions using the Chain Rule. This rule deals with
composite functions involving inverse trigonometric functions.
Example 3.2.2.1 Find the derivative of the following function:
(a)
Solution:
(b)
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.2.2.2 Use Chain Rule to find the derivative of the following:
(a)
Solution:
(b)
Solution:
3.2.3 Higher-Order Derivatives of Inverse Trigonometric Functions
In this section, we are to evaluate the higher-order derivatives of
inverse trigonometric functions.
Example 3.2.3.1 Find the second derivative of
Solution:
We have determined in Example 3.2.2.1 (a) its first derivative as
Now to find its 2nd derivative, apply the Quotient Rule
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.2.3.2 Find the third derivative of
Solution:
Its first derivative with respect to x is
Then its second derivative with respect to x is
So the third derivative is
3.2.4 Implicit Differentiation of Inverse Trigonometric Functions
In this section, we will apply the process of implicit differentiation
for equations involving inverse trigonometric functions.
Example 3.2.4.1 Use implicit differentiation to find the derivative of
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.2.4.2 Use implicit differentiation to find the derivative of
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise 3.2.
1. Verify each differentiation formula
(a)
(b)
(c)
2. True or False.
The
for all x in the domain. ans. True
Find the derivative of the following functions:
3.
ans.
4.
ans.
5.
ans.
6.
ans.
Find the indicated nth derivative of the following:
7. 2nd derivative of
ans.
8. 5th derivative of
ans.
Use implicit differentiation to find the derivative of:
9.
ans.
10. y
ans.
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3.3.
Logarithmic Functions
The definition of the logarithmic function that you encountered in
Chapter 1 was based on exponents, and the properties of logarithms were
then proved from the corresponding properties of exponents. In this
section, we will consider the derivatives of the natural logarithmic
function, denoted by ln, as well as the derivative of logarithmic function to
base .
3.3.1. Derivatives of Logarithmic Functions
The derivative of the logarithmic function is determined by the
following theorem.
Theorem 3.3 Chain Rule of the Derivative of Logarithmic Function
Let be a positive real number
function of x and
, then
and let u be a differentiable
The second formula is a special case of the first formula since the
natural logarithm
.
In the first formula we can set
thus
.
Example 3.3.1.1 Find the derivative of the following functions:
(a) y
Solution:
y
y
where
is constant
Solution:
y
y
y
(b)
Solution:
y
y
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MATH 401 – DIFFERENTIAL CALCULUS
(c)
Solution:
y
Another Solution:
Rewrite the given using a logarithmic property
y
,
y
y
(d)
y
Solution:
where
y
Solution:
y
y
y
y
y
(e) y
Solution:
Rewrite the given as:
y
Take note that
Use the Power Rule and Chain Rule to evaluate its derivative
y
Example 3.3.1.2 Find the derivative of the following functions:
(a) y
0
Solution:
y
ln 0
(b) y
Solution:
y
where
are constants
Solution:
y
y
ln
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MATH 401 – DIFFERENTIAL CALCULUS
y
ln
y
3.3.2. Higher-Order Derivatives of Logarithmic Functions
In this section, we will evaluate the nth derivatives of logarithmic
functions.
Example 3.3.2.1 Find the second derivative of the following functions:
(a) y
Solution:
y
y
1 x
x
(b)
Solution:
y
(c)
Solution:
y
Example 3.3.2.2 Find the third derivative of the following functions:
(a)
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
(b)
Solution: We can rewrite the given as
.
(c)
Solution:
3.3.3. Implicit Differentiation of Logarithmic Functions
In this section, we will apply the process of implicit differentiation
for equations involving logarithmic functions.
Example 3.3.3.1 Use implicit differentiation to find the derivative of
(a)
.
Solution: We can rewrite the second term applying the logarithmic
property
and
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MATH 401 – DIFFERENTIAL CALCULUS
Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
(b)
Solution: Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
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3.3.4. Logarithmic Differentiation
In this section we will evaluate the derivatives of some complicated
functions by using logarithms. That process is called logarithmic
differentiation which was developed in 1697 by Johann Bernoulli(16671748).
Let’s see how this works in the following examples:
Example 3.3.4.1 Find the derivative of the function
Solution:
The derivative of this function can be evaluated by using the Product Rule
and Quotient Rule but it is somewhat a complicated process. To simplify
the process, we can take the logarithms of both sides.
We need to use the properties of logarithms to expand the right side as
follows.
Perform implicit differentiation, we have
Multiply both sides by y and substitute the expression for y, we have
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MATH 401 – DIFFERENTIAL CALCULUS
This can still be simplified using algebraic rules.
Example 3.3.4.2 Find the derivative of the function
Solution:
The derivative of this function can be evaluated by using the Product Rule
Power Rule and Chain Rule but it is somewhat a complicated process. To
simplify the process, we can take the logarithms of both sides.
Use the properties of logarithms to expand the right side as follows.
Perform implicit differentiation, we have
Multiply both sides by f(z) and substitute the expression for f(z), we have
Simplifying this algebraically
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise 3.3.
Find the derivative of the following function:
1.
ans.
2.
ans.
3.
ans.
4.
ans.
5.
ans.
6.
ans.
Find the indicated nth derivative of the following:
7. 2nd derivative of
ans.
8. 5th derivative of
ans.
Use implicit and/or logarithmic differentiation to find the derivative y’ of:
9.
10.
ans.
ans.
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3.4.
Exponential Functions
Because the natural logarithmic function is increasing on its entire domain,
then by the inverse function theorem, it has an inverse that is also an increasing
function. The inverse of ln is called the natural exponential function, denoted
by exp. It is defined by
if and only if
. In general, the exponential
function to the base is defined by
, where is any positive number and
x is any real number.
In this section, we will explore the derivatives of these exponential
functions.
3.4.1. Derivatives of Exponential Functions
Theorem 3.4.1 Derivatives of Exponential Functions
Let
Proof :
Let
be a positive real number
then applying logarithms on both sides, we have
Apply implicit differentiation with respect to x on both sides
Multiply both sides by y
Replacing y by
, we obtain
The proof of the first formula is given as follows:
From the relationship between exponential and logarithmic
function that
, we set
and using the property of
logarithm
, then
Differentiating both sides with respect to x, we obtain
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Example 3.4.1.1
Exponential Functions
Derivatives
Before moving on to the next section we need to be aware of the
distinction between these two derivatives:
It is important to note that with the Power Rule the exponent n MUST be
constant and the base x MUST be a variable while for the derivative of an
exponential function, the exponent x MUST be a variable and the base
must be a constant.
In cases where both the exponent and the base involve variables will be
considered in a later section.
3.4.2. Chain Rule of the Derivatives of Exponential Functions
Theorem 3.4.2 Chain Rule of Derivatives of Exponential Functions
Let be a positive real number
differentiable function of x
and let u be a
Observe that the derivative of the function defined by
,
where k is a constant, is itself. The only other function we have previously
encountered that has this property is the constant function zero; actually,
this is the special case of
when
.
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Example 3.4.2.1 Apply the Chain Rule to find the derivative of the following
functions:
(b)
Solution:
Solution:
(d)
Solution:
Solution:
(e)
(f)
Solution:
Use the property of exponent
Solution:
Differentiate both sides
Example 3.4.2.2 Find the derivative of the following functions:
y
where
Solution:
y
(b) y
Solution:
ln
y
are constants
y
ln
y
y
0
k
Or
Solution:
y
ln 10
0
y
y
ln 10 0
ln 10 0
y
y
k
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3.4.3. Higher-Order Derivatives of Exponential Functions
In this section, we will evaluate the nth derivatives of exponential
functions.
Example 3.4.3.1 Find the second derivative of the following functions:
Solution:
(b)
Solution:
Example 3.4.3.2 Find the nth derivative of
Solution:
So the nth derivative of
, m is constant.
can be written as
3.4.4. Implicit Differentiation of Exponential Functions
In this section, we will apply the process of implicit differentiation
for equations involving exponential functions.
Example .4.4.1 Use implicit differentiation to find the derivative y’ of
(a)
.
Solution: We can rewrite the 3 rd term applying the rules of exponents
Since it is difficult to express y in terms of x only, differentiate implicitly
with respect to x.
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MATH 401 – DIFFERENTIAL CALCULUS
Combining like terms
(b)
Solution: Since it is difficult to express y in terms of x only, differentiate
implicitly with respect to x.
Combining like terms
3.4.5. Logarithmic Differentiation
Logarithmic differentiation can also be used to evaluate the
derivative of function of this form:
These are the cases where both the exponent and the base involve
variables.
Example 3.4.5.1 Differentiate the function
We have encountered the derivative of two similar functions like this,
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MATH 401 – DIFFERENTIAL CALCULUS
But neither of the two will work here because both the base and the
exponent are variables. Logarithmic differentiation can be used in this
case.
To simplify the process, we can take the logarithms of both sides.
Use the properties of logarithms to expand the right side as follows.
Perform implicit differentiation, we have
Multiply both sides by y and substitute
, we have
Now let us consider a more complicated example of this.
Example 3.4.5.2 Differentiate the function
Solution: Again the Power Rule and the derivative of exponential functions
will not work here because both the base and the exponent are variables.
Logarithmic differentiation can be used in this case.
To simplify the process, we can take the logarithms of both sides.
Use the properties of logarithms to expand the right side as follows.
Perform implicit differentiation, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Multiply both sides by y and substitute
, we have
Example 3.4.5.3 Differentiate the function
Solution: Again the Power Rule and the derivative of exponential functions
will not work here because both the base and the exponent are variables.
Logarithmic differentiation can be used in this case.
To simplify the process, we can take the logarithms of both sides.
Use the properties of logarithms to expand the right side as follows.
Perform implicit differentiation, we have
Multiply both sides by y and substitute
, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise 3.4.
Find the derivative of the following exponential functions:
1.
ans.
2.
ans.
3.
ans.
Find the indicated nth derivative of the following functions:
4.
2nd derivative of
ans.
5.
100th derivative of
ans.
Use implicit differentiation to find the derivative y’ of the following:
6.
7.
0
0
ans.
0
ans.
Use logarithmic differentiation to find the derivative of the following:
8.
9.
10.
ans.
ans.
ans.
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3.5.
Hyperbolic Functions
Because hyperbolic functions can be written in terms of the exponential
functions, you can easily derive rules for their derivatives.
In this section, we will explore the derivatives of the hyperbolic functions.
3.5.1. Derivatives of Hyperbolic Functions
Theorem3.5. Derivatives of Hyperbolic Functions
d
sinh x coshx
dx
d
cosh x sinh x
dx
d
tanhx sec h 2 x
dx
d
cot hx csc h 2 x
dx
d
sec hx sec hx tanhx
dx
d
csc hx csc hx cothx
dx
Proof:
Recall the definition of hyperbolic sine function
Differentiate both sides with respect to x
Next, for the hyperbolic tangent function, it is defined as
Differentiate both sides with respect to x
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MATH 401 – DIFFERENTIAL CALCULUS
Use Quotient Rule on the right side
Use the hyperbolic identity
Let’s have the following examples:
Example 3.5.1.1 Differentiate
Solution: We will need to use the Product Rule of differentiation on the
first term. Also the constant 2 will be considered part of the first function
in the product of the first term. Here is the derivative of the function y
Example 3.5.1.2 Find the derivative of
Solution: We will need to use the Product Rule of differentiation on the
second term. Be careful with the minus sign in front of the second term
and make sure it gets dealt with properly. There are two ways to deal with
this. One way is to make sure that you use a set of parentheses as follows:
Example 3.5.1.3 Find the derivative of
Solution: Just differentiate each term using the formula above.
.
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.5.1.4 Differentiate
Solution: We will use the quotient rule of differentiation to evaluate the
derivative of this function.
We can still simplify this by factoring out 3 in the last two terms in the
numerator and use the hyperbolic identity
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MATH 401 – DIFFERENTIAL CALCULUS
3.5.2. Chain Rule of the Derivatives of Hyperbolic Functions
In this section, we will determine the derivatives of hyperbolic
functions using the Chain Rule. This rule is applied on differentiating
composite functions involving hyperbolic functions.
The Chain Rule versions of the derivatives of the six hyperbolic functions
are as follows: Let u be a differentiable function of x
Example 3.5.2.1 Apply the Chain Rule to find the derivative of the following
functions:
(b)
Solution:
Solution:
(d)
Solution:
Solution:
(e)
(f)
Solution:
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.5.2.2 To understand the mathematical conventions regarding
parentheses, apply the Chain Rule to find the derivative of the following
functions:
Solution:
Function
Derivative
(a)
Example 3.5.2.3 Find the derivative of the following:
Solution: We must use the Product Rule and Chain Rule.
Solution: We must use the Quotient Rule along with the Chain Rule.
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MATH 401 – DIFFERENTIAL CALCULUS
Solution: Recognize here that we have
function
that is inside the
function; that is
. We
will start using the General Power Rule, then the Chain Rule by
approaching this step-by-step.
To express this in terms of hyperbolic function only, we can use the
hyperbolic identity
Solution:
Solution:
Solution:
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MATH 401 – DIFFERENTIAL CALCULUS
3.5.3. Higher-Order Derivatives of Hyperbolic Functions
Let’s have the following examples:
Example 3.5.3.1 Find the second derivative of the following:
(b)
Solution:
Solution:
( )
(d)
Solution:
Solution:
(e)
Solution:
Expressing this in terms of hyperbolic secant only, we can use the
hyperbolic identity
(d)
Solution: This a composite function, where
. Use Chain Rule
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MATH 401 – DIFFERENTIAL CALCULUS
Then differentiate to get the second derivative
Example 3.5.3.2 Find the third derivative of the following:
at
Solution:
at
(b)
Solution:
Use the Power Rule and Chain Rule
Use the Product Rule, Power Rule and Chain Rule
13
13 13
Using the hyperbolic identity
Combining like terms, we have
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MATH 401 – DIFFERENTIAL CALCULUS
(c)
Solution:
Use the Product Rule, Power Rule and Chain Rule
Use the Product Rule, Power Rule and Chain Rule in each term
(d)
Solution:
Use the Power Rule and Chain Rule
]
Using the hyperbolic identity
Differentiate each term to get the third derivative
Expressing this in terms of one function only, use the hyperbolic identity
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MATH 401 – DIFFERENTIAL CALCULUS
Combining like terms, we have
As an alternative, the final answer can be expressed in terms of hyperbolic
cotangent functions only.
Example 3.5.3.3 Find the 4th derivative of the following:
Solution:
(b)
Solution:
Remarks: The higher-order derivatives of sinh x and cosh x follow a
repeating pattern. Once we recognize the pattern of derivatives, we can
find any higher-order derivative by determining the step in the pattern to
which it corresponds. For example, every second derivative of sinh x
equals sinh x , so
while
The pattern of nth derivative for hyperbolic sine function sinh x is
For hyperbolic cosine function cosh x, the pattern for nth derivative is
Let us have the following examples:
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MATH 401 – DIFFERENTIAL CALCULUS
Example 3.5.3.4 Evaluate the indicated nth derivative of sinh x.
(a) Find the 50th derivative of sinh x.
Solution:
Using the pattern, since n = 50 and it is even, then
(b)Find the 111th derivative of sinh x.
Solution:
Using the pattern, since n = 111 and it is odd, then
Example 3.5.3.5 Evaluate the indicated nth derivative of cosh x.
(a)Find the 25th derivative of cosh x.
Solution:
Using the pattern, since n = 25 and it is odd, then
(b)Find the 3,350th derivative of cosh x.
Solution:
Using the pattern, since n =3.350 and it is even, then
3.5.4. Implicit Differentiation of Hyperbolic Functions
In this section, we will apply the process of implicit differentiation
for equations involving hyperbolic functions. Implicit differentiation is a
technique based on a Chain Rule that is used to find the derivative when
the relationship between the variables is given implicitly rather than
explicitly (solved for one variable in terms of the other).
Example 3.5.4.1 Find the
(a)
given that
.
Solution: Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
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MATH 401 – DIFFERENTIAL CALCULUS
Combining like terms,
(b)
Solution: Since it is difficult to express y in terms of x only, we will
differentiate implicitly with respect to x.
Example 3.5.4.2 Find the
given that
(a)
Solution : Using the result of of the Example 3.5.4.1 (b), we can evaluate
the 2nd derivative
as follows
By implicit differentiation with respect to x, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Substitute the expression of
(b)
Solution: Start with the implicit differentiation with respect to x
To evaluate the , perform another implicit differentiation with respect to
x. Use Product Rule and Chain Rule on the right side.
Substitute the expression for y’
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise 3.5.
Verify the following differentiation formula:
1.
2.
3.
4.
Find the derivative of the following functions:
5.
ans.
6.
ans.
7.
ans.
Find the indicated nth derivative of the following:
8. 25th derivative of
ans.
9. 44th derivative of
ans.
Use implicit differentiation to find the derivative of
10.
ans.
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Problem Set No. 3
Derivatives of Transcendental Functions
Find the derivative of the following functions:
1.
2.
3.
4.
5.
6.
Find the indicated nth derivative of the following:
7. 222nd derivative of
8. 115th derivative of
Use implicit and/or logarithmic differentiation to find the derivative y’ of:
9.
10.
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER IV
APPLICATIONS OF DERIVATIVES OF ALGEBRAIC AND
TRANSCENDENTAL FUNCTIONS
We have learned from Chapter II and Chapter III how to find the
derivatives of algebraic and transcendental functions by applying both the
definitions and theorems of differentiation. This chapter discusses several
applications of derivatives. We shall now learn the concepts and the process of
solving problems involving techniques on approximation, tangent line and
normal line to a given curve, curve sketching, optimization problems and related
rates. Definitions and theorems presented in this chapter are taken from [1]
Larson, R. (2010), [2] Leithold, L. (2002) and [3] Stewart, J. (2016)
At the end of this chapter, the student might be able to:
1. Apply the concept of differential in error propagation and in
approximation.
2. Apply differentiation to determine the equations of tangent and normal
lines to a graph of function at a given point
3. Apply the concepts of Relative Extrema, First Derivative Test, Second
Derivative Test, Concavity and Points of Inflection in drawing a sketch
of the graph of algebraic and transcendental functions
4. Apply the concepts of derivatives in solving optimization problems and
in related rates problems
4.1.
The Differential
Consider a function defined by y=f(x) where x is the independent variable.
In the three-step rule we introduced the symbol Δx to denote the increment of x.
Now we introduce the symbol dx which we call the differential of x. Similarly, we
shall call the symbol dy as the differential of y. To give separate meanings to dx
and dy, we shall adopt the following definitions of a function defined by the
equation y=f(x)
Definition4.1.1.a Differential of the Dependent Variable
If the function f is defined by the equation y f x , then the differential of y,
denoted by dy, is given by
dy f ' ( x)x
where x is in the domain of f ' and x is an arbitrary increment of x.
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MATH 401 – DIFFERENTIAL CALCULUS
In words, the differential of a function is equal to its derivative multiplied by the
differential of its independent variable.
Definition4.1.1.b Differential of the Independent Variable
If the function f is defined by the equation y f x , then the differential of x,
denoted by dx, is given by
dx x
where x is any number in the domain of f ' and x is an arbitrary increment of x.
In words, the differential of the independent variable is equal to the increment of
the variable
From Definitions 4.1 and 4.2, dy f ' ( x)dx
We emphasize that the differential dx is also an independent variable; it
may be assigned any value whatsoever. Therefore, from DEFINITION 4.1, we see
that the differential dy is a function of two independent variables x and dx. It
should also be noted that while dx = Δx, dy ≠ Δy in general.
Suppose dx≠0 and we divide both sides of the equation dy f ' ( x)dx by dx
then we get
dy
f ' x . Note that this time dy/dx denotes the quotient of two
dx
differentials, dy and dx . Thus the definition of the differential makes it possible to
define the derivative of the function as the ratio of two differentials. That is,
f ' x
dy the differential of y
dx the differential of x
Example1. Given f ( x) x x 1, find dy
2
Solution:
f ' x 2 x 1 0
f ' x 2 x 1
dy f ' x dx
dy 2 x 1dx
Example2. Find the differential of y x 5x 1
3
Solution:
y f x
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MATH 401 – DIFFERENTIAL CALCULUS
f x x 3 5 x 1
f ' x 3x 2 51 0
f ' x 3x 2 5
dy f ' x dx
dy 3x 2 5 dx
Example3. Find the differential of f ( x) x
2
x4 1
Solution:
f ( x) x 2 x 4 1
4x3 0
f ' ( x) x
x 4 12 x
4
2 x 1
2
4x5
2x5
4
f ' ( x)
2x x 1
2x x 4 1
4
4
2 x 1
x 1
f ' ( x)
f ' ( x)
2x
2x5 2x x 4 1
x4 1
5
2x5 2x
x4 1
4x5 2x
x4 1
dy f ' x dx
4x5 2x
dx
dy
4
x 1
Example4. Find
dy
by means of differentials if xy + sin x = ln y
dx
Solution:
x dy y dx cos x dx
1
dy
y
1
x dy y dx cos x dx dy y
y
xy dy y 2 dx y cos x dx dy
xy dy y 2 dx y cos x dx dy
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MATH 401 – DIFFERENTIAL CALCULUS
xy dy y dx y cos x dx dy dx1
2
dy
dy
y 2 y cos x
dx
dx
dy dy
xy
y 2 y cos x
dx dx
xy 1 dy y y cos x
dx
dy y y cos x
dx
xy 1
xy
4.1.1. Application of the Differential
Derivatives can also be used in finding the change in a certain
quantity. For instance in the function y f x , the change in x and y are
denoted by x and y while dx and dy are the differentials respectively.
Example1. Use differentials to approximate the change in the area of a
square if the length of its side increases from 6 cm to 6.23 cm
Solution:
Let x = length of the side of the square.
The area may be expressed as a function of x, where A = x2.
The differential dA is dA f ' x dx dA 2 x dx
Because x is increasing from 6 to 6.23, you find that Δ x = dx =0.23cm
hence, dA 26cm0.23cm
dA 2.76cm 2
The area of the square will increase by approximately 2.76cm 2 as its
side length increases from 6 to 6.23. Note that the exact increase in area
y is 2.8129cm2.
Example2. Use the local linear approximation to estimate the value of
3
26.55 to the nearest thousandth.
Solution:
Because the function we are applying is f x 3 x , choose a convenient
value of x that is a perfect cube and is relatively close to 26.55, namely
x 27 . The differential dy is
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MATH 401 – DIFFERENTIAL CALCULUS
dy f ' x dx
f x x
1
3
1
2
1 1
1
1
f ' x x 3 f ' x x 3 f ' x 2 3
3
3
3x
dy
1
dx
3x 2 3
Because x is decreasing from 27 to 26.55, then x dx 0.45
1
0.45 1 45 1 45 45 1 0.0167
23
39 100 27 100 2700 60
327
Hence, dy
which implies that
therefore,
3
26.55 will be approximately
1
less that
60
3
27 3
1
60
3
26.55 3
3
26.55 3 0.0167
3
26.55 2.9833
Example3. Use an appropriate local linear approximation to estimate the
value of cos 310
Solution:
Let y cos x then dy sin x dx
When x 30 , y cos x cos 30 0.8660
180
And when x 30 , x dx 1 1
0.01745 , then
dy sin x dx sin 30 0.01745 0.50.01745 0.008725
Therefore the required approximation is
y dy 0.8660 0.008725
y dy 0.8573
131
MATH 401 – DIFFERENTIAL CALCULUS
4.1.2. Approximate Formulas
The discussion on the geometric interpretation of the derivative
f or y is given as follows;
f f x x f x
f x x f x f
In the given figure the comparison of f and df will be shown
From the figure, it can be seen that dx x but dy y .
However if x is very small or when x 0 then dy will be
approximately equal to y . That is in symbols dy y or equivalently
df f .
Imposing
the
relation
df f
to
equation
f x x f x f we will have, f x x f x df . Since
df f ' ( x)dx the relation will finally given as
f x x f x f ' ( x)dx
Which is called the approximation formula
Example1. Given f x 3x x 1 , compute the value of f and df if
2
x 1 and x 0.01
Solution:
f f x x f x since f x 3x 2 x 1then
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MATH 401 – DIFFERENTIAL CALCULUS
f [3( x x) 2 ( x x) 1] [3x 2 x 1]
f {3[ x 2 2 xx (x) 2 ] ( x x) 1} [3x 2 x 1]
f 3x 2 6 xx 3(x) 2 x x 1 3x 2 x 1
f 6 xx 3(x) 2 x
since x 1 and x 0.01 then
f 6(1)(0.01) 3(0.01) 2 0.01
f 0.06 0.0003 0.01
f 0.0503
df f ' ( x)dx since f x 3x 2 x 1 , then
f ' ( x) 3(2 x) 1 0
f ' ( x) 6 x 1
df f ' ( x)dx
df (6 x 1)dx since dx 0.01 , then
df [6(1) 1][0.01]
df [5][0.01]
df 0.05
Example2. Given that f x
x
compute df and f when x 2
1 x
and x 0.01
Solution:
df f ' ( x)dx since f x
x
, then
1 x
(1 x)(1) ( x)(0 1) 1 x x
f ' ( x)
(1 x) 2
(1 x) 2
1
(1 x) 2
df f ' ( x)dx
f ' ( x)
df
1
dx since x dx 0.01 , then
(1 x) 2
133
MATH 401 – DIFFERENTIAL CALCULUS
1
[0.01]
(1 2) 2
0.01
df
9
df 0.001111
df
f f x x f x since f x
x
then
1 x
f
( x x)
x
(1 x x) (1 x)
f
( x x)(1 x) ( x)(1 x x)
(1 x x)(1 x)
f
x x 2 x xx x x 2 xx
(1 x x)(1 x)
f
x
(1 x x)(1 x)
since x 2 and x 0.01 then
f
0.01
0.01
(1 2 0.01)(1 2) (3.01)(3)
0.01
9.03
f 0.001107
f
Example3. Find the approximate value of
4
82 using the differential.
Solution:
4
Let y f ( x) x
Think of the nearest number to the given 82 which is a perfect power of
the index number 4, 81 is the nearest number to 82 which is a perfect
power of 4. Set x 81and used the approximation formula
f x x f x f ' ( x)dx , then
x x 82
x 82 x , since x 81 then
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MATH 401 – DIFFERENTIAL CALCULUS
x 82 81
x 1
f ( x) 4 x ( x)1 4
1
3
1 1 1
f ' ( x) x 4 x 4
4
4
1
f ' ( x) 4 3
4( x )
f x x f x f ' ( x)dx
f x x 4 x
1
dx
4(4 x )3
Since x 81 and x dx 1 then
f 81 1 4 81
f 82 3
1
(1)
3
4
4( 81)
1
4(3) 3
f 82 3
1
108
f 82 3 0.009259
f 82 3.009259
4.1.3. Error Propagation
We can also use differentials in Physics to estimate errors, say in
physical measuring devices. In these problems, we’ll typically take a
derivative, and use the “dx” or “dy” part of the derivative as the error.
Then, to get percent error, we’ll divide the error by the total amount and
multiply by 100.
The other thing to remember is that when we are solving for an
error, it can go either way, so we typically express our answers with a “±”
Example 1. The volume of a cube is 125 in3. If the volume measurement is
known to be correct to within 2.5 in3, estimate the error in the
measurement of a side of the cube.
135
MATH 401 – DIFFERENTIAL CALCULUS
Solution:
We first write down what the necessary information given in the problem.
We have V=125, and dV=2.5 ( “dx” is the error part of the equation).
We want the error in the side of the cube, so we want ds.
V s3
dV 3s 2 ds
Substitute and solve for ds. Note that since we know that V s ,
3
V 125 , we know a side s 5 then
dV 3s 2 ds
2.5 3(5) 2 ds
2.5
75
ds 0.0333in
ds
The error in the measurement of a side of the cube is ±0.0333in.
Example2. The radius of a sphere is measured to be 5 mm. If this
measurement is correct to within 0.05 mm,
a) Estimate the propagated error in the surface area of the sphere.
b) Estimate the propagated error in the volume of the sphere.
c) Estimate the percent error of the volume of the sphere.
Solution:
Given r = 5, and dr = 0.05 (“dx” is the error part of the equation). We need
the error in the surface area (dA) and the error in the volume (dV). We
have to remember the equations from Geometry.
a) Given the surface area of the sphere, differentiate it with respect to r:
A 4r 2
dA 8rdr
Since r = 5, and dr = 0.05 and π = 3.1416, then
dA 8rdr
dA 8(3.1416)(5)(0.05)
dA 6.2832mm2
The error in the measurement of the surface area of the sphere
is ±6.2832mm2.
b) Given the volume of the sphere, differentiate with respect to r:
136
MATH 401 – DIFFERENTIAL CALCULUS
4 3
r
3
4
dV (3r 2 )dr
3
dV 4r 2 dr
V
Since r = 5, and dr = 0.05 and π = 3.1416, then
dV 4(3.1416)(5) 2 (0.05)
dV 15.708mm3
The error in the measurement of the volume of the sphere is ±15.708mm3.
c) To get percent error:
Percent Error = ErrorVolume(100)
dV
dV
(100)
(
100
)
=
= 4 3
V
r
3
15.708
15.708
(
100
)
(100) (0.03)(100)
= 4
523
.
6
3
( )(5)
3
= 3%
The percent error in the measurement of the volume of the sphere is 3%.
Exercise 1. Find the derivative of the following functions by applying the
definition of derivatives.
1. Given the following, find a. df b. f
a. y x 3x 4
2
b. y 4 x 3x 1 , x 2 and x 0.2
2
c. f x
1
, x 2 and x 0.02
x2
2. Find the approximate value of
143 using differential
3. Find the approximate value of 3 128 using differential
4. The measurement of an edge of a cube is 15cm. If it is to be painted with
thickness of paint equal to 0.01cm. Find a) actual amount of paint; b)
approximate amount of paint.
137
MATH 401 – DIFFERENTIAL CALCULUS
5. The measurement of the base and altitude of a triangle
are 20 cm and 30 cm, respectively. The possible error in each
measurement is 0.3 cm.
a) Estimate the possible propagated error in computing the area of the
triangle.
b) Approximate the percent error in computing this area.
4.2.
Tangent Line and Normal Line to a curve (Algebraic and
Transcendental Functions)
The geometric interpretation of the derivative showed that the
derivative of a function y f ( x) at any point P( x, y) is equal to the slope of
the tangent line TL at the point P( x, y) . That is f '( x) at P( x, y) is equal to
the slope of TL at P( x, y) . For a particular point P0 ( x0 , y0 ) , the slope of the
tangent line TL will be f '( x0 ) . Knowing a point, P0 ( x0 , y0 ) and the slope,
f '( x0 ) of the tangent line we can use the point-slope form of the line to get
the equation of the tangent line as
y y0 f '( x0 )( x x0 )
Similarly, we can get the equation of the nomal line NL to the same
curve at the same point. Since the normal line is perpendicular to the
tangent line, the slope of the normal line is equal to the negative reciprocal
of the slope of the tangent line. So, if the slope of TL is f '( x0 ) then the
slope of NL will be
y y0
1
,
f '( x0 )
hence,
the
equation
NL
will
be
1
( x x0 ) .
f '( x0 )
Example1. Find the equations of TL and NL to the curve y 3x 2 2 x 1 at
(2,9)
Solution:
Given y f x 3x 2 2 x 1 , x0 2 and y 0 9
We first compute the slope of the tangent line
f x 3x 2 2 x 1
f ' x 6 x 2
f ' 2 6(2) 2 12 2
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MATH 401 – DIFFERENTIAL CALCULUS
f ' 2 10 Slope of Tangent Line
Equation of Tangent Line (TL)
Equation of Normal Line (NL)
1
x x0
f ' x0
y y0 f ' x0 x x0
y y0
y 9 10x 2
1
x 2
10
10 y 9 x 2
10 y 90 x 2
x 10 y 92 0
y 9 10 x 20
10 x 20 y 9 0
10 x y 11 0
y 9
Example2. Find the equations of T and N given the implicit equation of the
curve ( x 3 y)2 8x 12 0 at point (2,0) .
Solution:
Given ( x 3 y)2 8x 12 0 , x0 2 and y0 0
We first compute the slope of the tangent line using implicit differentiation
2( x 3 y)(1 3 y' ) 8 0 0
2( x 3xy '3 y 9 yy ' ) 8 0
2 x 6 xy '6 y 18 yy '8 0
(6 xy '18 yy ' ) (2 x 6 y 8)
y' (6 x 18 y) (2 x 6 y 8)
y'
(2 x 6 y 8)
then
(6 x 18 y )
y ' f ' x
(2 x 6 y 8)
(6 x 18 y)
since x 2 then
f ' 2
(2)(2) (6)(0) 8
(6)(2) (18)(0)
4 08
12 0
4
f ' 2
12
f ' 2
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MATH 401 – DIFFERENTIAL CALCULUS
f ' 2
1
Slope of Tangent Line
3
Equation of Tangent Line (TL)
Equation of Normal Line (NL)
1
x x0
f ' x0
y y0 f ' x0 x x0
y y0
1
y 0 x 2
3
1
x 2
1
3
y 3x 2
y 3x 6
y0
3 y ( x 2)
3 y x 2)
x 3y 2 0
3x y 6 0
2
Example3. Find the equation of the tangent line to the curve y 8x 4
which is perpendicular to the line x 2 y 5 .
Solution:
Let mTL as slope of Tangent Line
and mGL as slope of Given Line, then
mTL
1
mGL
from the equation of line x 2 y 5
compute the slope of the given line.
Using y mx b then
2 y x 5
y
mTL
1
5
1
x therefore mGL
2
2
2
1 1
2
mGL 1
2
then mTL 2 , Slope of Tangent Line
2
For y 8x 4 , compute y’ using implicit differentiation
,
2 yy , 8 y
8
4
4
,
y
2
m
y
'
since TL
, then
y
2y
y
140
MATH 401 – DIFFERENTIAL CALCULUS
y2
2
Substitute the value of y 2 to the equation y 8x 4 and solve for x
2
y 2 8x 4 (2) 8x 4 8x 4 4 8x 8
x 1
Therefore the value of ( x0 , y0 ) is 1,2
Equation of Tangent Line (TL)
y y0 f ' x0 x x0
y 2 2x 1
y 2 2x 2
2x 2 y 2 0
2x y 0
Example4. Find the equations of a Tangent and Normal line in a given
curve to the given value of x
1
a. y tan x , x = 1
Solution:
Solve for the point of tangency
y tan 1 x since x = 1
y tan 1 1
y
4
value of ( x0 , y0 ) is 1,
4
Find the Slope
f x y tan 1 x
f ' x
1
1 x2
1
1
f ' 1
2
1 1 1 1
141
MATH 401 – DIFFERENTIAL CALCULUS
f ' 1
1
slope of tangent line
2
Equation of Tangent Line (TL)
y y0 f ' x0 x x0
Equation of Normal Line (NL)
y y0
1
x 1
4 2
y
2 y ( x 1)
4
y
y
2y
2
x 1
x 2y 1
b.
2
y
0
4
4
4
1
x x0
f ' x0
1
x 1
1
2
2x 1
2 x 2
2x y 2
4
0
y x 2e x , x = 1
Solution:
Solve for the point of tangency
y x 2e x since x = 1
y (1)2 e1
y e1
1
value of ( x0 , y0 ) is 1, e
Find the Slope
f x y x 2e x
f ' x x 2 (e x ) (e x )(2 x)
f ' x x 2 e x 2 xe x
since x = 1, then
f ' 1 (1) 2 e 1 2(1)e 1
f ' 1 e 1 2e 1
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MATH 401 – DIFFERENTIAL CALCULUS
f ' 1 e 1 slope of tangent line
Equation of Tangent Line (TL)
y y0 f ' x0 x x0
Equation of Normal Line (NL)
y y0
1
x x0
f ' x0
y e 1 xe 1 e 1
1
x 1
e 1
e 1 ( y e 1 ) x 1
xe 1 e 1 y e 1 0
ye 1 e 2 x 1
xe 1 y 0
x ye 1 e 2 1 0
y e 1 e 1 ( x 1)
c.
y e 1
y ln( x 3)4 @ x = -2
Solution:
Solve for the point of tangency
y ln( x 3)4 since x = -2
y ln( 2 3)4
y ln(1)4
y ln 1
y0
value of ( x0 , y0 ) is 2,0
Find the Slope
f x y ln( x 3)4
f x 4 ln( x 3)
1
f ' x 4
(1)
x 3
4
f ' x
, since x = -2 , then
x3
4
4
f ' 2
23 1
143
MATH 401 – DIFFERENTIAL CALCULUS
f ' 2 4 slope of tangent line
Equation of Tangent Line (TL)
Equation of Normal Line (NL)
1
x x0
f ' x0
y y0 f ' x0 x x0
y y0
y 0 4x (2)
1
x 2
4
4 y x 2
4 y x 2
y0
y 4x 2
y 4x 8
4x y 8 0
x 4y 2 0
d. y sin x @ x
1
6
Solution:
Solve for the point of tangency
1
y sin x since x
6
1
y sin
6
y sin
sin(30 )
6
1
y
2
1 1
,
value of ( x0 , y0 ) is
2
6
Find the Slope
f x y sin x
f ' x cos x
f ' x cos x since x
1
, then
6
3
1
,
f ' x cos
cos
cos 30
6
6
2
f ' x
3
2
slope of tangent line
144
MATH 401 – DIFFERENTIAL CALCULUS
Equation of Tangent Line (TL)
y y0 f ' x0 x x0
1 3 1
y
x
2 6
2
Equation of Normal Line (NL)
1
x x0
y y0
f ' x0
1
1
y
2 3
2
y
1 3 1
x
2
2 6
3
y
1 3
3
x
2
2
12
3
3
2
x y
e. f x x
3
12
cosh x
1
0
2
1
x 6
1
1
y x
2
2
6
2
x
y
3
2
3
4
y
x
3
4
1
6
1
0
6
@ 1,1
Solution:
Find the Slope
Let y f x x
cosh x
y x cosh x
Apply ln to both sides
ln y ln x cosh x
ln y cosh x(ln x)
Differentiate implicitly with respect to x.
1 dy
1
cosh x ln xsinh x
y dx
x
dy
at 1,1
dx
1 dy
1
cosh 1 ln 1sinh 1
1 dx
1
dy
cosh 1 0sinh 1
dx
dy
cosh 1
dx
dy
f ' x
dx
Solve
145
MATH 401 – DIFFERENTIAL CALCULUS
f ' x cosh 1 slope of tangent line
Equation of Tangent Line (TL)
y y0 f ' x0 x x0
Equation of Normal Line (NL)
1
x x0
y y0
f ' x0
y 1 x cosh1 cosh1
1
( x 1)
cosh 1
(cosh1)( y 1) 1( x 1)
x cosh1 y cosh1 1 0
y cosh1 cosh1 x 1
y 1 cosh1( x 1)
y 1
x y cosh1 cosh1 1 0
Exercise 2. Find the equation of Tangent Line and Normal Line of the following
curves at the given point.
1. y x3 3x2 2 at x0 1 .
2. y 3x2 2 x 1 at (1, 2) .
3.
x2 xy 2 y 2 0 at x 2 .
4. Find the equation of the tangent line to the curve x2 4 y 2 8 and parallel
to the line x 2 y 8 .
5. Find the equation of the tangent line to the curve y x4 14x2 17 x 40
and perpendicular to the line x 7 y 4 .
6.
y arcsin 2 x , x = ¼
7.
y log x , x = e
8. y sec
9. y e
x
1
x , x = -2
ln 2 x , x = 0
10. y e 2 x 1 . x
1
2
146
MATH 401 – DIFFERENTIAL CALCULUS
4.3.
Relative Extrema
An important application of the derivative is to determine where a
function attains its maximum and minimum (extreme) values.
Definition4.3.1.a Relative Maximum Value
The Function f has a relative maximum value at the number c if there exists an
open interval containing c, on which f is defined, such that f(c) ≥ f(x) for all x in
this interval.
Figure below show a portion of the graph of a function having a
relative maximum value at c.
a
c
b
x
Figure 1
a
c
b
x
Figure 2
Definition4.3.1.b Relative Minimum Value
The Function f has a relative minimum value at the number c if there exists an
open interval containing c, on which f is defined, such that f(c) ≤ f(x) for all x in
this interval.
Figure below show a portion of the graph of a function having a
relative minimum value at c.
a
c
Figure 3
b
x
a
c
b
x
Figure 4
147
MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.1. If f(x) exists for all values of x in the open interval (a,b), and if f
has a relative extremum at c, where a < c < b, and if f ' c exists, then f ' c 0
2
Illustration1: Let f be a function defined by f x x 4 x 5
f ' x 2 x 4
f ' 2 2(2) 4 4 4
f ' 2 0
Because f ' 2 0 , f may have a relative
extremum at 2. Since f 2 1 and 1 f x
where either x 2 or x 2 , definition4.6.2,
guarantees that f has a relative minimum
value at 2. Figure 5 shows the graph of f, a
parabola whose vertex is at point (2,1) where
the graph has a horizontal tangent.
Note that f ' c 0 can equal zero even
Figure 5
if f does not have a relative extremum at c,
as shown in illustration 2.
3
Illustration2: Let f be a function defined by f x ( x 1) 2
f ' x 3( x 1) 2 (1) 0
f ' x 3( x 1) 2
f ' 1 3(1 1) 2 3(0) 2
f ' 1 0
Because f ' 1 0 , f may have a relative
extremum at 1. However, because f 1 2
and 2 f ( x) when x 1 and 2 f ( x) when
x 1, neither Definition 4.6.1 nor Definition
4.6.2 applies. So f does not have a relative
extremum at 1. The graph of this function is
shown in Figure 6 has a horizontal tangent at
the point (1,2), which is consistent with the
fact that the derivative is zero.
Figure 6
148
MATH 401 – DIFFERENTIAL CALCULUS
Definition4.3.1.c Critical Number
If c is a number in the domain of the function f , and if either f ' c 0 or f ' c
does not exists, then c is a critical number of f.
4
3
2
Example1. Find the critical numbers of f x x 4 x 2 x 12 x
Solution:
We compute f ' x , set it equal to zero and solve for x.
f x x 4 4 x 3 2 x 2 12 x
f ' x 4 x 3 12 x 2 4 x 12
f ' x 0
4 x 3 12 x 2 4 x 12 0
x 3 3x 2 x 3 0
( x 3 3x 2 ) ( x 3) 0
x 2 ( x 3) ( x 3) 0
( x 3)( x 2 1) 0
( x 3)( x 1)( x 1) 0
( x 3) 0
x 3
( x 1) 0
x 1
( x 1) 0
x 1
We have confirm that the critical numbers are -3, 1, and -1
2 x
Example2. Find the critical numbers of f x x e
Solution:
We compute f ' x , set it equal to zero and solve for x.
f x x 2 e x
f ' x x 2 (e x ) e x (2 x)
f ' x x 2 e x 2 xe x
f ' x 0 then x 2e x 2 xe x 0
e x ( x 2 2 x) 0
e x 0
( x 2 2 x) 0
x(x 2) 0
x0
(x 2) 0
x2
We have confirm that the critical numbers are 0 and 2
149
MATH 401 – DIFFERENTIAL CALCULUS
4.3.1. Increasing and Decreasing Functions and the First-Derivative
Test
The term increasing, decreasing, and constant are used to describe
the behavior of a function as we travel left to right along its graph. An
example is shown below.
Definition4.3.2.a Increasing Function
A function f defined on an interval is increasing on that interval if and only
if f x1 f x2 whenever x1 x2 , where x1 and x2 are any numbers in the
interval.
150
MATH 401 – DIFFERENTIAL CALCULUS
Definition4.3.2.b Decreasing Function
A function f defined on an interval is decreasing on that interval if and only if
f x1 f x2 whenever x1 x2 , where x1 and x2 are any numbers in the interval.
Definition4.3.2.c Constant Function
A function f defined on an interval is constant on that interval if and only if
f x1 f x2 for all points x1 and x2 .
151
MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.2.a Let the function f be continuous on the closed interval [a,b] and
differentiable on the open interval (a,b):
If f ' x 0 for every value of x in (a,b), then f is increasing on [a,b]
(i)
If f ' x 0 for every value of x in (a,b), then f is decreasing on [a,b]
(ii)
(iii)
If f ' x 0 for every value of x in (a,b), then f is constant on [a,b]
2
Illustration1. Find the intervals on which f x x 4 x 3 is increasing and
decreasing.
Solution:
Graph the equation.
Take the derivative of
f x x 2 4 x 3
f ' x 2 x 4
f ' x 0 when x 2 f is decreasing on - ,2
thus
f ' x 0 when x 2 f is increasing on 2,
3
Illustration2. Find the intervals on which f x x is increasing and decreasing.
Solution:
Graph the equation.
Take the derivative of
f x x 3
f ' x 3x 2
f ' x 0 when x 0 f is increasing on - ,0
thus
f ' x 0 when x 0 f is increasing on 0,
152
MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.2.b Suppose that f is a function defined on an open interval
containing the point x0 . If f has a relative extremum at x x0 , then x x0 is a
critical point of f ; that is, either f ' ( x0 ) 0 or f is not differentiable at x0 .
In general, we define a critical point for a function f to be a point in the
domain of f at which either the graph of f has a horizontal tangent line or f is
not differentiable (line is vertical). To distinguish between the two types of
critical points we call x a stationary point of f if f ' ( x) 0 .
3
Illustration1. Find all critical points of f ( x) x 3x 1
Solution:
We compute f ' x set it equal to zero
3
and solve for x. Since f ( x) x 3x 1
2
then f ' x 3x 3
2
let f ' x 0 then 3x 3 0
3( x 2 1) 0 3( x 1)( x 1) 0
Thus, 3 0 , ( x 1) 0 x 1
( x 1) 0
x 1
If x 1 y 3 and If x 1 y 1
Therefore the critical points are (1,3) and (1,1)
2
Illustration2: Let f be a function defined by f x x 4 x 5
Solution:
We compute f ' x set it equal to zero
2
and solve for x. Since f x x 4 x 5
then f ' x 2 x 4
let f ' x 0 then 2 x 4 0
2( x 2) 0
Thus, 2 0
( x 2) 0
x2
If x 2 y 1
Therefore the critical point is (2,1)
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MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.2.b asserts that the relative extrema must occur at
critical points, but it does not say that a relative extremum occurs at every
critical point. A function has a relative extremum at those critical points
where f ' changes sign.
Theorem4.3.2.c The First-Derivative Test
Suppose that f is a continuous at a critical point x0
(i) If f ' ( x) 0 on an open interval extending left from x0 and f ' ( x) 0 on an
open interval extending right from x0 , then f has a relative maximum at x0 .
(ii) If f ' ( x) 0 on an open interval extending left from x0 and f ' ( x) 0 on an
open interval extending right from x0 , then f has a relative minimum at x0 .
(iii) If f ' ( x) has the same sign on an open interval extending left from x0 as it
does on an open interval extending right from x0 , then f does not have a
relative extremum at x0 .
154
MATH 401 – DIFFERENTIAL CALCULUS
The above Theorem4.3.2.c simply say that for a continuous function,
relative maxima occur at critical points where the derivative changes from
(+) to (–) and relative minima where it changes from (–) to (+).
3
2
Illustration1. Given f ( x) x 6 x 9 x 3 , find the relative maximum
and relative minimum point and sketch the graph using the first derivative
test.
Solution:
Plot the graph of the function and
compute the derivative of f to
determine the critical values.
f ( x) x 3 6 x 2 9 x 3
f ' ( x) 3x 2 12 x 9
f ' x 3( x 2 4 x 3)
f ' x 3( x 1)( x 3)
The only critical numbers are those
For which f ' ( x) 0 and f ' ( x) does
not exists.
3( x 1)( x 3) 0
30
( x 1) 0 x 1
( x 3) 0 x 3
The critical numbers are 1 and 3. To determine whether f has a
relative extremum at these numbers, we apply the first-derivative test and
summarize the results in the table.
Interval
f ( x) x 3 6 x 2 9 x 3
x 1
x 1
1
1 x 3
x3
x3
-3
f ' x 3( x 1)( x 3)
Conclusion
+
f is increasing
0
f has a relative maximum
-
f is decreasing
0
f has a relative minimum
+
f is increasing
The sign of f ' x changes sign from + to – at x 1 relative maximum
155
MATH 401 – DIFFERENTIAL CALCULUS
The sign of f ' x changes sign from - to + at x 3 relative minimum
5
2
Illustration2. Given f ( x) 3x 3 15 x 3 , find the relative maximum and
relative minimum point and sketch the graph using the first derivative test.
Solution:
Plot the graph of the function and
compute the derivative of f to
determine the critical values.
5
3
f ( x) 3x 15 x
2
3
5 5 1
2 2 1
f ' ( x) 3 x 3 15 x 3
3
3
2
3
f ' ( x) 5 x 10 x
1
3
1
f ' ( x) 5 x 3 x 2
The only critical numbers are those
For which f ' ( x) 0 and
f ' ( x) does not exists.
1
5 x 3 x 2 0
1
3
5 x 0 x 0 and
( x 2) 0 x 2
The critical numbers are 0 and 2. To determine whether f has a
relative extremum at these numbers, we apply the first-derivative test and
summarize the results in the table.
Interval
5
3
f ( x) 3x 15 x
x0
x0
0
0 x2
x2
x2
93 4
2
3
f ' ( x) 5 x
1
3
x 2
Conclusion
+
f is increasing
Does not exists
f has a relative maximum
-
f is decreasing
0
f has a relative minimum
+
f is increasing
The sign of f ' x changes sign from + to – at x 0 relative maximum
156
MATH 401 – DIFFERENTIAL CALCULUS
The sign of f ' x changes sign from - to + at x 2 relative minimum
2 x
Illustration3. Given f ( x) x e , find the relative maximum and relative
minimum point and sketch the graph using the first derivative test.
Solution:
Plot the graph of the function and
compute the derivative of f to
determine the critical values.
f ( x) x 2 e x
f ' ( x) x 2 (e x ) e x (2 x)
f ' ( x) x 2 e x 2 xe x
f ' ( x ) e x ( x 2 2 x)
The only critical numbers are those
For which f ' ( x) 0 and f ' ( x) does
not exists.
e x ( x 2 2 x) 0
ex 0
( x 2 2 x) 0 x(x 2) 0
x0
(x 2) 0 x 2
The critical numbers are 1 and 3. To determine whether f has a
relative extremum at these numbers, we apply the first-derivative test and
summarize the results in the table.
Interval
f ( x) x 2 e x
x0
x0
0
0 x2
x2
x2
4e 2
f ' ( x ) e x ( x 2 2 x)
Conclusion
-
f is decreasing
0
f has a relative minimum
+
f is increasing
0
f has a relative maximum
-
f is increasing
157
MATH 401 – DIFFERENTIAL CALCULUS
The sign of f ' x changes sign from - to + at x 0 relative minimum
The sign of f ' x changes sign from + to - at x 2 relative maximum
Exercise 3. Solve the following: (a) find the relative extrema of the function using
the first-derivative test; (b) determine the values of x at which the relative
extrema occur; (c) determine the intervals on which the function is increasing
and decreasing; (d) Sketch the graph of the function from your answers.
3
2
1. f ( x) x 3x 3x
3
2
2. y x 3x 5
x 2 4x 5
3. f ( x)
x2
2
4. f ( x) 2 x ln x
x
5. y e sin x
158
MATH 401 – DIFFERENTIAL CALCULUS
4.3.2. Concavity, Points of Inflection, and the Second-Derivative Test
Although the sign of the derivative of f reveals where the graph of f
is increasing or decreasing, it does not reveal the direction of the
curvature. Figure4.3.3 below suggests two ways to characterize the
concavity of a differentiable f on an open interval:
f is concave up on an open interval if its tangent lines have
increasing slopes on that interval and is concave down if they have
decreasing slopes.
f is concave up on an open interval if its graph lies above its tangent
lines and concave down if it lies below its tangent lines.
Figure4.3.3
Definition4.3.3.a Concave Upward and Downward
If f is differentiable on an open interval, then f is said to be concave
upward on the open interval if f ' is increasing on that interval, and f is said to
be concave downward on the open interval if f ' is decreasing on that interval.
Since the slopes of the tangent lines to the graph of a differentiable
function f are the values of its derivative f ' , it follows from
Theorem4.3.2.a (applied to f ' rather than f ) that f ' will be increasing
on intervals where f ' ' is positive and that f ' will be decreasing on
intervals where f ' ' is negative. Thus we have the following theorem
159
MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.3.a Let f be twice differentiable on an open interval.
(i) If f ' ' 0 for every value of x in the open interval, then f is concave
upward on that interval.
(ii) If f ' ' 0 for every value of x in the open interval, then f is concave
downward on that interval.
2
Illustration1. Let f x x 4 x 5 , determine where the graph is concave
upward and concave downward.
Solution:
2
Given f x x 4 x 5
f ' x 2 x 4
f ' ' x 2
f ' x 0 then 2 x 4 0 , solve for critical numbers
2( x 2) 0 Thus, 2 0 and ( x 2) 0 x 2
The critical number is 2, summarized the results in Table below and graph
Interval
f (x)
x2
x2
x2
f ' x
-
1
0
+
Conclusions
f is decreasing
f has a relative
minimum
f is increasing
f ' ' x
Conclusions
+
f is concave upward
+
f is concave upward
+
f is concave upward
160
MATH 401 – DIFFERENTIAL CALCULUS
Illustration2. Let
f ( x) x 3 3x 1 , determine where the graph is
concave upward and concave downward.
Solution:
3
Given f ( x) x 3x 1
f ' x 3x 2 3
and
f ' ' x 6 x
f ' x 0 then 3x 2 3 0 , solve for critical numbers
3( x 2 1) 0 3( x 1)( x 1) 0
Thus, 3 0 , ( x 1) 0 x 1
( x 1) 0 x 1
The critical numbers are -1 and 1, summarized the results in Table below
and graph.
Interval
f (x)
x 1
x 1
+
3
1 x 1
x 1
x 1
f ' x
0
-
-1
0
+
Conclusions
f is increasing
f has a relative
maximum
f is decreasing
f has a relative
minimum
f is increasing
f ' ' x
Conclusions
-
f is concave downward
-
f is concave downward
+
f is concave downward
then concave upward
+
f is concave upward
+
f is concave upward
161
MATH 401 – DIFFERENTIAL CALCULUS
Points where the curve changes from concave upward to concave
downward or vice-versa are called points of inflection.
Definition4.3.3.b Points of Inflection
If f is continuous on an open interval containing a value x0 , and if f
changes the direction of its concavity at the point x0 , f x0 , then we say that
f has an inflection point at x0 , and we call the point x0 , f x0 on the graph of
f an inflection point of f (see figure below)
Figure4.3.3.b
3
2
Example1. Given the equation f x x 3x 1 determine the intervals
on which f is increasing, decreasing, concave upward and concave
downward. Locate all inflection points and confirm that your conclusions
are consistent with the graph.
Solution:
3
2
Given f x x 3x 1
f ' x 3x 2 6 x
f ' ' x 6 x 6
f ' x 0 then 3x 2 6 x 0 , solve for critical numbers
3x( x 2) 0
Thus, 3x 0 x 0
( x 2) 0 x 2
The critical numbers are 0 and 2, summarized the results in the Table and
graph.
162
MATH 401 – DIFFERENTIAL CALCULUS
Interval
x0
x0
0 x2
x2
x2
Interval
x0
x0
0 x 1
x 1
1 x 2
x2
x2
f (x)
1
-3
f (x)
1
-1
-3
f ' x
+
0
0
+
f ' ' x
0
+
+
+
Table1
Conclusions
f is increasing
f has a relative maximum
f is decreasing
f has a relative minimum
f is increasing
Table2
Conclusions
f is concave downward
f is concave downward
f is concave downward
f has a point of inflection
f is concave upward
f is concave upward
f is concave upward
Table2 shows that there is a point of inflection at x 1 , since f
changes from concave downward to concave upward at that point. The
point of inflection is (1,-1).
163
MATH 401 – DIFFERENTIAL CALCULUS
4
3
Example2. Given the equation f x x 4x determine the intervals on
which f is increasing, decreasing, concave upward and concave downward.
Locate all inflection points and confirm that your conclusions are
consistent with the graph.
Solution:
4
3
Given f x x 4x
f ' x 4 x 3 12 x 2 and f ' ' x 12 x 2 24 x
f ' x 0 then 4 x 3 12 x 2 0 , solve for critical numbers
4 x 2 ( x 3) 0 Thus, 4 x 2 0 x 0 and ( x 3) 0 x 3
Thus critical numbers are 0 and 3
Table2
Interval
f (x)
x0
x0
f ' x
-
0
0
0 x2
x2
2 x3
-16
-
x3
-27
0
x3
+
Conclusions
f is decreasing
f has neither a relative
max nor min
f is decreasing
f is decreasing
f is decreasing
f has a relative
minimum
f is increasing
f ' ' x
Conclusions
+
f is concave upward
0
f has a point of inflection
0
+
f is concave downward
f has a point of inflection
f is concave upward
+
f is concave upward
+
f is concave upward
Table2 shows that there is a point of inflection at x 0 and x 2
since f changes from concave upward to concave downward and vice versa
at that point. The point of inflection are (0,0) and (2,-16).
164
MATH 401 – DIFFERENTIAL CALCULUS
Theorem4.3.3.b Suppose the function f is differentiable on some open interval
containing c, and c, f c is a point of inflection of the graph of f . Then if f ' ' c
exists, f ' ' c 0
4
Illustration1. Given the equation f x x determine the intervals on
which f is increasing, decreasing, concave upward and concave downward.
Locate all inflection points if any and confirm that your conclusions are
consistent with the graph.
Solution:
4
Given f x x
f ' x 4 x 3 and f ' ' x 12 x 2
f ' x 0 then 4 x 3 0 4 x 3 0 x 0
The critical number is 0, summarized the results in the Table and graph.
Interval
f (x)
x0
x0
x0
f ' x
-
0
0
+
Table1
Conclusions
f is decreasing
f has a relative
minimum
f is increasing
f ' ' x
Conclusions
+
f is concave upward
0
f has no inflection point
+
f is concave upward
Table1 shows that there is no change in concavity and hence no
inflection point at x 0 , even though f ' ' x 0
165
MATH 401 – DIFFERENTIAL CALCULUS
x
Illustration2. Let f ( x) xe , determine the intervals on which f is
increasing, decreasing, concave upward and concave downward. Locate all
inflection points if any and confirm that your conclusions are consistent
with the graph.
Solution:
Given
f ( x) xe x
f ' x xe x e x
f ' x x(e x ) e x (1)
f ' ' x ( x)(e x ) e x (1) (e x )
f ' x xe x e x
f ' ' x xe x e x e x
f ' x 0 then xe x e x 0 ,
f ' ' x xe x 2e x
solve for critical numbers
f ' ' x 0
xe x e x 0 e x ( x 1) 0
xe x 2e x 0 e x ( x 2) 0
Thus, e x 0 , (x 1) 0 x 1
Thus, e x 0 , ( x 2) 0 x 2
(1, e 1 ) critical point
(2,2e 2 ) possible inflection point
Interval
f (x)
x 1
x 1
+
e 1
1 x 2
x2
x2
f ' x
2e
2
0
-
Conclusions
f is increasing
f has a relative
maximum
f is decreasing
f is decreasing
f is decreasing
f ' ' x
Conclusions
-
f is concave downward
-
f is concave downward
0
+
f is concave downward
f has a point of inflection
f is concave upward
166
MATH 401 – DIFFERENTIAL CALCULUS
There is another test for relative extrema that is based on the following
geometric observation:
• A function f has a relative maximum at stationary point if the graph
of f is concave downward on an open interval containing that point.
• A function f has a relative minimum at stationary point if the graph
of f is concave upward on an open interval containing that point.
Theorem4.3.3.b Second Derivative Test
Suppose that f is twice differentiable at the point x0
(i) If f ' x0 0 and f ' ' x0 0 , then f has a relative minimum at x0 .
(ii) If f ' x0 0 and f ' ' x0 0 , then f has a relative maximum at x0 .
(iii) If f ' x0 0 and f ' ' x0 0 , then the test is inconclusive; that is, f may
have a relative maximum, a relative minimum, or neither at x0 .
5
3
Illustration1. Given f x 3x 5x Find the relative extrema of f by
applying the second-derivative test. Use this information to sketch the
graph of f.
Solution:
5
3
Given f x 3x 5x
f ' x 15x 4 15x 2 and f ' ' x 60 x 3 30 x
f ' x 0 then
15x 4 15x 2 0 15x 2 ( x 2 1) 0 15x 2 ( x 1)( x 1) 0
15x 2 0 , ( x 1) 0 , ( x 1) 0
x0,
x 1 ,
x 1
167
MATH 401 – DIFFERENTIAL CALCULUS
Interval
f (x)
f ' x
f ' ' x
x 1
2
0
-
x0
0
0
0
x 1
-2
0
+
Table1
Conclusions
f has a relative maximum
f has neither a relative maximum nor
minimum
f has a relative minimum
3
2
Illustration2. Given 3 y x 3x 9 x 3 Find the relative extrema of f
by applying the second-derivative test. Use this information to sketch the
graph of f.
Solution:
3
2
Given 3 y x 3x 9 x 3
1
y x 3 x 2 3x 1
3
y f x
f x
1 3
x x 2 3x 1
3
f ' x x 2 2 x 3 and f ' ' x 2 x 2
f ' x 0 then
x 2 2 x 3 0 ( x 3)( x 1) 0
( x 3) 0 , ( x 1) 0
x 3 ,
x 1,
168
MATH 401 – DIFFERENTIAL CALCULUS
Interval
f (x)
x 3
10
2
3
x 1
Table1
f ' x
f ' ' x
0
-
f has a relative maximum
0
+
f has a relative minimum
4.3.3. Sketching Graphs (Curve
Transcendental Functions
Conclusions
Tracing)
of
Algebraic
and
We learned in the previous section how properties of graphs of
functions can be determined from their derivative. Now, we summarized
the steps incorporating the properties discussed in this chapter that we
should follow when sketching the graph of a function f.
1. Find any x and y intercepts.
2. Compute for the first and second derivative.
3. Determine the critical numbers of f. These are the values of x in the
domain of f for which either f ' x 0 or f ' x does not exists.
4. Determine the relative extremum using the first-derivative test or
second-derivative test.
5. Determine the intervals on which f is increasing or decreasing.
6. Find critical numbers of f ' , that is, the values of x for which f ' ' x does
not exists or f ' ' x 0 to obtain possible inflection points.
169
MATH 401 – DIFFERENTIAL CALCULUS
7. Check for concavity of the graph.
Example1. Analyze and trace the curve of y
4x
4 x2
Solution:
The x and y intercept at the origin (0,0)
4x
Since y f x then f x
4 x2
(4 x 2 )(4) (4 x)(2 x) 16 4 x 2 8 x 2 16 4 x 2
f ' x
(4 x 2 ) 2
(4 x 2 ) 2
(4 x 2 ) 2
f ' x
16 4 x 2
let f ' x 0
(4 x 2 ) 2
16 4 x 2
0 16 4 x 2 0 4(4 x 2 ) 0 4(2 x)(2 x) 0 then
2 2
(4 x )
(2 x) 0 x 2 ,
( 2 x) 0 x 2
4 0,
Critical values are -2 and 2. Critical points are (-2,-1) and (2,1)
f ' ' x
(4 x 2 ) 2 (8 x) (16 4 x 2 )(2)(4 x 2 )(2 x)
(4 x )
2 2 2
f ' ' x
(8 x)(4 x 2 ) 2 4 x(16 4 x 2 )(4 x 2 )
(4 x 2 ) 4
(4 x)(4 x )(2)(4 x
f ' ' x
2
2
) (16 4 x 2 )
(4 x 2 ) 4
f ' ' x
(4 x)8 2 x 2 16 4 x 2 )
(4 x 2 ) 3
(4 x)(2 x 2 24) (4 x)(2)( x 2 12)
f ' ' x
(4 x 2 ) 3
(4 x 2 ) 3
f ' ' x
(8 x)( x 2 12)
let f ' ' x 0
(4 x 2 ) 3
(8 x)( x 2 12)
0 (8x)( x 2 12) 0 then
2 3
(4 x )
8x 0 x 0
( x 2 12) 0 x 2 12 x 2 3
3
3
and 2 3 ,
Possible point of inflections are 0,0 , 2 3 ,
2
2
170
MATH 401 – DIFFERENTIAL CALCULUS
Interval
f (x)
x 2 3
x 2 3
3
2
2 3 x 2
x 2
f ' x
Conclusions
f ' ' x
Conclusions
-
f is decreasing
-
f is concave downward
-
f is decreasing
0
f has inflection point
-
f is decreasing
+
f is concave upward
+
f is concave upward
+
0
-
f is concave upward
f has inflection point
f is concave downward
-
f is concave downward
-1
0
2 x 0
x0
0 x2
0
+
+
+
x2
1
0
2 x2 3
3
2
x2 3
x2 3
f has a relative
minimum
f is increasing
f is increasing
f is increasing
f has a relative
maximum
-
f is decreasing
-
f is concave downward
-
f is decreasing
0
f has inflection point
-
f is decreasing
+
f is concave upward
5
4
Example2. Analyze and trace the curve of f x 3x 5x
Solution:
5
4
Given f x 3x 5x
f ' x 15x 4 20 x 3 let f ' x 0
15x 4 20 x 3 0 5x 3 (3x 4) 0 then
5x 3 0 x 0 ,
(3x 4) 0 x
4
3
171
MATH 401 – DIFFERENTIAL CALCULUS
4
. Critical points are (0,0) and
3
Critical values are 0 and
f ' ' x 60 x 3 60 x 2
4 256
,
3 81
let f ' ' x 0
60 x 3 60 x 2 0 (60 x 2 )( x 1) 0 then
60 x 2 0 x 0
( x 1) 0 x 1
Possible point of inflections are 0,0 , and 1,2
Interval
x
f (x)
4
3
x 4 3
256
81
4 3 x 1
x 1
1 x 0
x0
x0
2
0
f ' x
Conclusions
f ' ' x
Conclusions
+
f is increasing
-
f is concave downward
0
f has a relative maximum
-
f is concave downward
0
+
f is decreasing
f is decreasing
f is decreasing
f has a relative minimum
f is increasing
0
+
0
+
f is concave downward
f has inflection point
f is concave upward
f is concave upward
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MATH 401 – DIFFERENTIAL CALCULUS
Exercise 4. Solve the following: (a) find the relative extrema of the function using
the second-derivative test; (b) determine the values of x at which the relative
extrema occur; (c) determine the intervals on which the function is increasing
and decreasing; (d) concave upward or concave downward; (e) Locate inflection
points if any and (f) Sketch the graph of the function from your answers.
3
2
1. f ( x) 2 x 3x 12 x 7
2. y
2
x 4
2
4
3
2
3. y 3x 4 x 12 x
4.
y
x
1 x 2
5.
y
1 5 2 3
x x
5
3
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MATH 401 – DIFFERENTIAL CALCULUS
4.4.
OPTIMIZATION PROBLEMS
One of the most common applications of calculus involves the
determination of minimum and maximum values. As a would-be engineer, you
will encounter terms like greatest strength, greatest voltage, greatest profit,
greatest distance, optimum size, least size, least time, and least cost. Finding
solution to these situations are called Optimization Problems. To solve this, we
need to be familiar on the application of calculus in solving these optimization
problems.
Here are the guidelines for solving maximum and minimum problems.
GUIDELINES FOR SOLVING OPTIMIZATION PROBLEMS
1. Identify all given quantities and all quantities to be determined. If
possible, make a sketch.
2. Write a primary equation for the quantity that is to be maximized or
minimized. (Review useful formulas from Geometry)
3. Reduce the primary equation to one having a single independent
variable. This may involve the use of secondary equations relating the
independent variables of the primary equation.
4. Determine the feasible domain of the primary equation. That is, determine
the values for which the stated problem makes sense.
5. Determine the desired maximum or minimum value by the calculus
techniques discussed in the previous Section 4.2.
TAKE NOTE:
When performing Step 5, recall that to determine the maximum or
minimum value of a continuous function f on a closed interval, you should
compare the values of f at its critical numbers with the values of f at the
endpoints of the interval. Another way is to apply the Second Derivative Test.
4.4.1 Optimization Problems involving Algebraic Functions
The following examples illustrate the applications of calculus in solving
optimization problems involving algebraic functions:
Example 4.4.1.1
Maximum Volume of a Box
An engineering firm wants to design an open box having a square base and a
surface area of 192 square inches, as shown in the figure. What dimensions will
produce a box with maximum volume?
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MATH 401 – DIFFERENTIAL CALCULUS
Solution:
Step 1: Given a box with a square base, let x be its length and width and y
be its height. Given its surface area
.
Surface Area = 192
Figure 1. Open box with a square base
Step 2: Since the box has a square base, its Volume is
This is a primary equation because it gives a formula for the quantity to be
maximized.
Step 3: Write V as a function of just one variable, say x. To do this we need
a secondary equation which is the surface area of the box.
Solve this equation for h in terms of x to obtain
Substituting into the primary equation for V produces
.
Step 4: Determine the feasible domain of x that will yield a maximum value
of V. That is, what values of x make sense in this problem? We all know
that the volume
, and x must be nonnegative and that the area of the
base of the box
is at most 192. So, the feasible domain is
.
Step 5: To maximize V, find the critical numbers of the volume function on
the interval
.
To find the critical values of x, set
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MATH 401 – DIFFERENTIAL CALCULUS
These are the critical numbers. We will not consider the
is not part of the feasible domain. To verify if this
maximum volume, apply the Second Derivative Test.
because it
will give a
Substituting
gives
Since
, the V has a relative maximum. Therefore we conclude that
the dimensions of the box with maximum volume is when its length or
width
.
When
the box is
, the height of the box
. The maximum volume of
.
Example 4.4.1.2
Minimum Material for a Piece of Paper
A rectangular page is to contain 24 square inches of print. The margins at
the top and bottom of the page are to be 1.5 inches, and the margins on the
left and right are to be 1 inch. What should the dimensions of the page be
so that the least amount of paper is used?
Solution:
Step 1: Start by labeling a drawing with the known and unknown
quantities.
1.5
1
.
5
1.5
Figure 2. Rectangular Page
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: Let A be the area which is the quantity to be minimized. The area of
the whole page which is the primary equation is given as
Step 3: Write A as a function of one variable only, let us say x. To do this we
need a secondary equation which is the area of print; 24 square inches.
This is given by:
Solving this equation for y gives
.
Now, substituting it into the primary equation produces
Determine the feasible domain of x that will yield a maximum value
of A. We all know that the area A
, and x must be nonnegative, that is
Step 5: To minimize A, find the critical values of x by differentiating A with
respect to x
Set
These are the critical numbers. We will not consider the
is not part of the feasible domain. To verify if this
minimum area, apply the Second Derivative Test.
because it
will give a
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MATH 401 – DIFFERENTIAL CALCULUS
Substituting
gives
Since
, then A has a relative minimum when
Therefore, we
concluded that the dimensions of the page with minimum area should be
by
Example 4.4.1.3
Endpoint Maximum
Four feet of wire is to be used to form a square and a circle. How much of
the wire should be used for a square and how much should be used for the
circle to enclosed the maximum total area?
Solution:
Step 1: Given the total length of the wire is 4 ft. This is equal to the sum of
perimeter of a square and the circumference of a circle.
Perimeter = 4x
x
Area = x2
4 feet
x
.
r
Circumference = 2πr
Area = πr2
Figure 3. A square and a circle
Step 2: Let A be the total area which is the quantity to be maximized. The
rimary equation is given as
Step 3: Write A as a function of one variable only, let us say x. To do this we
need a secondary equation which is the total length of the wire = 4ft. This
is
given
by:
Solving for r,
Now, substituting it into the primary equation produces
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MATH 401 – DIFFERENTIAL CALCULUS
Determine the feasible domain of x that will yield a maximum value
of A. The area A
, and
restricted by the square’s perimeter.
Step 5: To maximize A, find the critical values of x by differentiating A with
respect to x
Set
This is the only critical number in the feasible domain.
To verify if this
will give a maximum area, apply the Second
Derivative Test.
Since
, then A has a relative minimum when
But what we need is the value x that will give a maximum area. So, using
the endpoints in the feasible domain and the critical value, the respective
total area are given as:
Therefore we concluded that the value of x that will give maximum area is
when
That is, all the wire is used for the circle.
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MATH 401 – DIFFERENTIAL CALCULUS
Example 4.4.1.4
Maximum Illumination
A Norman window consists of a rectangle surmounted by a semicircle. If
the perimeter of a Norman window is to be 40 ft, determine what should
be the radius of the semicircle and the height of the rectangle such that the
window will admit the most light?
Solution:
Step 1: Draw a picture and label the variables:
x = length of a rectangle
x = diameter of a semicircle
x/2 = radius of a semicircle
y = height of a rectangle
Figure 4. The Norman window
Step 2: The primary equation for this problem is area - the area for this
window is the rectangular section plus the semicircular section. It is given
by:
Step 3: Write A as a function of one variable only, let us say x. To do this we
need a secondary equation which is the perimeter - three sides of the
rectangle plus the semicircle. It is given by:
Solving for y in terms of x, it will probably make things easier if we
multiplied both sides by 2 first to get rid of that fraction on the right side:
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MATH 401 – DIFFERENTIAL CALCULUS
Substituting this into the primary equation gives
Determine the feasible domain of x that will yield a maximum value
of A. The area A
, and x must be nonnegative.
Step 5: To maximize A, find the critical values of x by differentiating A with
respect to x
Set
This is the only critical number in the feasible domain.
To verify if this
will give a maximum area, apply the Second
Derivative Test.
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MATH 401 – DIFFERENTIAL CALCULUS
Since
, then A has a relative maximum. We can now concluded that
the radius of the semicircle is
, and the height of the
rectangle is
Example 4.4.1.5
Minimum Surface Area
A 120-m3 closed aluminum tank is to be in the form of a right-circular
cylinder, determine the base radius of the tank if the least amount of
aluminum is to be used in its manufacture.
Solution:
Step 1: Draw the figure of a right-circular cylindrical tank. We are asked to
determine the base radius for which the total surface area of the tank is a
minimum.
r = base radius
h = height
V = Volume = 120 m3
Figure 5. Cylindrical Tank
Step 2: The primary equation for this problem is total surface area which
is equal to the lateral surface area
and the total area of both the top
and bottom (each of
. If S is the total surface area, then
Step 3: Write S as a function of one variable only, let us say r. To do this we
need a secondary equation which is the volume of a right-circular cylinder.
It is given by:
V
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MATH 401 – DIFFERENTIAL CALCULUS
Solving this equation for h, we have
, and substituting into
the primary equation, we obtain S as a function of r:
Determine the feasible domain that will yield a minimum value of
S. From the equation defining S, r cannot be 0. Theoretically, r may be any
positive number. Therefore, the feasible domain of S is
.
Step 5: To minimize S, find the critical values of r by differentiating S with
respect to r, we have
Set
This is a critical number in the feasible domain.
To verify if this
Derivative Test.
will give a minimum value of S, apply the Second
If we substitute the critical value of r, then we will get
. Therefore, S
has a relative minimum value. We can now conclude that the least amount
of Aluminum will be used in the manufacture of the tank when the base
radius is
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MATH 401 – DIFFERENTIAL CALCULUS
Example 4.4.1.6
Minimum Cost of Box Material
A closed box with a square base is to have a volume of 4,000 cubic
centimeters. The material for the top and the bottom of the box is to cost 6
cents per square centimeter and the material for the sides is to cost 3 cents
per square centimeter. Find the dimensions of the box so that the total
cost of the material is least.
Solution:
Step 1: Draw the figure and label the quantities:
x = length of a side of the square base
y = depth of the box
y
x
V = Volume = 4,000 cm3
Figure 6. A closed Box
x
Step 2: The primary equation for this problem is total cost of the material
which is 6 cents x combined area of the top and bottom
plus 3 cents
x area of sides(4xy . The number of cents in the total cost of the material is
Step 3: Write C as a function of one variable only, let us say x. To do this we
need a secondary equation which is the volume of box that is equal to area
of the base and the depth.
V
Solving this equation for y, we have
, and substituting into
the primary equation, we obtain C as a function of x:
Determine the feasible domain that will yield a minimum value of
C. From the equation defining C, x cannot be 0. Theoretically, x may be any
positive number. Therefore, the feasible domain of C is
.
Step 5: To minimize C, find the critical values of x by differentiating C with
respect to x, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Set
This is a critical number in the feasible domain.
To verify if this
the Second Derivative Test.
will give a minimum value of C, apply
If we substitute the critical value of
, then we will get
.
Therefore, C has a relative minimum value. We can now conclude that the
cost of the material will be least when side of a square base is
and the depth is
Example 4.4.1.7
Greatest Volume
Find the dimensions of the right-circular cylinder of greatest volume that
can be inscribed in a right-circular cone with a radius of 15cm and a height
of 36cm.
Solution:
Step 1: Draw the figure and label the quantities:
For the cylinder:
x = radius
y = height
h-y
F
For the cone:
r = base radius = 15 cm
h = height = 36 cm
Figure 7. Cylinder inscribed in a cone
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: The primary equation for this problem is volume of the cylinder
which
is
Step 3: Write V as a function of one variable only, let us say x. To do this we
need a secondary equation involving x and y. From Figure 7, and by similar
triangles,
and
,
Solving this equation for y, we have
, and substituting into the
primary equation, we obtain V as a function of x:
Determine the feasible domain that will yield a maximum value of
V. From the equation defining C, the feasible domain of C is
.
Step 5: To maximize V, find the critical values of x by differentiating V with
respect to x, we have
Set
Solve for x
These are the critical values of x, both of which are in
.
To verify which of these values will give a maximum value of V, apply the
Second Derivative Test.
Substitute the critical values:
and
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MATH 401 – DIFFERENTIAL CALCULUS
When
,
. Therefore, V has a relative maximum value. Also
the corresponding height is
We can
now conclude that the greatest volume of an inscribed cylinder in a given
cone is
, which occurs when the
radius is 10 cm and the height is 12 cm.
Example 4.4.1.8
MINIMUM Distance
Which point on the graph of
is closest to the point
Solution:
Step1: Sketch a graph of the function and label the points:
Figure 8
Step 2: Since the function is symmetrical, and the point (0, 1) is in the
middle, there are two points that have the same minimum distance. For
this problem, we are minimizing distance, so for our primary equation we
will use the formula for distance between two points, those points being
(0,1) and (x, y):
Step 3: Write d in terms of one variable only, say x. We need here the
secondary equation which is the original function,
It is already
solved for y, so we can replace the y in the primary equation with it:
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MATH 401 – DIFFERENTIAL CALCULUS
Step 4: Determine the feasible domain that will yield a minimum value of
d. Because d is smallest when the expression inside the radical is smallest,
we need only to find the critical numbers of
.
The domain of this f(x) is the entire real line.
Step 5: To minimize d, find the critical values of the function by
differentiating f(x) with respect to x:
Set
Solving for x values, we have
and
, then
and
Substituting this to the original function, the corresponding y values are:
From looking at the graph, we can see that x = 0 yields a relative
maximum, which is not what is required in the problem. To verify if the
critical values yield a minimum distance d, apply the Second Derivative
Test
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MATH 401 – DIFFERENTIAL CALCULUS
Substituting the critical values, we have
When
,
Since
, then the d has a relative maximum. This means that the
point
is the farthest point from (0,1) which is not what is required
in the problem.
When
,
When
,
Since
, then the d has a relative minimum. So, the closest points
from point (0,1) are
and
.
4.4.2 Optimization Problems involving Transcendental Functions
Example 4.4.2.1
Largest Lateral Surface Area
A right-circular cylinder is to be inscribed in sphere of given radius. Find
the ratio of the height to the base radius of the cylinder having the largest
lateral surface area.
Solution:
Step 1: Draw the figure and label all the quantities:
r = constant radius of sphere
R = radius of cylinder
h = height of cylinder
θ
θ = angle at the center of the
sphere subtended by R
Figure 9. Cylinder inscribed in a sphere
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: The primary equation for this problem is lateral surface area (S) of
the cylinder,
W
v
θ T
need secondary equations expressing R and h in terms of θ F
figure, consider a right triangle ABO, applying SOHCAHTOA, we have
w
Substituting these into the primary equation, we obtain S as a function of
θ:
Determine the feasible domain that will yield a maximum value of
S. From the equation defining S, the feasible domain is
.
T
z
v
θ
differentiating S with
θ take note here that r is a constant radius of a sphere, we have
Set
This is a critical number in the feasible domain
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MATH 401 – DIFFERENTIAL CALCULUS
To verify if this
Derivative Test.
will give a maximum value of S, apply the Second
If we substitute the critical value of
, then we will get
. Since
then S has
a relative maximum value.
When
, the height h and the base radius R is
We can now conclude that for the cylinder having the largest lateral
surface area, the ratio of the height to the base radius is 2.
Example 4.4.2.2
Best View
An engineer designed a sports arena where the television screen is vertical
and .4 m high. The lower edge is 8. m above an observer’s eye level. If
the best view of the screen is obtained when the angle subtended by the
screen at eye level is a maximum, how far from directly below the screen
must the observer be?
Solution:
Step 1: Draw the figure and label all the quantities.
angle subtended by the screen at eye level
x = distance between the point directly below the
screen and the observer
Figure 10. Television screen
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: The primary equation for this problem is the angle subtended by
the screen at eye level. From the figure, it is
Step 3: Write as a function of one variable only, let us say x. To do this we
need secondary equations that will relate and with x. Consider two
right triangles in the figure and apply SOHCAHTOA.
This gives
Substituting these into the primary equation, we obtain
x:
as a function of
Determine the feasible domain that will yield a minimum value of
. From the equation defining , x cannot be 0. Theoretically, x may be any
positive number. Therefore, the feasible domain is
.
Step 5: To maximize , find the critical values of x by differentiating
respect to x, we have
with
Combining the right side by getting the LCD, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Set
These are the critical numbers. We will disregard the
is not in the feasible domain.
To verify if this
Derivative Test.
For
For
For
will give a maximum value of
, the
, the
, so
, so
, the
because it
, apply the First
is increasing
has a relative maximum value
, so
is decreasing
We can now conclude that to have the best view of the screen, the observer
should stand approximately 9.63 m from the point directly below the
screen.
Example 4.4.2.3
Largest Rectangle
A computer is programmed to inscribe a series of rectangles in the first
quadrant under the curve of
. What is area of the largest rectangle
that can be inscribed?
Solution:
Step 1: Sketch the graph and label the quantities.
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MATH 401 – DIFFERENTIAL CALCULUS
Figure 11. One of the possible rectangles under the curve.
Step 2: The primary equation for this problem is area - the area of the
rectangle which is given by:
Step 3: Write A as a function of one variable only, let us say x. To do this we
need a secondary equation which is the exponential curve
.
Substituting this into the primary equation gives
Determine the feasible domain of x that will yield a maximum value
of A. The area A
, and x must be nonnegative.
Step 5: To maximize A, find the critical values of x by differentiating A with
respect to x. Use Product Rule
Set
This expression only equals zero when
.
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MATH 401 – DIFFERENTIAL CALCULUS
To verify if this
Derivative Test.
will give a maximum area, apply the Second
Use Product Rule to get the 2nd derivative
If substitute the critical number
, we get
. Since
then A has a relative maximum. The length of the largest rectangle is
then its height is
.
,
,
We can now conclude that the area of the largest rectangle that can be
inscribed is
4.4.3. Number Problems
In this section, we are to apply optimization in finding two positive
numbers that satisfy the given requirements.
Example 4.4.3.1
The product is 100 and the sum is a minimum.
Solution:
Let x be the first number and y be the second number.
Their product is
while their sum
Express S in terms of one variable only, say x. To do this, solve for y in the
primary equation
. Substitute this in their sum S, we have now S as
a function of x:
To minimize S, find the critical numbers by differentiating S with respect
to x.
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MATH 401 – DIFFERENTIAL CALCULUS
Set S’ = 0
Since the requirement are two positive numbers, the numbers that will
give a product of 100 and the sum is minimum are both 10.
Example 4.4.3.2
The sum is 12 and their product is a maximum. Ans. 6
Solution:
Let x be the first number and y be the second number.
Their sum is
while their product is
Express P in terms of one variable only, say x. To do this, solve for y in the
primary equation
. Substitute this in their product P, we have
now P as a function of x:
To maximize P, find the critical numbers by differentiating P with respect
to x.
Set P’ = 0
then
Since the requirement are two positive numbers, the numbers that will
give a sum of 12 and the product is maximum are both 6.
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MATH 401 – DIFFERENTIAL CALCULUS
Exercises 4.4
Solve the following optimization problems:
1. A funnel of specific volume is to be in a shape of a right-circular cone. Find
the ratio of the height to the base radius if the least amount of material is
to be used in its manufacture.
Ans.
2. A right-circular cone is to be circumscribed about a sphere of given radius.
Find the ratio of the altitude to the base radius of the cone of least possible
volume.
Ans.
3. Find the volume of the largest right-circular cylinder that can be inscribed
in a right-circular cone having a radius of 4 cm and a height of 8 cm.
Ans.
4. A piece of wire 10 ft long is cut into two piece. One piece is bent into the
shape of a circle and the other into a shape of a square. How should a wire
be cut so that (a) the combined area of the two figures is as small as
possible; (b) the combined area of the two figures is as large as possible?
Ans. (a) radius of circle is
(b) radius of circle is
length of side of square is
there is no square
5. A piece of wire 20 cm long is cut into two pieces, and each piece is bent
into a shape of a square. How should a wire be cut so that the total area of
the two squares is as small as possible?
Ans. Cut in half
6. If R meters is the range of a projectile, then
,
, where
ft/s is the initial velocity, m/s2 is the acceleration due to gravity, and
is the radian measure of the angle that the launcher makes with the
horizontal. Find the value of that makes the range maximum.
Ans.
7. Which points on the graph of
are closest to the point (0,2)?
Ans.
and
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MATH 401 – DIFFERENTIAL CALCULUS
8. A photographer is taking a picture of a painting hung in an art gallery. The
height of a painting is 4 ft. The camera lens is 1 ft below the lower edge of
the painting. How far should the camera be from the painting to maximize
the angle subtended by the lens camera?
Ans. 2.24 ft
Find two positive numbers that satisfy the given requirements.
9. The second number is the reciprocal of the first number and the sum is
minimum.
Ans. 1
10. The sum of the first number squared and the second number is 54 and the
product is a maximum.
Ans. 3 and 36
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MATH 401 – DIFFERENTIAL CALCULUS
4.5
RELATED RATES
A problem in related rates is one involving rates of change of related
variables. In real-world applications involving relates rates, the variables have a
specific relationship for values of time t. This relationship usually expressed in
the form of an equation which represents a mathematical model of the situation.
The table below lists examples of mathematical models involving rates of
change.
Verbal Statement
The velocity of a car after traveling for
1 hour is 120 kilometers per hour.
Mathematical Model
when
Water is being pumped into a
swimming pool at a rate of 10, 000 L
per hour
A wheel is revolving at a rate of 50
revolutions per minute.
An electric charge is flowing in a wire
at a rate of 2 Coulombs per second.
Here are the guidelines for solving Related-Rate Problems
GUIDELINES FOR SOLVING RELATED-RATE PROBLEMS
1. GIVEN. Identify all given quantities and quantities to be determined.
Make a sketch and label the quantities.
2. MODEL. Write an equation involving the variables whose rates of change
either are given or are to be determined. This is a mathematical model of
the situation.
3. CALCULUS. Using the Chain Rule, implicitly differentiate both sides of the
equation with respect to time t.
4. SIMPLIFY. Substitute into the resulting equation all known values for the
variables and their rates of change. Then solve for the required rate of
change.
5. CONCLUSION. Write a conclusion, consisting of one or more complete
sentences, that answers the questions of the problem. Be sure your
conclusion contains the correct units of measurement.
Note: When using these guidelines, be sure to perform Step 3 before Step 4.
Substituting the known values of the variables before differentiating will produce
an inappropriate derivative.
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MATH 401 – DIFFERENTIAL CALCULUS
4.5.1. Related-Rate problems involving algebraic functions
Example 4.5.1.1
Ladder Problem
A 20-ft ladder leaning against a vertical wall starts to slide. The bottom of
the ladder is pulled horizontally away from the wall at 4 ft/s. Determine
how fast the top of the ladder is sliding down the wall when the bottom is
12 ft from the wall.
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
20 ft
y
x
Figure 1 . Ladder leaning against a wall
t = time that has elapsed since the ladder started to slide down the wall
x = distance from the bottom of the ladder to the wall at time t
y = distance from the ground to the top of the ladder at time t
rate at which the bottom of the ladder moves horizontally from the wall
rate at which the top of the ladder slides down the wall
Step 2: MODEL
Because the bottom of the ladder is pulled horizontally away from the wall
at 4 ft/s,
. In order to determine
when
12, write an
equation to relate x and y. From the Pythagorean Theorem,
. Solving for y in terms of x yields
Step 3: CALCULUS
Using the Chain Rule, implicitly differentiate both sides with respect to t,
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MATH 401 – DIFFERENTIAL CALCULUS
Step 4: SIMPLIFY
Substitute the known values of x, y, and
. When
for
Because
in the above equation and solve
12,
,
The minus sign indicates that y decreases as t increases.
Step5: CONCLUSION
The top of the ladder is sliding down the wall at the rate of 3 ft/s when the
bottom of the ladder is 12 ft from the wall
Example 4.5.1.2
An Inflating Balloon
A spherical balloon is being inflated at a rate of 10 ft 3/min. How fast the
radius of the balloon is expanding when the radius is 1 foot.
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
Figure 2. Spherical Balloon
Step 2: MODEL
We need an equation which relates V and r for a sphere; that is the volume
of the sphere
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MATH 401 – DIFFERENTIAL CALCULUS
Step 3: CALCULUS
Using the Chain Rule, implicitly differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known values of r and
in the above equation, we have
.
The positive sign indicates that r increases as t increases.
Step5: CONCLUSION
When the radius of the balloon is 1 ft, it is expanding at a rate of
approximately 0.8 ft3/min.
Example 4.5.1.3
Flow Rate in a Cylindrical Tank
Water is flowing into a 4-ft radius vertical cylindrical tank at the rate of
24ft3/min. How fast is the surface of water rising?
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
h = level of water in the tank
r = radius of tank = 4
Figure 3. Cylindrical Tank
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: MODEL
We need an equation which relates V and h ; that is the volume of a
cylinder
Step 3: CALCULUS
Using the Chain Rule, implicitly differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known value of
in the above equation, we have .
The positive sign indicates that h increases as t increases.
Step5: CONCLUSION
When the radius of the tank is 4 ft, the water surface is rising at a rate of
approximately 0.48 ft/min.
Example 4.5.1.4
Flow Rate in a Conical Tank
A water tank in the shape of an inverted cone is leaking water at a rate of 4
ft3/hour. The base radius of the tank is 5 ft and the height of the tank is 15
ft. When the depth of water is 6 ft, (a) at what rate is the depth of water in
the tank changing, (b) at what rate is the radius of the top of water in the
tank changing?
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
h = depth of water in the
tank
r = radius of top of water
in the tank
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MATH 401 – DIFFERENTIAL CALCULUS
Figure 4. Conical Tank
Step 2: MODEL
We need an equation which relates V , r and h ; that is the volume of water
in the tank at any time
To answer (a) we need to express V in terms of h only. To do this, we need
a secondary equation that will relate r and h. Consider two similar
triangles, the ratios of any two sides are equal. From the figure, we have
Substitute this into the volume formula gives
Step 3: CALCULUS
Using Chain Rule, implicitly differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known value of
(negative because water is leaking
out), and the depth h = 6 in the above equation, we have
The negative sign indicates that h decreases as t increases.
To answer part (b), use the equation of r in terms of h,
Differentiate both sides with respect to time gives
the value of
.
. Substituting
, we have
The negative sign indicates that r decreases as t increases.
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MATH 401 – DIFFERENTIAL CALCULUS
Step5: CONCLUSION
When the depth of water in the tank is 6 ft, (a) the depth of water in the
tank decreases at a rate approximately 0.28 ft/hour ; and (b) the radius of
the top of water in the tank decreases at a rate of approximately 0.10
ft/hour.
Example 4.5.1.5
Flow Rate in a Trapezoidal Trough
A trapezoidal trough is 10 ft long, 4 ft wide at the top, 2 ft wide at the
bottom and 2 ft deep. If water flows in at a rate of 10 ft 3/min, find how fast
the water surface is rising, when the water is 1 ft deep.
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
y = depth of water in the trough
For a trapezoid:
b1 = 2
b2 = x +2+x = 2x +2
h =y
Figure 5. Water flowing in a trapezoidal trough
Step 2: MODEL
We need an equation which relates V and y ; that is the volume of water in
the trough at any time. Recall volume of a container is
, with the
area of a trapezoid
. The volume of water in the
trapezoidal trough at any time is
. We need to express this volume V in terms of y only. To do this,
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MATH 401 – DIFFERENTIAL CALCULUS
express x in terms of y. Consider two similar triangles, the ratios of any
two sides are equal. From the figure, we have
Substitute this into the volume formula gives
Step 3: CALCULUS
Using Chain Rule, implicitly differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known value of
(positive because water flows in),
and the depth y = 1 in the above equation, we have
The positive sign indicates that y increases as t increases.
Step5: CONCLUSION
When the depth of water in the tank is 1 ft, the water surface is rising at a
rate of 1/3 ft/min.
4.5.2 Related Rates Problems involving Transcendental Functions
Example 4.5.2.1
Rate of Separation
Two people are 50 ft apart. One of them starts walking north at a rate so
that the angle shown in the diagram below is changing at a constant rate of
0.01 rad/min. At what rate is distance between the two people changing
when
?
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MATH 401 – DIFFERENTIAL CALCULUS
Moving Person
x
θ
Stationary Person
50ft
Figure 6
Solution:
Step 1: GIVEN
Let x be the distance between the two person at any given time.
Step 2: MODEL
We need an equation which relates x and θ. From the figure, consider a
right triangle, apply SOHCAHTOA, we have
Step 3: CALCULUS
Using Chain Rule, implicitly differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known value of
we have
and
in the above equation,
The positive sign indicates that x increases as t increases.
Step5: CONCLUSION
When
, the rate at which the distance between two people is
changing is approximately 0.71 ft/min.
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MATH 401 – DIFFERENTIAL CALCULUS
Example 4.5.2.2
Changing Angle of Elevation
A camera is on the ground, filming a rocket launch. The rocket is rising
according to the position function
, where y is measured in feet
and t in seconds. Find the rate of change of the camera’s angle at 1
seconds after the rocket initially launches. The camera is 1,000 feet away
from the rocket.
Solution:
Step 1: GIVEN
Draw the figure and label all the quantities:
Figure 7. Rocket Launching
Step 2: MODEL
We need an equation which relates s and θ. From the figure, consider a
right triangle, apply SOHCAHTOA, we have
Step 3: CALCULUS
Using Chain Rule, implicitly differentiate both sides with respect to t,
Differentiating the function
with respect to t gives
.
To solve for
, consider the figure, and apply Pythagorean theorem and
SOHCAHTOA, we have
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MATH 401 – DIFFERENTIAL CALCULUS
Step 4: SIMPLIFY
Substitute the known values of
, we have
and
Then
The positive sign indicates that θ increases as t increases.
Step5: CONCLUSION
The rate of change of the camera’s angle at 1 seconds after the rocket
initially launches is approximately 0.02 rad/s.
Example 4.5.2.3
Ladder Problem
A 20-ft ladder leaning against a vertical wall starts to slide. If the top of the
ladder is slides down at 4 ft/s. How fast is the angle of elevation of the
ladder changing when the lower end of the ladder is 12 ft from the wall.
Solution:
Step 1: GIVEN
Draw the figure and label all quantities
20 ft
y
θ
x
Figure 8 . Ladder leaning against a wall
t = time that has elapsed since the ladder started to slide down the wall
x = distance from the bottom of the ladder to the wall at time t
y = distance from the ground to the top of the ladder at time t
angle of elevation
rate at which the top of the ladder slides down the wall
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MATH 401 – DIFFERENTIAL CALCULUS
Step 2: MODEL
In order to determine
when
12, write an equation to relate y and θ.
From the figure, consider the right triangle and used SOHCAHTOA
Step 3: CALCULUS
Differentiate both sides with respect to t,
Step 4: SIMPLIFY
Substitute the known values of x, y, and
for
. When
Because
in the above equation and solve
12,
,
The minus sign indicates that θ decreases as t increases.
Step5: CONCLUSION
The angle of elevation of the ladder decreases by
end of the ladder is 12 ft from the wall.
when the lower
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MATH 401 – DIFFERENTIAL CALCULUS
4.5.3. Motion Problems
In this section, we will consider the application of the derivative of a
function in analyzing the motion of a particle on a line. Such motion is
called rectilinear motion.
Let s be the directed distance(displacement) of the particle from
the origin O at time t. Then f is the function defined by the equation
The instantaneous velocity of the particle at t units of time is given as
The instantaneous rate of change of velocity is called the
instantaneous acceleration given as
or
Example 4.5.3.1
A particle is moving along a horizontal line according to the equation
Determine the intervals of time when the particle is moving to the right
and when it is moving to the left. Also determine the instant when the
particle is at rest. What is the acceleration of the particle at t = 1 s?
Solution:
To determine the instantaneous velocity, differentiate s with respect to t,
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MATH 401 – DIFFERENTIAL CALCULUS
Determine the sign of various v for various intervals of t, and the results
are given below:
t-2
0
+
+
+
t–6
0
+
V
+
0
0
+
Conclusion
particle is moving to the right
particle is at rest
particle is moving to the left
particle is at rest
particle is moving to the right
For the instantaneous acceleration, differentiate the velocity v with respect
to time t,
At t = 1 s,
Since at t = 1 s, the particle is moving to the right, the negative acceleration
indicates that its velocity is decreasing.
Example 4.5.3.2
A ball is thrown vertically upward from the ground with an initial velocity
of 64 ft/s. The equation of motion is given by
(a) How high the ball go and how many seconds it takes for the ball to
reach its highest point?
(b) Find the instantaneous velocity of the ball at 1 s and 3 s.
(c) Find the instantaneous velocity of the ball when it reaches the ground.
(d) What is the acceleration of the ball at any time t?
Solution:
(a)At the highest point, the instantaneous velocity is zero, that is
Differentiate s with respect to t, then set
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MATH 401 – DIFFERENTIAL CALCULUS
Substitute this t into the equation of motion s, we have
Therefore, the ball reaches a highest point of 64 ft above the starting point
at 2 s.
(b) The instantaneous velocity of the ball:
At the end of 1 s, the ball is rising with an instantaneous velocity of 32 ft/s.
At the end of 3 s, the ball is falling with an instantaneous velocity of -32
ft/s.
(c) Since the ball reaches the highest point at t = 2 s, then it takes a total of
4 s to reach the ground. Substituting it to the instantaneous velocity, we
have
At the end of 4 s, the ball will reach the ground with an instantaneous
velocity of -64 ft/s.
(e) The instantaneous acceleration of the ball is
It means that the particle is constantly accelerating at 32 ft/s 2 downward.
That -32 ft/s2 is also the acceleration due to gravity.
Example 4.5.3.3
A particle is moving along a line according to the equation of motion
, with
. Find the value of t for which the measure
velocity is
(a) 0, (b) 1; and (c) 2.
Solution:
To determine the instantaneous velocity, differentiate s with respect to t,
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MATH 401 – DIFFERENTIAL CALCULUS
(a) When v = 0
(b) When v = 1
Since there is no negative time t, we only consider
.
(c) When v = 2
There is no time t possible when v = 2.
Example 4.5.3.3
A particle is moving on a line according to the equation of motion
where s centimeters is the directed distance of the particle from the origin
at t seconds. Express the velocity and acceleration as a function of time t.
Solution:
To determine the velocity of a particle as a function of time t, differentiate
s with respect to t.
To determine the acceleration of a particle as a function of time t,
differentiate v with respect to t.
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MATH 401 – DIFFERENTIAL CALCULUS
Comparing the acceleration
and the directed distance s, we have
Since the acceleration and the position s are oppositely directed, then the
motion is called simple harmonic.
Example 4.5.3.4
A C800 airplane take off from an airport at sea level and its altitude (in
feet) and time t (in minutes) is given by
What is the rate of climb at t = 3min?
Solution:
To determine the rate of climb (vertical velocity) of the airplane,
differentiate h with respect to t, we have
At t = 3 min,
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MATH 401 – DIFFERENTIAL CALCULUS
Example 4.5.3.5
A particle is moving on a line according to the equation of motion
(3 sinh t + 4 cosh t)
where s centimeters is the directed distance of the particle from the origin
at t seconds. Express the velocity and acceleration as a function of time t.
Solution:
To determine the velocity of a particle as a function of time t, differentiate
s with respect to t.
To determine the acceleration of a particle as a function of time t,
differentiate v with respect to t.
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MATH 401 – DIFFERENTIAL CALCULUS
Exercises 4.5
Solve the following related-rate and motion problems:
1. Water is flowing at the rate of 2 m 3/min into a tank in the form of an
inverted cone having an altitude of 16 m and a radius of 4 m. How fast is
the water level rising when the water is 5 m deep?
Ans. 40.74 cm/min
2. Two cars, one is going due East at the rate of 90km/hr and the other is
going due South at the rate of 60 km/hr, are traveling toward the
intersection of two roads. At what rate are the cars approaching each other
at the instant when the first car is 0.2 km and the second car is 0.15 km
from the intersection?
Ans. 108 km/hr
3. If a body of weight W pounds is dragged along a horizontal floor at
constant velocity by means of a force of magnitude F pounds and directed
at an angle of radians with the plane of the floor, then F is given by the
equation
where is a constant called the coefficient of friction. If
instantaneous change of F with respect to θ when
.
, find the
Ans. 2W
4. Boyle’s Law for the expansion of gas is
, where P is the number of
pounds per square unit of pressure, V is the number of cubic units of
volume of a gas, and C is a constant. At a certain instant the pressure is
3000 lb/ft2, and the volume is 5 ft3, and the volume is increasing at the rate
of 3 ft3/min. Find the rate of change of pressure at this instant.
Ans. 1,800lb/ft2 per min
5. If P pounds per square foot is the atmospheric pressure at a height of h
feet above sea level, then
Find the time rate of change of the atmospheric pressure outside an
airplane that is 5,000 ft high and rising at the rate of 160 ft/s?
Ans. -9.17lb/ft2 per second
6. A picture 40 cm high is placed on a wall with its based 30 cm above the
level of the eye of an observer. If the observer is approaching the wall at
the rate of 40 cm/s, how fast is the measure of the angle subtended at the
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MATH 401 – DIFFERENTIAL CALCULUS
observer’s eye by the picture changing when the observer is 1 m from the
wall?
Ans. 0.08 rad/s
7. Suppose we have two resistors with resistances
and
measured in
ohms(Ω) connected in parallel. The equivalent resistance, R, iis then given
by
Supposed
is increasing at a rate of 0.4Ω/min and
is decreasing at a
rate of 0.7Ω/min . At what rate is R changing when
and
?
Ans. -0.00 Ω/min
8. A man 6 feet tall walks at a rate of 5 feet per second away from the light
that is 15 feet above the ground. When he is 10 feet from the base of the
light,
(a) At what rate is the tip of his shadow moving?
Ans.
(b) At what rate is the length of his shadow moving?
Ans.
9. A particle is moving along a line according to the equation of motion
Where s meters is the directed distance of the particle from the origin at t
seconds. Find the t, s, and velocity v when the acceleration a = 0.
Ans.
10. A stone is dropped from a building 256 ft high. How long it takes the stone
to reach the ground? What is the velocity of the stone when it reaches the
ground?
Ans.
Problem Set No 4. Solve the following problems
1. Given y
1
, find df.
x 1
2. Given that f x 1 x 2 x
2
compute df and f when x 1 and
x 0.02 .
3. Find the actual and approximate volume of a spherical shell whose inner
radius is 3cm and whose thickness is 1/8cm.
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MATH 401 – DIFFERENTIAL CALCULUS
4. How much variation dr in the radius of a coin can be tolerated if the
volume of the coin is to be within 11000 of their ideal volume?
5. Given f x
1 4 2 3
x x 2 x 2 : (a) find the relative extrema of the
2
3
function using the second-derivative test; (b) determine the values of x at
which the relative extrema occur; (c) determine the intervals on which the
function is increasing and decreasing; (d) concave upward or concave
downward; (e) Locate inflection points if any and (f) Sketch the graph of
the function from your answers.
6. A piece of wire 10 ft long is cut into two piece. One piece is bent into the
shape of an equilateral triangle and the other piece is bent into a shape of a
square. How should a wire be cut so that (a) the combined area of the two
figures is as small as possible; (b) the combined area of the two figures is
as large as possible?
7. A water tank in the form of an inverted cone is being emptied at the rate of
6m3/min. The altitude of the cone is 24 m, and the radius is 12 m. Find
how fast the water level is lowering when the water is 10 m deep?
8. A right-circular cone is to be inscribed in a sphere of given radius. Find the
ratio of the altitude to the base radius of the cone of largest possible
volume.
9. Find two positive numbers such that their sum is 12 and the sum of their
squares is a minimum.
10. A ball is thrown vertically upward from the ground with an initial velocity
of 50 m/s. The equation of motion is given by
a. How high the ball go and how many seconds it takes for the ball to
reach its highest point?
b. Find the instantaneous velocity of the ball at 1 s and 3 s.
c. Find the instantaneous velocity of the ball when it reaches the
ground.
d. What is the acceleration of the ball at any time t?
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MATH 401 – DIFFERENTIAL CALCULUS
CHAPTER V
PARTIAL DIFFERENTIATION
The concept of a function of a single variable is extended to a multi –
valued functions. In this chapter, we shall generalize the notion of function of a
function expressed in two or more independent variable. For instance, the
function f defined by f x, y, z x 2 y 2 z 2 25 is a multi – valued function with
x, y and z as independent variables. To facilitate learning on multi – valued
functions, this chapter presents the definition of derivatives of multi – valued
function also known as partial differentiation. Also, chain rule of partial
differentiation will be discussed on this chapter. Definitions, theorems an some
examples and illustrations presented in this chapter are taken from [1] Larson, R.
(2018), [2] Leithold, L. (2002) and [3] Stewart, J. (2016)
At the end of this chapter, the learners might be able to:
1. Determine the partial derivatives of functions of two or more variables;
and
2. Evaluate the higher-order partial derivatives of functions of two or
more variables.
5.1.
DEFINITION OF PARTIAL DERIVATIVES OF A FUNCTION
The process of finding the partial derivatives of a function is called partial
differentiation. Let f be a function in terms of two variables say x and y written
as f x, y , then the partial derivatives of f are denoted by the following:
Notations
D1 f ,
f1 , f x ,
f
x
Meaning
D1 f read as “D sub 1 of f ” and f 1 read as “ f sub 1”
denote the partial derivative of f with respect to the
first variable. In the case of f x, y , that is with respect
to x
f x read as “ f sub x ” denotes the partial derivative of f
with respect to x .
f
is read as “partial derivative of f with respect to x ”
x
D2 f read as “D sub 2 of f ” and f 2 read as “ f sub ”
D2 f ,
f
f2 , f y ,
y
denote the partial derivative of f with respect to the
second variable. In the case of f x, y , that is with
respect to y
f y read as “ f sub y ” denotes the partial derivative of f
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MATH 401 – DIFFERENTIAL CALCULUS
with respect to y .
f
is read as “partial derivative of f with respect to y ”
y
These notations can be used interchangeably to denote partial derivatives of a
function.
Definition 5.1 gives the formal definition of the partial derivative of a multi–
valued function.
Definition 5.1. [2] Let f be a function of two variables, x and y . The partial
derivative of f with respect to x is that function, denoted by D1 f , such that its
function value at any point x, y in the domain of f is given by
D1 f x, y lim
x 0
f x x, y f x, y
x
if this limit exists. Similarly, the partial derivative of f with respect to y is that
function, denoted by D2 f , such that its function value at any point x, y in the
domain of f is given by
D2 f x, y lim
y 0
f x, y y f x, y
y
if this limit exists.
Example 5.1. Apply the definition of partial differentiation to find D1 f and D2 f if
f x, y 2 x 2 3xy y 2 .
Solution:
D1 f x, y f x lim
2x x
x 0
lim
x 0
2x
2
2
3x x y y 2 2 x 2 3xy y 2
. Simplifying,
x
2 xx x 3xy 3xy y 2 2 x 2 3xy y 2
x
2
2 x 2 4 xx 2x 3xy 3xy y 2 2 x 2 3xy y 2
x 0
x
2
lim
4 xx 2x 3xy
lim
x 0
x
2
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MATH 401 – DIFFERENTIAL CALCULUS
x4 x 2x 3 y
x 0
x
lim
lim 4 x 2x 3 y Applying the theorem on limits
x 0
4 x 20 3 y
D1 f x, y f x 4 x 3 y
2x
2
2x
lim
2
D2 f x, y f y lim
y 0
y 0
3x y y y y 2 x 2 3xy y 2
. Simplifying,
y
2
3xy 3xy y 2 2 yy y 2 x 2 3xy y 2
y
2
2 x 2 3xy 3xy y 2 2 yy y 2 x 2 3xy y 2
y 0
y
2
lim
3xy 2 yy y
y 0
y
2
lim
y 3x 2 y y
y 0
y
lim
lim 3x 2 y y Applying the theorem on limits
y 0
3x 2 y 0
D2 f x, y f y 3x 2 y
Example 5.2. Let f be a function defined by f x, y x 2 y , find
f
f
and
.
x
y
Solution:
By definition 5.1. we have
x x 2 y x 2 y
f
lim
x
0
x
x
lim
x x 2 y x 2 y
x
x 0
lim
x 0
x x 2 y x 2 y
x x x 2 y x 2 y
x x 2 y x 2 y
x x 2 y x 2 y
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MATH 401 – DIFFERENTIAL CALCULUS
lim
x x x 2 y x 2 y
x 0
x x 2 y x 2 y
x 0
1
1
lim
x
x 0 2y x 2y
1
2 x 2y
f
1
x 2 x 2 y
x 2 y y x 2 y
f
lim
y y 0
y
x 2 y y x 2 y
lim
y
y 0
x 2 y y x 2 y
x 2 y y x 2 y
x 2 y 2y x 2 y
y 0
y x 2 y y x 2 y
lim
lim
y 0
y
f
y
x 2 y y x 2 y
y 0
x 2 y y x 2 y
2
lim
2y
2
x 2 y 0 x 2 y
2
2 x 2y
1
x 2y
We shall now extend the concept of Definition 1.1 into a function of n variables.
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MATH 401 – DIFFERENTIAL CALCULUS
Definition 5.2. [2] Let Px1 , x2 ,...xn1 , xn be a point in R n , and let f be a function
of n variables, x1 , x2 ,...xn1 , xn . The partial derivative of f with respect to x k is
that function, denoted by Dk f , such that its function value at any point P in the
domain of f is given by
Dk f x1 , x2 ,...xn1 , xn lim
x 0
f x1 , x2 ,...xn1 , xn xk ,...xn f x1 , x2 ,...xn1 , xn
xk
if this limit exists.
For instance, if f is a function in three variables x, y and z , then the partial
derivatives of f are given by
f
f x x, y, z f x, y, z
lim
x x0
x
f
f x, y y, z f x, y, z
D2 f x, y, z
lim
y y 0
y
f
f x, y, z z f x, y, z
D3 f x, y, z
lim
z
0
z
z
D1 f x, y, z
Exercise 5.1. Find the indicated partial derivatives of the following
functions by applying the definition of partial derivatives.
1. f x, y
x y
x y
2. z x 2 y 2
3. f x, y, z x 2 y 2 z 2
4. f x, y
5. f x, y
1
x y
1
x y
f
f
and
y
x
z
z
and
y
x
f x , f y and f z
f
f
and
y
x
f 1 and f 2
References: [2] Leithold, L. (2002) [4] Peterson, T. (1964)
5.2.
PARTIAL DERIVATIVES BY FORMULAS OF DIFFERENTIATION
Theorems and formulas in finding the derivatives of a function presented
in Chapter 2 and 3 of this module can be used in finding the derivatives of multi –
valued functions. For instance, if we let f be a function defined by f x, y, z , if we
look for the partial derivative of f in terms of x , then we shall hold y and z
constant. Similarly, x and z will be treated constant if we look for the partial
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MATH 401 – DIFFERENTIAL CALCULUS
derivative of f in terms of y ; and finally x and y will be treated constant if we
look for the partial derivative of f in terms of z . For example, if f is a function
defined by f x, y, z 2 x 2 3 y 2 4 z 2 then the following are the partial derivatives
of f :
f
f x, y, z 2 x 2 3 y 2 4 z 2 By applying the formulas of
x x
x
x
x
Differentiation
f
f x, y, z 4 x 0 0
x x
f
f x, y, z 4 x
x x
f
f x, y, z 2 x 2 3 y 2 4 z 2 By applying the formulas of
y y
y
y
y
Differentiation
f
f x, y, z 0 6 y 0
y y
f
f x, y, z 6 y
y y
f
f x, y, z 2 x 2 3 y 2 4 z 2 By applying the formulas of
z z
z
z
z
f
f x, y, z 0 0 8 z
z z
f
f x, y, z 8 z
z z
Differentiation
Consider the following examples:
Example 5.3. Let f be a function defined by f x, y, z 2 x 2 y 3 y 2 z 4 z 2 x , find
a . f b . f and c . f .
y
x
z
Solution:
f x, y, z 2 x 2 y 3 y 2 z 4 z 2 x
f
(a.)
2 x 2 y 3 y 2 z 4 z 2 x
x x
x
x
f
2
2 y x 3 y 2 z 4 z 2 x
x
x
x
x
f
2 y2 x 0 4 z 2 1
x
f
4 xy 4 z 2
x
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MATH 401 – DIFFERENTIAL CALCULUS
(b.)
(c.)
f
y
f
y
f
y
f
y
f
z
f
z
f
z
f
z
2 x 2 y 3 y 2 z 4 z 2 x
y
y
y
2 x 2 y z 3 y 2 4 z 2 x
y
y
y
2 x 2 1 z 6 y 0
2 x 2 6 yz
2 x 2 y 3 y 2 z 4 z 2 x
z
z
z
2 x 2 y 3 y 2 z x 4 z 2
z
z
z
0 3 y 2 1 x8 z
3 y 2 8 xz
Example 5.4. Given f x, y, z x 2 y yz 2 z 3 , show that xf x yf y zf z 3 f x, y, z .
Solution:
To show that xf x yf y zf z 3 f x, y, z , we first find f x , f y and f z
f x, y, z x 2 y yz 2 z 3
f x x, y, z x 2 y yz 2 z 3
x
x
x
x
f x 2 xy
f y x, y, z x 2 y yz 2 z 3
y
y
y
y
f y x2 z 2
f z x, y, z x 2 y yz 2 z 3
z
z
z
z
2
f z 2 yz 3z
xf x yf y zf z 3 f x, y, z
x2 xy yx 2 z 2 z2 yz 3z 2 3x 2 y yz 2 z 3
2 x 2 y x 2 y yz 2 2 yz 2 3z 3 3x 2 y yz 2 z 3
3x 2 y 3 yz 2 3z 3 3x 2 y yz 2 z 3
3x 2 y yz 2 z 3 3x 2 y yz 2 z 3
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MATH 401 – DIFFERENTIAL CALCULUS
Example 5.5. Find f 1 and f 2 if f x, y e x sin y ln xy .
Solution:
f
and
x
f
f 2 means partial derivative of f in terms of y or
y
f1
f x, y e x sin y ln xy
x
x
x
f1
f x, y sin y e x ln xy
x
x
x
y
f1
f x, y sin ye x
x
xy
f
1
f1
e x sin y
x
x
f 1 means partial derivative of f in terms of x or
y
f2
y
f2
y
f2
y
f2
x
e sin y lnxy
y
y
sin y ln xy
f x, y e x
y
y
x
f x, y e x sin y
xy
1
f x, y e x sin y
y
f x, y
Example 5.6. Let z coshx 2 y sinh2 x y 2 , find z x and z y .
Solution:
coshx 2 y sinh2 x y 2
x
x
z x sinh x 2 y x 2 y cosh2 x y 2 2 x y 2
x
x
2
z x sinhx 2 y 1 cosh2 x y 2
zx
z x sinhx 2 y 2 cosh2 x y 2
coshx 2 y sinh2 x y 2
y
y
z y sinh x 2 y x 2 y cosh2 x y 2 2 x y 2
y
y
2
z y sinhx 2 y 2 cosh2 x y 2 y
zy
z y 2 sinhx 2 y 2 y cosh2 x y 2
227
MATH 401 – DIFFERENTIAL CALCULUS
Example 5.7. If f x, y ln x 2 y xy 2 , find f x 3,2 and f y 3,2 .
Solution:
For f x 3,2 , we first find f x .
ln x 2 y xy 2
x
2 xy y 2
. Substitute x 3 and y 2
fx 2
x y xy 2
fx
f x 3,2
232 2
2
3 2 32
f x 3,2
2
2
12 4 8 4
18 12 6 3
4
3
For f y 3,2 , we first find f y .
ln x 2 y xy 2
y
x 2 2 xy
. Substitute x 3 and y 2
fy 2
x y xy 2
fy
2
3 232
f y 3,2 2
3 2 322
2
f y 3,2
9 12
4
2
18 12
6
3
3
4
4
Example 5.8. Find Fx 0, and Fy 0, if F x, y e x tanx y .
Solution:
tanx y tanx y e x
x
x
x
2
x
Fx e sec x y e tanx y
Fx e x
Fx 0, e 0 sec 2 0 e 0 tan 0
4
4
4
Fx 0, 1sec 2 1 tan
4
4
4
2
Fx 0, 2 1
4
Fx 0, 1
4
tanx y tanx y e x
y
y
x
2
Fy e sec x y tanx y 0
Fy e x
Fy e x sec 2 x y
228
MATH 401 – DIFFERENTIAL CALCULUS
Fy 0, e 0 sec 2 0
4
4
2
Fy 0, 1 2
4
Fy 0, 2
4
Example 5.9. Find the slopes of the surface given by f x, y 1 x 12 y 12 at
the point 1,2 in the x direction and y direction.
Solution:
We first find the partial derivatives f x and f y . That is
f x, y 1 x 1 y 1
2
2
f x 2x 1
at the point 1,2 ; f x 1,2 21 1 0
Therefore, the slope of the surface at the point 1,2 on the x
direction is 0.
f x, y 1 x 1 y 1
f y 2 y 1
2
2
at the point 1,2 ; f x 1,2 22 1 2
Therefore, the slope of the surface at the point 1,2 on the y
direction is 2 .
Exercise 5.2. Find the first partial derivatives of the following.
1. f x, y
x y
x y
2. z x 2 y 2
3. f x, y, z x 2 y 2 z 2
4. f x, y
5. f x, y
1
x y
1
x y
6. z ln x x 2 y 2
7. f x, y Sin 1 1 x 2 y 2
8. If z
z
z
x2 y2
, show that x y 3z .
x
y
x y
229
MATH 401 – DIFFERENTIAL CALCULUS
e x y
u u
u.
, show that
x
y
x y
e e
10. If u x 2 y y 2 z z 2 x , show that u x u y u z x y z 2 .
9. If u
5.3.
HIGHER – ORDER PARTIAL DERIVATIVES
Since the partial derivatives of a function f x, y are expressed in terms of x and
y , they can still be differentiated with respect to the same variables. For instance
f
f
or
can still be differentiated with respect to x and
y
x
f
f
y i.e.)
is the second partial derivative of f in terms of x ;
is the
x x
y x
f
partial derivative of
with respect to y , so that is the second partial
x
derivatives of f .
Let f be a multi – valued function defined by f x, y, z , the following are
the first partial derivatives of f :
f
f
f
; fy
and f z
fx
y
x
z
The second partial derivatives of f are the partial derivatives of f x , f y and f z .
the partial derivatives
So we have
f 2 f
f xx
x x x 2
f 2 f
f xy
y x yx
f 2 f
f xz
z x zx
f 2 f
f yx
x y x y
f 2 f
f yy
y y y 2
f 2 f
f yz
z y z y
f 2 f
f zx
x z x z
f 2 f
f zy
y z y z
f 2 f
f zz
z z z 2
Example 5.10. Find the second partial derivatives of f in Example 5.3.
Solution: f in Example 5.3 is defined by
f
4 xy 4 z 2 ;
x
f
2 x 2 6 yz
y
and
f x, y, z 2 x 2 y 3 y 2 z 4 z 2 x and
f
3 y 2 8 xz . So the second partial
z
derivatives of f are:
f 2 f
f xx 4 y
x x x 2
f 2 f
f xy 4 x
y x yx
f 2 f
f xz 8 z
z x zx
f 2 f
f yx 4 x
x y x y
f 2 f
f 2 f
2 f yy 6 z
f yz 6 y
y y y
z y z y
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MATH 401 – DIFFERENTIAL CALCULUS
f 2 f
f 2 f
f 2 f
f zx 8 z
f zy 6 y
f zz 8 x
x z x z
z z z 2
y z y z
Example 5.11. Find f12 , f 21 , f11 and f 22 if f x, y e x sin y ln xy .
Solution: From Example 5.5, we see that
1
f
1
f x, y e x sin y
e x sin y and f 2
y
y
x
x
2
2
f f
f f
and f 21
. Then we have
f12
y x yx
x y xy
f1
x
1
e sin y e x cos y
y
x
1
f 21 e x sin y e x sin y
x
y
f12
f11
f 2 f
f 2 f
and
f
22
x x x 2
y y y 2
x
1
1
e sin y e x sin y 2
x
x
x
1
1
e x sin y e x cos y 2
y
y
y
f11
f 22
Example 5.12. If z ln x 2 y 2 , show that
2z 2z
0.
x 2 y 2
Solution:
If z ln x 2 y 2 , then the first partial derivatives of z are
z
2y
z
2x
2
2
and
.
2
y x y 2
x x y
Differentiating these with respect to x and y respectively, we get the
second partial derivatives as follows:
2 z x 2 y 2 2 2 x2 x 2 x 2 2 y 2 4 x 2 2 y 2 2 x 2
x 2
x 2 y 2 2
x 2 y 2 2
x 2 y 2 2
2 z x 2 y 2 2 2 y2 y 2 x 2 2 y 2 4 y 2 2 x 2 2 y 2
y 2
x 2 y 2 2
x 2 y 2 2
x 2 y 2 2
2z 2z
0
x 2 y 2
2 y 2 2x 2 2x 2 2 y 2
0
x 2 y 2 2 x 2 y 2 2
00
We now show that
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MATH 401 – DIFFERENTIAL CALCULUS
2z
2z
y
Example 5.13. If z Sin 1 , show that
.
xy
x
yx
Solution:
y
If z Sin 1 , then the first partial derivatives of z are
x
y
y
2
2
z
x
x
2
2
x
y
y
1 2
1
x
x
1
1
z
x
x
2
2
y
y
y
1 2
1
x
x
y
x2
x2 y2
x2
y
y
x2
x2 y2
x x2 y2
x
(1)
1
x
2
x y2
x2
1
x
2
x y2
x
(2)
1
x2 y2
Differentiating (1) with respect to y and (2) with respect to x , we get the
second partial derivatives of z .
2z
yx
2z
yx
2z
yx
x
1
1
x 2 y 2 1 y x x 2 y 2 2 2 y
2
x
xy 2
x x2 y2
x
x y
2
2
2
2
xx 2 y 2 xy 2
x
x2 y2
x2 y2
2
z
2x
xy
2 x2 y2
xx 2 y 2 y 2
x2
x2 y2
3
(3)
3
2
x2 y2
x2 y2
x
x2 y2
3
x
x2 y2
3
It can be seen that (3) and (4) are equal, therefore
(4)
2z
2z
.
xy yx
Exercise 5.3. Find all the second partial derivatives of the following
functions.
1. z 3xy 2
2. f x, y x 2 2 xy 3 y 2
3. z x 2 y 2
4. f x, y, z e x tan y cos z
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MATH 401 – DIFFERENTIAL CALCULUS
5. f x, y 2 xe y 3 ye x
x y
2z
2z
, show that
.
x y
xy yx
2z
2z
7. If z ln x x 2 y 2 , show that
.
xy yx
2z 2z
x
y
z
e
sin
y
e
sin
x
8. If
, show that 2 2 0 .
x
y
2
2
z z
y
9. If z Tan 1 , show that 2 2 0 .
x
y
x
3
u
3u
3
10. If u x 2 y 2 , verify that 2
.
x y yx 2
6. If z
References: [2] Leithold, L. (2002) [4] Peterson, T. (1964)
5.4.
TOTAL DERIVATIVES
Let z be a function in terms of x and y , z f x, y , and let x and y be
the arbitrary increments of x and y respectively, then we can write an
increment in z as
(1)
z f x x, y y f x, y
Also, we can find the total derivatives of z by applying the following definition.
Definition 5.3. [1] If z f x, y and x and y are the increments of x and y , then
the differentials of the independent variables x and y are
dx x and dy y
and the total derivatives of the independent variable z is given by
dz
z
z
dx dy f x x, y dx f y x, y dy .
x
y
(2)
This definition can be extended into functions of more than two variables, say
w f x, y, z . The total derivative of w , denoted by dw is given by
dw
w
w
w
dx
dy
dz
x
y
z
(3)
Example 5.14. Find the total derivative of z x 2 e 2 y ye x .
Solution:
Applying Definition 5.3, we have dz
dz 2 xe 2 y ye x dx 2 x 2 e 2 y e x dy
z
z
dx dy
x
y
233
MATH 401 – DIFFERENTIAL CALCULUS
Example 5.15. Find the total derivative of w 3x 2 y 2 2 z 2 .
Solution:
Following (2), the total derivative of w is dw
w
w
w
dx
dy
dz , so we
x
y
z
have dw 6 xdx 2 ydy 4 zdz
Definition 5.3 can be extended into a case where x and y are continuous
functions express in terms of the third variable, say t ; that is
x f t and y g t
(4)
If the values of (4) is substituted to z f x, y , z becomes a function expressed
in terms of t . To find the total derivative of z in terms of t , we shall only divide
(2) by the differential of t , that is dt , thus
dz z dx z dy
dt x dt y dt
(5)
Similarly if x, y and z in (3) are continuous functions in terms of t , then the total
derivative of w is
dw w dx w dy w dz
dt
x dt y dt z dt
Example 5.16. If z x ln y ; x 2 u 2 and y e u find
(6)
dz
.
du
Solution:
dz z dx z dy
z x dx
z
dy
. We have
;
ln y ,
2u and
eu
du x du y du
y y du
x
du
x
dz
ln y 2u e u
du
y
By (5)
2 u2
dz
ln e u 2u u
du
e
dz
2u 2 2 u 2
du
dz
3u 2 2
du
u
e
Simplifying
234
MATH 401 – DIFFERENTIAL CALCULUS
Example 5.17. [4] The height of a right circular cylinder is 50 inches and
decreases at the rate of 4 inches per second, while the radius of the bases is 20
inches and increases at the rate of 1 inch per second. At what rate is the volume
changing?
Solution: Let V r 2 h where r is the radius and h is the height. Differentiating V
partially in terms of the time t , we have
dV
dr
dh
2rh r 2
dt
dt
dt
dh
dr
Since h 50 , r 20 ;
4 and
1 , then
dt
dt
dV
2
220501 20 4
dt
dV
400 .
dt
Thus, the volume is increasing at a rate of 400 cubic inches per second.
Example 5.18. Find
du
if u Tan 1 xy ; x sin t and y sec t
dt
Solution:
du u dx u dy
dt x dt y dt
du
y
x
cos t
sec t tan t
2 2
2 2
dt 1 x y
1 x y
du
sec t
sin t
cos t
sec t tan t
2
2
2
2
dt 1 sin t sec t
1 sin t sec t
du
1
sin t
sec t tan t
2
dt 1 tan t 1 tan 2 t
du
1
sin t sin t
2
dt sec t sec t cos t
du
cos 2 t sin 2 t
dt
du
1
dt
2
Example 5.19. Find the total derivative of z if z x 2 2 xy y 2 ; x t 1 and
y t 1
2
Solution:
dz z dx z dy
dt x dt y dt
235
MATH 401 – DIFFERENTIAL CALCULUS
dz
2 x 2 y 2t 1 2 x 2 y 2t 1
dt
dz
2
2
22 x 2 y t 1 t 1 4 t 1 t 1 2
dt
dz
8t 2 2t 1 t 2 2t 1 84t
dt
dz
32t
dt
Exercise 5.4.
du
in each of the following.
dt
y
1. u Tan 1
x ln t
x
2. u xy xz yz x t cos t
xt
3. u
x ln t
yt
du
Find
in each of the following
dx
x y
y x 2
4. u
1 y
Find
5. u ln x 2 y 2 z 2
y et
y t sin t
1
y ln
t
y x sin x
z t
z x cos x
References: [2] Leithold, L. (2002) [4] Peterson, T. (1964)
5.5.
CHAIN RULE OF PARTIAL DIFFERENTIATIONS
Suppose x and y in (4) are expressed in two variables, say r and t , i.e.
x f r , t and y g r , t , then we can now express z as a function of r and t or
z f r , t . By applying the chain rule of partial differentiation we get
z z x z y
t x t y t
z z x z y
r x r y r
Similarly if x, y and z in (3) are continuous functions in terms of r and t , then
the partial derivative of w is given by
w w x w y w z
t
x t y t z t
236
MATH 401 – DIFFERENTIAL CALCULUS
w w x w y w z
r x r y r z r
Definition 5.4. [2] Suppose that u is a differentiable function of the n variables
x1 , x2 ,...xn1 , xn and each of these variables is in turn a function of the m
variables y1 , y 2 ,... y m1 , y m . Suppose further that each of the partial derivatives
x i
where i 1,2,3,..., n 1, n and j 1,2,3,..., m 1, m ; exists. Then u is a function
x j
of y1 , y 2 ,... y m1 , y m and
u
u x1 u x2
u xn
...
y1 x1 y1 x2 y1
xn y1
u
u x1 u x 2
u x n
...
y 2 x1 y 2 x 2 y 2
x n y 2
u
u x1
u x 2
u x n
...
y m x1 y m x 2 y m
x n y m
Example 5.20. Let u x 2 yz ; x r sin t ; y r cos t and z r sin 2 t find
u
u
and
.
r
t
Solution: By Definition 5.4, we have
u u x u y u z
u u x u y u z
and
r x r y r z r
t x t y t z t
u
r
u
r
u
r
u
r
2 x sin t z cos t y sin 2 t
2r sin t sin t r sin 2 t cos t r cos t sin 2 t
2r sin 2 t 2r sin 2 t cos t
2r sin 2 t 1 cos t
u
2 x r cos t z r sin t y 2r sin t cos t
t
u
2r sin t r cos t r sin 2 t r sin t r cos t 2r sin t cos t
t
u
2r 2 sin t cos t r 2 sin 3 t 2r 2 sin t cos 2 t
t
237
MATH 401 – DIFFERENTIAL CALCULUS
u
r 2 sin t 2 cos t sin 2 t 2 cos 2 t
t
y
x
Example 5.21. Given u e ; x 2r cos t and y 4r sin t , find
u
u
and
.
r
t
Solution:
u u x u y
r x r y r
1 xy
u y xy
e 2 cos t e 4 sin t
r x 2
x
sin t
4 r sin t
1
u
4r sin t 24rr cos
t 2 cos t
2 r cos t
2
e
e
2r cos t
r 4r cos 2 t
u
2 sin t 2 tan t 2 sin t 2 tan t
e
e
r
r cos t
r cos t
u
0
r
4 sin t
u u x u y
t x t y t
1 y
u y xy
2 e 2r sin t e x 4r cos t
t x
x
sin t
4 r sin t
1
u
4r sin t 24rr cos
t 2r sin t
2 r cos t
2
e
e
2r cos t
t 4r cos 2 t
u 2 sin 2 t 2 tan t
e
2e 2 tan t
2
t
cos t
u
2 tan 2 te 2 tan t 2e 2 tan t
t
u
2e 2 tan t tan 2 t 1
t
4r cos t
Example 5.22. Let u r 2 s 2 ; r x y and s x y , find
Solution:
u
x
u
x
u
x
u
x
u r u s
r x s x
2r 1 2s 1
2r s
2x y x y
u
y
u
y
u
y
u
y
u
u
and
.
y
x
u r u s
r y s y
2r 1 2s 1
2r s
2x y x y
238
MATH 401 – DIFFERENTIAL CALCULUS
u
4x
x
u
4y
x
Implicit Partial Derivatives
If z f x, y and y is a function of x , it follows from Section 5.4 that
dz f f dy
dx x y dx
For z 0 , identically, we have
0
Solving for
dz
0 ; hence
dx
f f dy
x y dx
dy
, we have
dx
f
dy
f
x where
0.
f
dx
y
y
Example 5.23. If x sin y y cos x 3 , find
dy
.
dx
Solution:
If we first let f x, y x sin y y cos x 3 0 . SO we have
f
dy
sin y y sin x
x
f
dx
x cos cos x
y
If
z is defined as an implicit function of x and y by the equation
F x, y, z 0 , then
Fy
F
F
z
y
z
x
and
x where Fz 0
x
Fy
x
Fz
y
Fz
Example 5.24. If x 3 y 3 z 3 3xyz 2 , find
z z
y
,
and
.
x y
x
Solution:
We first write x 3 y 3 z 3 3xyz 2 as F x, y, z x 3 y 3 z 3 3xyz 2 0
239
MATH 401 – DIFFERENTIAL CALCULUS
F
z
3x 2 3 yz
x 2 yz
x 2
2
x
Fz
3z 3xy
z xy
Fy
z
3 y 2 3xz
y 2 xz
2
2
y
Fz
3z 3xy
z xy
F
y
3x 2 3 yz
x 2 yz
x 2
2
x
Fy
3 y 3xz
y xz
Example 5.25. Let x y y x 1 , find
y
.
x
Solution:
F
y
x
x
Fy
We first take the logarithm of both side of the equation.
Applying the properties of logarithmic function
ln x y y x ln 1
ln x y ln y x 0
y ln x x ln y 0
We shall now let F y ln x x ln y 0
y
y x ln y
ln y
F
y
y y x ln y
x
x x
x
y ln x x
x
Fy
x y ln x x
ln x
y
y
Exercise 5.5. Find the indicated partial derivatives of the following
functions by applying the chain rule of partial differentiation.
p x2;
1. u p qr ;
2. w xyz ;
xst;
3. w x cos yz ; x s 2 ;
4. w x 2 y 2 z 2 ;
y
x
q xy ;
r y2 ;
y s t ;
z st 2 ;
y t2;
z s 2t ;
x t sin r ; y t cos r ; z rt 2
5. u Tan 1 ;
x r cos ; y r sin ;
6. ye 2 x xe 2 y 0 ;
dy
dx
u
x
w
s
w
s
w
r
u
r
and
and
and
and
and
u
y
w
t
w
t
w
t
u
z z
y
,
and
x y
x
z z
y
8. ln x 2 y 2 e z 1; ,
and
x y
x
7. x sin y y cos z 3z sin x 0 ;
240
MATH 401 – DIFFERENTIAL CALCULUS
z z
y
,
and
x y
x
y
10. 4 x 2 2 xy y 2 1 ;
x
9.
x y 1
;
x y z
References: [2] Leithold, L. (2002) [4] Peterson, T. (1964)
CHAPTER TEST
Find the indicated partial derivatives of the following functions by applying the
definition of partial derivatives.
1. f x, y x 2 2 xy 3 y 2
2. z x y
3. f x, y, z x 2 y 2 z 2 2 xz
f
f
and
x
y
z
z
and
y
x
f x , f y and f z
Find the first partial derivatives of the following by applying the formulas of
differentiation.
4. f x, y x 2 2 xy 3 y 2
5. f x, y xe y ye 2 x
6. z
1
ln x y 2
2
x
y
z
x
y
z
7. If u e e e , show that xu x yu y zu z 0 .
8. If z e y cosx y , show that z x z y z 0
9. If z e y cos x e x cos y , find
x
y
10. If z Sin 1 , find
11. If u x 3 y 3 , find
2
2z
2z
and
y 2
x 2
2z
2z
and
.
y 2
x 2
3u
3u
and
x 2 y
yx 2
Find the indicated derivatives
12. ex e y e z e x y z ;
z z
y
,
and
x y
x
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MATH 401 – DIFFERENTIAL CALCULUS
13. z yTan 1 xz ;
z z
y
,
and
x y
x
x
y
14. If u Tan 1 ; x e t ; y ln t find
du
.
dt
15. If u xy xz yz ; x t sin t ; y t cos t and z t , find
du
.
dt
16. If w x 3 y 3 z 3 ; x t sin r ; y t cos r ; and z rt 2 , find
17. If w x 2 y 2 z 2 ;
x s t ; y s t ; and z st ; find
w
w
and
.
r
t
w
w
and
.
s
t
18. Find
x y
du
if u
and y 2 x 2
1 y
dx
19. Find
du
if u ln x 2 y 2 z 2 ; y x sin x ; and z x cos x .
dx
20. Find
du
if u ln x 2 y 2 z 2 ; y x sin x ; and z x cos x .
dz
References: [2] Leithold, L. (2002) [4] Peterson, T. (1964)
242
MATH 401 – DIFFERENTIAL CALCULUS
References
[1] Larson, R. (2018), Calculus. 11th Edition. Cengage Learning Asia Pte.
Ltd.
[2] Leithold, Louis (2002). The Calculus 7. Pearson Education Asia Pte. Ltd
[3] Stewart, James. (2016). Calculus: Early Transcendentals. 8th Edition.
Cengage Learning
[4] http://www.copingwithcalculus.com/DifferentiationApplications.html
[5] https://themathpage.com/aCalc/applied.htm
[6] https://mathalino.com/reviewer/differential-calculus
[7] https://www.intmath.com/differentiation-transcendental/4applications-derivatives-trigonometric.php
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