Study Guides in Mathematics
School of Mathematics
University of the Witwatersrand
CALCULUS
First Semester Study Guide
for
MATH1036A
MATH1036A
This book
belongs to
NAME
STUDENT NUMBER
Contents
1
Functions and limits
1
1.1 The idea of a limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Infinite limits, limits at infinity and asymptotes . . . . . . . . . . . . . . . . . . . . . . 10
1.4 Limits through One-sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.5 The Sandwich Theorem & Trigonometric Limits . . . . . . . . . . . . . . . . . . . . . 11
2
Continuity of Functions
2.1 Continuity at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Left and Right Continuity at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Continuity on an Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Theorems on Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
15
17
17
18
20
3
Differentiation
3.1 First Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Non-vertical tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Vertical tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Differentiability vs. Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 The derivative as a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Derivative Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Trigonometric Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . .
3.9 Linear Approximation and Differentials . . . . . . . . . . . . . . . . . . . . . . . . . .
23
23
25
25
25
26
27
27
32
35
37
39
4
Introduction to Integration
4.1 Antidifferentiation and the Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . .
4.2 The Definite Integral & FTC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
41
45
46
48
5
Differentiation Techniques
5.1 Higher Order Derivatives and Taylor Polynomials . . . . . . . . . . . . . . . . . . . . .
5.2 Continuity and Differentiation of Inverse Functions . . . . . . . . . . . . . . . . . . . .
5.3 Inverse Trigonometric Derivatives and Integrals . . . . . . . . . . . . . . . . . . . . . .
5.4 Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
49
50
51
54
i
1st Semester Calculus Study Guide
ii
5.5
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Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
6
Applications of Differentiation
6.1 Indeterminate Forms and l’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Extrema and the Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Rolle’s Theorem and the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . .
6.5 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5.1 Increasing and Decreasing Functions and The First Derivative Test . . . . . . . .
6.5.2 Concavity and The Second Derivative Test . . . . . . . . . . . . . . . . . . . .
6.5.3 A summary of Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
58
60
62
65
68
68
71
75
76
7
Hyperbolic Functions
79
Acknowledgements
This study guide has been developed by many lecturers from the calculus teams over many years. Particular thanks goes to Prof. Inderasan Naidoo who typeset this guidebook for the first time.
Chapter 1
Functions and limits
LEARNING OUTCOMES:
On completion of this chapter you should (tick the checkbox when you have mastered the skill)
□ 1. understand the concept of a limit of a function at a point intuitively,
□ 2. be able to estimate the value of a limit of a function at a given point by calculating function
values close to the given point,
□ 3. be able to evaluate the limit of a function at a given point graphically,
□ 4. understand and be able to calculate one-sided limits,
□ 5. know the behaviour of a function with infinite limits,
□ 6. understand and be able to determine vertical and horizontal asymptotes of a function,
□ 7. be able to compute limits of exponential functions,
□ 8. be able to use the appropriate Limit Laws to find the limit of a function at a given point,
□ 9. know how to prove that the limit of a function at a given point exists using one-sided limits,
□ 10. be able to use factorization and rationalization to calculate limits of rational functions,
□ 11. be able to use the Squeeze Theorem to calculate limits,
□ 12. know, understand and be able to prove the theorems on trigonometric limits given in these notes
on this section and
□ 13. know how to solve the examinable tutorial problems and the worked out problems from these
notes on this chapter.
The single most important concept in all of calculus is that of a limit. Every single notion of calculus is
encapsulated in one sense or another to that of a limit. In this course, we will only consider an intuitive
and informal discussion of a limit of a function at a point. The precise approach to limits, using the
standard ε-δ definition is reserved for the second year Analysis course and will not be considered here.
1.1
The idea of a limit
Consider a function f defined near the point x = a, not necessarily at a itself. We wish to investigate
the behaviour of f for values x in the domain of f that are close to a. We first consider this numerically,
then graphically.
1
1st Semester Calculus Study Guide
2
MATH1036
x−9
Worked Example 1.1.1. Consider the function f (x) = √
. What happens to the values of f (range
x−3
values of f ) when x (domain values of f ) is close to 9?
Solution. A table of values will conjecture the behaviour of f . We want to approach 9 and can do this in
a direction closely from the left (x < 9, i. e. from values below 9) and a direction closely from the right
(x > 9, i. e. from values above 9).
x<9
x−9
f (x) = √
x−3
8
8.5
8.9
8.95
8.99
8.999
8.9999
8.99999
It appears that as x approaches 9 from below (the left), the closer f (x) is to
, the limiting value of f .
To say “x approaches 9 from below” or “x approaches 9 from the left” we write symbolically x → 9− .
We then write lim− f (x) (called the left-hand limit) for the value that f (x) approaches as x → 9− . In the
x→9
above calculations we thus have
lim f (x) =
x→9−
.
Approaching 9 from the right gives the following table of values.
x>9
x−9
f (x) = √
x−3
9.1
9.05
9.001
9.0001
9.00001
Again, it can be observed that the closer that x is to 9 from above, the closer f (x) is to
. We write
x → 9+ to denote “x approaches 9 from above” or “x approaches 9 from the right”. We then write
lim+ f (x) (called the right-hand limit) for the value that f (x) approaches as x → 9+ . In our calculations
x→9
lim f (x) =
x→9+
.
1st Semester Calculus Study Guide
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3
lim f (x) and lim+ f (x) are one-sided limits. We have
x→9−
x→9
lim f (x) =
= lim+ f (x)
x→9−
x→9
and we phrase this combined result as “the limit of f (x) as x approaches 9 exists and is equal to
and write
lim f (x) =
.
”
x→9
Both the left-hand limit lim− f (x) and the right-hand limit lim+ f (x) of a function must exist and
x→a
x→a
must both be equal for the limit of the function lim f (x) to exist i. e.,
x→a
lim f (x) = L ⇔ lim− f (x) = L = lim+ f (x).
x→a
x→a
x→a
Graphically: The one-sided limits can be determined easily from the graph. Consider the graph of
x−9
f (x) = √
(sketch the graph in the space provided below).
x−3
It is clear that x = 9 is inadmissible since f (9) does not exist. However lim− f (x) =
both exist, easily “read off” from the graph of f so that lim f (x) =
x→9
x→9
= lim+ f (x)
x→9
.
Worked Example 1.1.2. (The number e) Let
1
f (x) = (1 + x) x .
Use the tabular method to estimate lim f (x) (if it exists) to five decimal places.
x→0
Hint: Solve the limit from the left at x = 0 and also from the right at x = 0.
Note.
1. The function y = e x is called the natural exponential function.
2. Recall that the exponential functions given as y = a x with a > 0, a , 1, have the inverse y =
loga x. Particularly, the logarithmic function to the base e, loge x, is called the natural logarithmic
function y = loge x which we denote in future as y = ln x (logarithmus naturalis), i. e.,
ln x = loge x.
1st Semester Calculus Study Guide
4
MATH1036
3. y = ln x has all the logarithmic properties:
(a) y = ln x ⇔ x = ey ,
(b) eln x = x and ln e x = x,
(c) ln xr = r ln x,
loge x ln x
=
,
(d) loga x =
loge a ln a
(e) ln xy = ln x + ln y,
Å ã
x
(f) ln
= ln x − ln y,
y
(g) ln 1 = 0 and ln e = 1.
You will encounter y = ln x in particular in the chapters on integration. Note the properties above
remembering that ln x = loge x.
Note: If f does not approach a unique value as x approaches a, then lim f (x) does not exist, i. e.,
x→a
lim f (x) , lim+ f (x) ⇒ lim f (x) DOES NOT EXIST.
x→a−
x→a
x→a
It is obvious that it is tedious to determine lim− f (x), lim+ f (x) and eventually the existence of lim f (x)
x→a
x→a
x→a
using the tabular approach. More importantly, the tabular approach does not prove the existence
of the limit nor does it give the exact value of the limit. The graphical way (if we know how to
sketch the graph) is more succinct. A computer algebra program (such as GEOGEBRA, which is freely
downloadable) may assist you in graphical sketching.
Worked Example 1.1.3. Sketch the graph of the given function and determine graphically whether the
corresponding limit exists; if so, find the limit.
√

 −x

1. Let g(x) = 3 − x



(x − 3)2
if x < 0,
if 0 ≤ x < 3,
if x > 3.
Find lim g(x) and lim g(x) if they exist.
x→0
2. h(x) =
x→3
|x|
. Find lim h(x) if it exists.
x→0
x
3. Consider the greatest integer function (also called the “floor” function) defined as
⌊x⌋ = greatest integer less than or equal to x.
Find lim ⌊x⌋ if it exists.
x→4
The graphs of the remaining problems have been sketched with a drawing program. We evaluate the
limits of the given functions below accordingly using the graphs. All trigonometric functions are defined
in radian measure (refer to the Algebra Lecture Guide).
1st Semester Calculus Study Guide
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5
Å ã
1
. Find lim f (x). (In this example, using the tabular
Worked Example 1.1.4. 1. Let f (x) = sin
x→0
x
method to conjecture the value of the limit is misleading.)
1
0.5
−1
−0.5
0
0
1
0.5
−0.5
−1
Hint: The function f (x) is fluctuating between 1 and −1 when x → 0 .
2. Let g(x) =
sin x
. Find lim g(x).
x→0
x
1
0.8
0.6
0.4
0.2
−10
−5
5
−0.2
10
1st Semester Calculus Study Guide
6
3. Let h(x) =
MATH1036
1 − cos2 x
. Find lim h(x).
x→0
x
0.8
0.6
0.4
0.2
−10
−5
10
5
−0.2
−0.4
−0.6
−0.8
Tutorial 1.1.1. 1. [C1] §2.2, pp. 156–158: 46–49, 55–58, 62–64, 68–70.
2. If f is a function such that f (2) = 4, can you conclude anything about lim f (x)? Explain your answer
x→2
and provide a graph for f to substantiate.
3. If g is a function such that lim g(x) = 4, can you conclude anything about g(2)? Explain your answer
x→2
and provide a graph for g to substantiate.
4. Let f (x) = x − ⌊x⌋. For each integer n, find lim− f (x) and lim+ f (x) if they exist.
x→n
x→n
Note: In the above examples we showed that either a limit of a function f exists at x = a ( f (x) approaches a unique value L ∈ R) or it does not exist ( f (x) does not approach a unique value).
Infinite Limits
The limit of a function f can also be unbounded, i. e., as x → a+ or x → a− , the values of f (x)
become positively large or negatively large. We use the infinity symbol ∞ to denote unboundedness.
+∞, or shortly ∞, will denote positively large values and −∞ will denote negatively large values. ∞ is
a symbolic representation of having no bound/limit and is not a real number. You cannot algebraically
operate with ∞ as if it were an element of the reals.
The following infinite limit cases arise
1. lim− f (x) = ∞ = lim+ f (x) ⇒ lim f (x) = ∞ means that f (x) is arbitrarily positively large for
x→a
x→a
x→a
x sufficiently close to a. lim f (x) in fact does not exist here but the infinity notation describes the
x→a
behaviour of f close to a.
2. lim− f (x) = −∞ = lim+ f (x) ⇒ lim f (x) = −∞ means that f (x) is arbitrarily negatively large
x→a
x→a
x→a
for x sufficiently close to a. lim f (x) in fact does not exist here but the infinity notation describes the
x→a
behaviour of f close to a.
1st Semester Calculus Study Guide
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7
3. lim− f (x) = ∞ and lim+ f (x) = −∞ (or vice versa). lim f (x) in fact does not exist here but the infinity
x→a
x→a
x→a
notation describes the behaviour of f close to a− and a+ .
The following examples illustrate infinite (unbounded) limits graphically. In such cases lim f (x) DOES
x→a
NOT EXIST.
1
Worked Example 1.1.5. 1. Let f (x) = . Find lim− f (x), lim+ f (x), lim f (x).
x→0
x→0
x→0
x
1
2. Let g(x) = − 2 . Find lim− g(x), lim+ g(x), lim g(x).
x→0
x→0
x→0
x
3. Find limπ − sec x and limπ + sec x.
x→ 2
x→ 2
Vertical asymptotes arise when limits are infinite. These are auxiliary vertical lines that the function
values approach (but never attain). We say that the line x = a is a vertical asymptote of f if at least one
of the following limits hold
lim f (x) = ∞ or
x→a−
lim f (x) = ∞ or
x→a+
(or
lim f (x) = ∞
x→a
lim f (x) = −∞
x→a−
or
or
lim f (x) = −∞
x→a+
lim f (x) = −∞).
x→a
The previous worked example illustrates the concept of a vertical asymptote.
Limits at infinity
Horizontal asymptotes arise when we consider the limit of a function f for unbounded values of x, i. e.,
lim f (x) and lim f (x). If, for arbitrary positively or negatively large values of x, the function values
x→∞
x→−∞
approach a unique value b ∈ R, i. e., if lim f (x) = b or lim f (x) = b, then the line y = b is called a
x→∞
x→−∞
horizontal asymptote.
Horizontal and vertical asymptotes are useful in sketching graphs, as you will discover in Chapter 6 of
this guide.
1
Worked Example 1.1.6. 1. Let f (x) = 2 + . Find lim f (x) and lim f (x) from the graph of f .
x→∞
x→−∞
x
1
2. Let g(x) = 2 . Find lim g(x) and lim g(x) from the graph of g.
x→∞
x→−∞
x
2x2
3. Let h(x) =
. From the graph evaluate lim− h(x), lim+ h(x), lim − h(x), lim + h(x), lim h(x) and
x→∞
x→3
x→−3
x→3
x→−3
9 − x2
lim h(x). What are the horizontal and vertical asymptotes (if any)?
x→−∞
f (x)
, we may possibly argue with the behaviour of the numerator and denomg(x)
f (x)
inator near a to calculate lim
.
x→a g(x)
For functions of the form
Worked Example 1.1.7. Evaluate the following limits without the use of a sketch.
2
2
1. lim+
and lim−
.
x→3 x − 3
x→3 x − 3
x−1
2. lim + 2
.
x→−2 x (x + 2)
Tutorial 1.1.2. [C1] §4.6, pp. 436–437: 252, 257, 258, 259, 260, 261, 271, 279, 283.
1st Semester Calculus Study Guide
8
1.2
MATH1036
Limit Laws
We now will make use of the following Limit Laws to calculate a limit of a function rather than the
approaches used in Section 1.1.
Theorem 1.1 (Limit Laws). Let c ∈ R and suppose that lim f (x) = L and lim g(x) = M both exist. Then
x→a
x→a
1. lim [ f (x) + g(x)] = lim f (x) + lim g(x) = L + M.
x→a
x→a
x→a
2. lim [ f (x) − g(x)] = lim f (x) − lim g(x) = L − M.
x→a
x→a
x→a
3. lim [c f (x)] = c lim f (x) = cL.
x→a
x→a
h
4. lim [ f (x)g(x)] = lim f (x)
x→a
ih
x→a
i
lim g(x) = LM.
x→a
lim f (x)
f (x) x→a
L
=
= .
x→a g(x)
lim g(x) M
5. if M , 0, lim
x→a
6. if L , 0 and M = 0, lim
f (x)
x→a g(x)
does not exist.
in
h
7. if n ∈ Z+ , lim f (x)n = lim f (x) = Ln .
x→a
x→a
8. lim c = c.
x→a
9. lim x = a.
x→a
10. if n ∈ Z+ , lim xn = an .
x→a
√
√
11. if n ∈ Z+ , lim n x = n a. If n is even, we assume that a > 0.
x→a
p
√n
»
12. if n ∈ Z+ , lim n f (x) = n lim f (x) = L.
x→a
x→a
13. lim |h(x)| = 0, if and only if lim h(x) = 0.
x→a
x→a
Worked Example 1.2.1. Use the above Limit Laws to calculate the following limits if they exist.
1. lim (5x2 − 9x − 8).
x→4
√
2
x3 + 3 x
2. lim
.
x→8 4 − 16
x
4x − 3
3. lim 2
.
x→2 x − 4
Recall that a polynomial function is of the form
f (x) = bn xn + bn−1 xn−1 + · · · + b2 x2 + b1 x + b0
with bi ∈ R for i = 0, 1, 2, . . . , n and n any non-negative integer. A rational function is of the form
p(x)
f (x) =
with p(x) and q(x) polynomials. We then have the following as a consequence of Theorem
q(x)
1.1 (1,8,10,5).
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9
Theorem 1.2. If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) =
x→a
f (a).
If f (x) = bn xn + bn−1 xn−1 + · · · + b2 x2 + b1 x + b0 , then
lim f (x) = f (a) = bn an + bn−1 an−1 + · · · + b2 a2 + b1 a + b0 .
x→a
2x2 + 1
.
x→2 x2 + 6x − 4
Worked Example 1.2.2. Evaluate lim
For rational functions f (x) =
p(x)
where both the numerator and denominator tend to zero (said to be
q(x)
0
of the indeterminate form ), we use factorization to simplify the limit and then use the Limit Laws.
0
Worked Example 1.2.3. Evaluate the following limits if they exist.
Hint: Use factorisation to simplify the limits where substitution fails.
x2 − 4
1. lim
.
x→2 x − 2
x3 + 1
.
x→−1 x + 1
2. lim
3. lim
x+1
x→−1 (2x2 + 7x + 5)2
4. lim
x→2
.
x4 − 16
.
x3 − 8
0
Certain quotients are also of the form where the Quotient Limit Law cannot be applied immediately.
0
However, rationalizing the numerator or denominator may allow us to proceed with the Limit Laws.
Worked Example 1.2.4. Evaluate the following limits.
√
x+1−1
1. lim
.
x→0
x
x
2. lim √
.
x→0
1 + 3x − 1
Tutorial 1.2.1. 1. [C1] §2.3, p. 176: 83–86, 88, 91, 107–112.
2. Given that lim f (x) = 2, lim g(x) = 0 and lim h(x) = −3 find the limits that exist. If the limit does not
x→a
x→a
x→a
exist, explain why.
p
a. lim [ f (x) − h(x)], b. lim ( f (x))3 , c. lim 3 2 f (x) − 4h(x),
x→a
x→a
x→a
g(x)
f (x)
d. lim
, e. lim
.
x→a h(x)
x→a g(x)
3. Find the limit, if it exists.
√
√
1
1
x2 + 4 − 5
3 + x
a. lim
, b. lim √
,
x→−3 3 + x
x→1
x+3−2
Å
c. lim
x→0
4. [C1] pp. 211–212: 213, 214, 217–219, 221.
ã
1
−
,
√
x 1+x x
1
x2 − 1
d. lim √
√ ,
x→1
2x + 1 − 3
1st Semester Calculus Study Guide
10
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5. If you are given that lim g(x) = π1 , what can you say about g(e)?
x→e
6. If you are given that lim g(x) = π1 , what can you say about lim− g(x)?
x→e
x→e
7. If you are given that f (−5) D.N.E., what can you say about lim f (x)?
x→−5
1.3
Infinite limits, limits at infinity and asymptotes
Horizontal asymptotes of graphs of functions correspond to limits of functions as x → ∞ and x → −∞.
The following theorems and techniques allow calculating the limits algebraically.
Theorem 1.3 (Horizontal aymptote: x-axis). If r is any positive rational number, then
1
lim
x→∞ xr
=0
and
lim
1
x→−∞ xr
= 0,
where in the case of x → −∞ we have to assume that xr is defined for negative x, i. e., r can be written
p
as r = where p and q are positive integers with q odd.
q
Worked Example 1.3.1.
1
1. lim 3 = 0.
x→∞ x
1
2. lim √3 = 0.
x→−∞
x2
If f is a rational function, then lim f (x) and lim f (x) may be found by first dividing the numerator
x→∞
x→−∞
and denominator by the highest power of x in the denominator and then apply Theorem 1.3.
Worked Example 1.3.2. Evaluate the following limits.
2x − 3
1. lim 2
.
x→∞ x + 7x − 1
2.
√
9x2 + 2
lim
.
x→∞ 3 − 4x
Theorem 1.4 (Vertical asymptote: x = a).
1. If n is an even positive integer, then lim
1
x→a (x − a)n
= ∞.
1
1
= ∞ and lim−
= −∞.
n
x→a (x − a)
x→a (x − a)n
2. If n is an odd positive integer, then lim+
1
1
and lim−
.
3
x→3 (x − 3)
x→3 (x − 3)3
Worked Example 1.3.3. Evaluate lim+
Theorem 1.5 (Limits of exponential functions). For the exponential function a x ,
1. if a > 1, then
lim a x = ∞
and
lim a x = 0.
x→∞
x→−∞
2. if 0 < a < 1, then
lim a x = 0
x→∞
and
lim a x = ∞.
x→−∞
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11
Worked Example 1.3.4. Evaluate the following exponential limits.
1. lim e x and lim e x .
x→∞
x→−∞
x
−x
e −e
e x − e−x
and
lim
.
x→∞ e x + e−x
x→−∞ e x + e−x
2. lim
Tutorial 1.3.1. 1. [C1] §4.6, p. 437: 262, 263, 265, 269, 270.
2. Find √
the limit, if
p
p
p
p
√ it exists.
a. lim x + 2 − x + 5, b. lim
x2 + 5 − x2 + 3, c. lim
x2 + 5x − x2 + 3x
x→∞
x→−∞
3. Find the limit, if it exists.
x3 + πx
3
a. lim 3/2 , b. lim
,
x→∞ (x + 1)2
x→−∞ x
1.4
x→∞
2x
.
x→−∞ 3 x
c. lim
Limits through One-sided Limits
The following theorem allows us to prove algebraically the existence (or non-existence) of limits and to
calculate their value. We used this theorem graphically in Section 1.1.
Theorem 1.6. lim f (x) = L ⇔ lim− f (x) = L = lim+ f (x).
x→a
x→a
x→a
Worked Example 1.4.1. 1. Prove that lim |x| = 0.
x→0
Remember that |x| = x if x ≥ 0, and |x| = −x if x < 0. Then solve for the left-hand limit and the
right-hand limit.
|x + 4|
exists.
x→−4 x + 4
2. Determine algebraically whether lim
Tutorial 1.4.1. For the following functions f (x), determine if the one-sided limits at the indicated point
a exist, find whether the limit at a exists, and compare with the value of f (a) if f is defined at a.
x2 − 1
1 1
at a = 0, b. f (x) =
at a = 1, c. f (x) = ⌊x⌋ + ⌊−x⌋ at a = −3.
a. f (x) = −
x |x|
|x − 1|
1.5
The Sandwich Theorem & Trigonometric Limits
The next important theorem allows us to compute the limits of functions that are bounded (squeezed,
pinched or sandwiched) between two other functions that have the same limit.
Theorem 1.7 (Sandwich Theorem). If f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and
lim f (x) = L = lim h(x), then lim g(x) = L.
x→a
x→a
x→a
Note. The Sandwich Theorem is also called Squeeze Theorem.
Worked Example 1.5.1.
1. Use the Sandwich theorem to prove that lim sin θ = 0 and hence conclude that lim cos θ = 1. (Note
θ→0
that θ is in radian measure.)
1
2. Prove that lim x sin = 0 (note that x is in radians).
x→0
x
Remember that −1 ≤ sin t ≤ 1 for all t ∈ R
sin x
3. Determine lim
.
x→∞ x
θ→0
1st Semester Calculus Study Guide
12
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sin x
1
1
≤ , i. e., the graph of
Solution. Since −1 ≤ sin x ≤ 1, it follows that for x > 0 (x → ∞), − ≤
x
x
x
Å ã
sin x
1
1
1
1
is bounded by those of − and . Since lim −
= 0 and lim = 0, we have by the Sandwich
x→∞
x→∞ x
x
x
x
x
Theorem that
sin x
lim
= 0.
x→∞ x
The Sandwich Theorem has fundamental applications in trigonometric limit calculations as seen in the
examples above. We will now consider evaluating trigonometric limits with all calculations based on
radian measure.
We recall from the above example, part 1, that
lim sin θ = 0
θ→0
lim cos θ = 1
and
θ→0
which will be used in Theorem 1.8.
The next theorem is a crucial result that will be used in Chapter 3 of these notes. You will need to recall
the Area of Sector calculation in the Algebra Guide. The area of a sector is given by
1
A = r2 θ
2
where θ is the central angle of the sector with radius r. Again we must stress that all angle measurement
is in radians.
sin θ
= 1.
θ→0 θ
Theorem 1.8. lim
Proof. Consider the sector OAC in the first quadrant (0 < θ < π2 ) of a circle of radius 1.
1
B
tan θ
sin θ
A
θ
O
D
θ>0
C
1st Semester Calculus Study Guide
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13
We see that
Area of △OAC < Area of sector OAC < Area of △OBC
1
1
1
⇒ (OC)(AD) < (OC)2 θ < (OC)(BC)
2
2
2
1
1 2
1
⇒ (1) sin θ < (1) θ < (1) tan θ
2
2
2
1
1
1
⇒ sin θ < θ < tan θ
2
2
2
⇒ sin θ < θ < tan θ
sin θ
⇒ sin θ < θ <
cos θ
θ
1
⇒1<
<
sin θ cos θ
π
we have sin θ > 0, and so dividing by sin θ leaves the inequalities unchanged. Taking
2
sin θ
> cos θ or equivalently
reciprocals we have 1 >
θ
since for 0 < θ <
cos θ <
sin θ
< 1.
θ
(∗)
Now if − π2 < θ < 0, then 0 < −θ < π2 so that by the result just established
cos(−θ) <
sin(−θ)
< 1,
−θ
which reduces again to (∗) since sin(−θ) = − sin θ and cos(−θ) = cos θ. Thus the inequality (∗) also
holds for − π2 < θ < 0. Hence for all θ ∈ (− π2 , π2 ) except θ = 0,
cos θ <
sin θ
< 1.
θ
Since lim cos θ = 1 and lim 1 = 1, by the Sandwich Theorem lim
θ→0
θ→0
θ→0
sin θ
= 1.
θ
By the above
sin A
=1
A→0 A
lim
or
A
=1
A→0 sin A
lim
Together with the above theorem and the Limit Laws we also have the following important limit.
1 − cos θ
= 0.
θ→0
θ
Theorem 1.9. lim
□
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Proof.
ãÅ
ã
Å
1 + cos θ
1 − cos θ
1 − cos θ
= lim
lim
θ→0
θ→0
θ
θ
1 + cos θ
2
1 − cos θ
= lim
θ→0 θ(1 + cos θ)
sin2 θ
= lim
θ→0 θ(1 + cos θ)
ãÅ
ã
Å
sin θ
sin θ
= lim
θ→0
θ
1 + cos θ
Å
ãÅ
ã
sin θ
sin θ
= lim
lim
θ→0 θ
θ→0 1 + cos θ
Å
ã lim sin θ
sin θ
θ→0
= lim
θ→0 θ
lim (1 + cos θ)
θ→0
sin θ
θ→0 θ
Å
= lim
ã
lim sin θ
θ→0
1 + lim cos θ
θ→0
0
= (1)
1+1
= 0.
Worked Example 1.5.2. Evaluate the following trigonometric limits.
sin 5x
1. lim
.
x→0
x
sin2 x
.
2. lim
x→0
x
sin(x2 )
3. lim
.
x→0
x
sin2 x
4. lim 2 .
x→0 x
sin θ
5. lim
.
θ→0 θ + tan θ
√
√
Tutorial 1.5.1. 1. a. Show that 3 − 2x2 ≤ 3 − x2 for −1 ≤ x ≤ 1.
p
p
b. Show that lim 3 − 2x2 = lim 3 − x2 .
x→0
x→0 √
√
c. Let f be a function that satisfies 3 − 2x2 ≤ f (x) ≤ 3 − x2 for −1 ≤ x ≤ 1. Show that lim f (x)
x→0
exists and find the limit.
2. Evaluate the following limits if they exist.
sin 3x
tan 5x
1 − cos 3x
a. lim
, b. lim
, c. lim x2 cot 2x cosec 3x, d. lim
x→0 sin 7x
x→0
x→0
x→0 sin2 4x
x
3. Evaluate the following limits if they exist.
sin(x2 − 1)
sin(sin x)
, b. lim
a. lim
x→1
x→0
5x
x−1
(
x if x is rational,
4. Let f (x) =
Show that lim f (x) = 0.
x→0
0 if x is irrational.
Chapter 2
Continuity of Functions
LEARNING OUTCOMES:
On completion of this chapter you should (tick the checkbox when you have mastered the skill)
□ 1. know and understand the limit definitions of continuity of a function at a point, continuity of a
function over an interval and a continuous function,
□ 2. be able to establish if a given function is continuous at a point using the definition,
□ 3. know and understand left-continuity (continuous from the left) and right-continuity (continuous
from the right),
□ 4. know when and be able to test whether a function is continuous over an interval,
□ 5. know and understand the various theorems on continuity given in these notes of this section,
□ 6. know the statement of and understand the Intermediate Value Theorem (IVT),
□ 7. apply the IVT in determining existence of roots of a given function over an interval,
□ 8. know, understand and identify the various types of discontinuities and
□ 9. know how to solve the examinable tutorial problems and the worked out problems from these
notes on this chapter.
The usual meaning of the word continuous in everyday language is that of going on all the time or
proceeding without a break or interruptions and without abrupt changes.
From a mathematics viewpoint, this can also be taken as a loose meaning of continuity. A curve is
continuous between two points a and b if you can draw it in one go from a without lifting up your pencil
until you reach b.
We will now present the precise meaning of continuity, which is closely related to the concept of the
limit described in the previous chapter.
2.1
Continuity at a Point
We have seen in the previous chapter that it is not necessary for a function f to be defined at a point a
in order for f to have a limit at a as in Worked Example 1.1.3.3 or all the examples of Worked Example
1.1.4. That is, lim f (x) may exist even though a is not in the domain of f .
x→a
On the other hand, if f (a) exists, i. e., f is defined at a, and lim f (x) exists it may not be the case that
x→a
both f (a) and lim f (x) are equal as illustrated in the following example.
x→a
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Worked Example 2.1.1. Let
f (x) =
(
1
if x , 0,
2
if x = 0.
Find f (0) and lim f (x).
x→0
The function f is continuous at the point a if and only if f (a) = lim f (x).
x→a
Equivalently (called the slot-machine test),
f is continuous at a if and only if the following three conditions are satisfied:
1. f (a) is defined, i. e., a is in the domain of f ,
2. lim f (x) exists, i.e., lim− f (x) = lim+ f (x), and
x→a
x→a
x→a
3. f (a) = lim f (x).
x→a
In most cases we will use the above three conditions to determine continuity at a point algebraically. If
just one of the three conditions above fails, then f is not continuous at the given point. If such is the
case, then f is said to be discontinuous at the given point.
Worked Example 2.1.2. 1. Determine whether

 |x + 3|
f (x) =
x+3
1
if x , −3,
if x = −3,
is continuous at x = −3.
2. Is


x2 + 4


g(x) = 5


 3
x
if x < 2,
if x = 2,
if x > 2,
continuous at x = 2?
3. Show that


sin x


h(x) = 0



x cos x
if x > 0,
if x = 0,
if x < 0,
is continuous at x = 0.
4. Let


ax − b


f (x) = 3x


 2
bx − a
if x ≤ 1,
if 1 < x < 2,
if x ≥ 2.
Find the value(s) of a and b so that f is continuous at x = 1 but not continuous at x = 2.
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17
Tutorial 2.1.1. 1. [C1] p. 212: 226, 227.
2. Find the values of a and b which make the function


x−1
if x ≤ −2,


f (x) = ax2 + b if − 2 < x < 1,



x+1
if x ≥ 1,
continuous at x = −2 and x = 1.
2.2
Left and Right Continuity at a Point
From the examples in the previous section it is apparent that we can speak of continuity from the left or
right at a point. We thus have
A function f is continuous from the right at a if lim+ f (x) = f (a).
x→a
and
A function f is continuous from the left at a if lim− f (x) = f (a).
x→a
Worked Example 2.2.1. For these worked examples, check if the conditions of continuity from left or
right at a point are satisfied.
1. Let

 |x| + x if x , 0,
2x
f (x) =

0
if x = 0.
Determine the right and left continuity of f at x = 0.
2. Show that at each integer n, the greatest integer function f (x) = ⌊x⌋ is right-continuous at n, but
discontinuous from the left at n.
It can be observed that
f is continuous at a if and only if f is right-continuous and left-continuous at a.
Tutorial 2.2.1. Consider the function

 ⌊x⌋
f (x) =
x
−1
if x , 0,
if x = 0.
Investigate continuity from the left and the right at x = 0, x = π and x = 1.
2.3
Continuity on an Interval
We can also now consider continuity on an interval.
f is continuous on an interval if and only if f is continuous at each point in the interval.
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In particular,
1. f is continuous on an open interval (a, b) if and only if f is continuous at each point a < x < b.
2. f is continuous on the interval [a, b) if and only if f is continuous on the open interval (a, b) and f is
right-continuous at a.
3. f is continuous on the interval (a, b] if and only if f is continuous on the open interval (a, b) and f is
left-continuous at b.
4. f is continuous on the closed interval [a, b] if and only if f is continuous on the open interval (a, b)
and f is right-continuous at a and left-continuous at b.
5. f is a continuous function if and only if f is continuous at each point of its domain.
6. f is continuous everywhere if and only if f is continuous at each point in R, i. e., f is continuous on
the open interval (−∞, ∞).
√
Worked Example 2.3.1. Show that f (x) = x − 4 is continuous on the interval [4, 8].
2.4
Theorems on Continuity
By the definition of continuity, f is continuous at a provided that lim f (x) = f (a). Thus where direct
x→a
substitution applies in calculating a limit of a function, the function is necessarily continuous at the
given point. We have the following with regards to the algebra of functions that are continuous at a
given point a.
Theorem 2.1. If f and g are continuous at a and c ∈ R, then
1. the sum f + g,
2. the difference f − g,
3. the product f g,
4. the quotient
f
if g(a) , 0 and
g
5. the scalar multiple c f
are functions that are also continuous at a.
Theorem 2.2. The following functions are continuous on their domains.
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 , ai ∈ R, n ∈ N0 .
2. Rational functions
p(x)
, p and q , 0 polynomials.
q(x)
3. Root functions.
4. The trigonometric functions sin x, cos x, tan x, cosec x, sec x and cot x.
5. The exponential functions a x and the natural exponential function e x .
6. The logarithmic functions loga x and the natural logarithmic function ln x.
7. The absolute value function |x|.
1st Semester Calculus Study Guide
MATH1036
Worked Example 2.4.1. 1. For c, d ∈ R, let


4x


h(x) = cx + d



−5x
19
if x ≤ −1,
if − 1 < x ≤ 2,
if x > 2.
Determine the values of c and d so that h is a continuous function. For these values, sketch h.
2. Describe the intervals on which the function f (x) = tan x is continuous.
Theorem 2.3 (Composite functions). If g is continuous at a and f is continuous at g(a), then the composite function f ◦ g is continuous at a and
lim ( f ◦ g)(x) = lim f (g(x)) = f (lim g(x)) = f (g(a)).
x→a
x→a
x→a
The next example illustrates the continuity of the composite of functions together with the above theorems on continuity.
Worked Example 2.4.2. Use the above theorems on continuity to evaluate the following limits.
1. lim |1 − 2x − 3x2 |.
x→3
2. lim
x→2
x2 − 4
.
x−2
3. limπ sin2 x.
x→ 3
4. limπ ecos 2x .
x→ 6
√
√3
5. lim ln( 2x + 3 4 − 3x).
x→0
Theorem 2.4 (Intermediate Value Theorem (IVT)). Suppose that f is continuous on the closed interval
[a, b] and let N be a number between f (a) and f (b), where f (a) , f (b). Then there exists a number c in
the open interval (a, b) such that f (c) = N.
Worked Example 2.4.3. 1. Show that the equation cos x = x has a solution in the closed interval [0, π2 ].
2. The equation x3 − 4x2 + 2x + 2 = 0 has three solutions. Find intervals between successive integers
that contain the solutions.
Tutorial 2.4.1. 1. [C1] §2.4, pp. 192–193: 161–167.
∗ 2. Show that, at any given time, there are antipodal (diametrically opposite) locations on Earth’s equator
which have the same temperature. You may assume that the temperature depends continuously on the
location.
3. The equation x3 − 12x + 8 = 0 has three solutions. Find intervals between successive integers that
contain these solutions.
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2.5
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Discontinuities
If one of the three conditions for continuity of a function f (the slot-machine test) fails at a given point a,
we know then that f is discontinuous at a. The following are the various types of discontinuities.
1. Removable discontinuity
If lim f (x) exists but f fails to be continuous at x = a because either
x→a
(a) f (a) is not defined or
(b) f (a) is defined, but a number different from lim f (x),
x→a
then f is said to have a removable discontinuity at x = a.
a
a
This discontinuity at a can be removed by redefining f (a) = lim f (x).
x→a
2. Jump discontinuity
If lim+ f (x) and lim− f (x) exist but f is not continuous at a because lim f (x) does not exist, then
x→a
x→a
x→a
lim− f (x) , lim+ f (x) and f is said to have a jump discontinuity at x = a. The function jumps from one
x→a
x→a
value to another one.
a
3. Infinite discontinuity
If f is not continuous at a because lim− f (x) = ±∞ or lim+ f (x) = ±∞, then f is said to have an infinite
x→a
x→a
discontinuity at a.
Below are some illustrations for infinite discontinuities.
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1st Semester Calculus Study Guide
21
a
a
a
a
a
a
Worked Example 2.5.1. Let
 3

x




x2 − 2







3 − x
f (x) = 4x − 1


x−1





15




7−x



5x + 2
if x ≤ −1,
if − 1 < x < 0,
if 0 ≤ x < 2,
if 2 ≤ x < 4,
if 4 < x < 7,
if x ≥ 7.
Fill out the following table and then discuss the continuity of f at a. At each point of discontinuity,
classify the discontinuity. List the intervals on which f is continuous.
1st Semester Calculus Study Guide
22
a
f (a)
lim f (x)
x→a−
lim f (x)
x→a+
−1
0
2
4
7
List the intervals on which f is continuous.
Tutorial 2.5.1. [C1] §2.4, p. 191: 131–142.
lim f (x)
x→a
Continuos or
discontinuous at a?
MATH1036
Discontinuity
type
Chapter 3
Differentiation
LEARNING OUTCOMES:
On completion of this chapter you should (tick the checkbox when you have mastered the skill)
□ 1. know the limit definition of the derivative of a function at a point,
□ 2. be able to find the derivative of a function using the definition (first principles),
□ 3. understand and be able to explain the concept of differentiability on an interval,
□ 4. be able to interpret and find the derivative at a point as the slope of a tangent line at that point,
□ 5. be able to interpret graphically non-differentiability of functions at a point,
□ 6. understand the relationship between differentiable and continuous functions,
□ 7. know how to find algebraically the derivative function f from the given formula of f ,
□ 8. know how to find and sketch the derivative function f from the graph of f ,
□ 9. know the proofs of the Derivative Rules,
□ 10. know and be able to calculate the derivative of a function using the Derivative Rules,
□ 11. know the derivatives of the trigonometric functions and know how to prove these derivatives,
□ 12. know the derivatives of the exponential functions a and the logarithmic functions log x,
□ 13. know the derivative of the natural exponential function e and the natural logarithmic function
ln x,
□ 14. know and be able to apply the Chain Rule in differentiating composite functions,
□ 15. know how to approximate function values using linearization,
□ 16. know the concept of and know how to evaluate differentials,
□ 17. know how to solve the examinable tutorial problems and the worked out problems from these
′
′
x
a
x
notes on this chapter.
3.1
First Principles
In evaluating limits, functional notation plays a convenient and important role.
Worked Example 3.1.1. Let f (x) =
1
. Evaluate
(x + 1)2
f (5 + h) − f (5)
.
h→0
h
lim
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The derivative of the function f at a, denoted f ′ (a) is defined as
f ′ (a) = lim
h→0
f (a + h) − f (a)
h
(if it exists)
or equivalently,
f ′ (a) = lim
x→a
f (x) − f (a)
x−a
(if it exists)
provided the limit exists. The above limit calculations provide the derivative of the function f at a by
first principles.
Worked Example 3.1.2. Find the derivative of the following by first principles.
1
1. f ′ (5) if f (x) =
.
(x + 1)2
√
2. g′ (4) if g(x) = x2 + 9.
If f ′ (a) exists we say that the function f is differentiable at a or f has a derivative at a.
Now given a function f and the point a we define the left-hand derivative of f at a as
h→0
f (a + h) − f (a)
h
f+′ (a) = lim+
f (a + h) − f (a)
.
h
f (x) − f (a)
x−a
and
f−′ (a) = lim−
and the right-hand derivative of f at a as
h→0
Equivalently,
f−′ (a) = lim−
x→a
f+′ (a) = lim+
x→a
f (x) − f (a)
.
x−a
Then certainly
f is differentiable at a provided that f−′ (a) and f+′ (a) both exist and f−′ (a) = f+′ (a)
so that f ′ (a) = f−′ (a) = f+′ (a).
Worked Example 3.1.3. 1. Show that f (x) = |x| is not differentiable at x = 0.
2. Is the function
g(x) =
(
4x
2x2 + 2
if x < 1,
if x ≥ 1,
differentiable at x = 1?
We say that f is differentiable on an open interval (a, b) if f is differentiable for all c ∈ (a, b). In
like manner, we refer to functions that are differentiable on intervals of the form (a, ∞), (−∞, b) or
(−∞, ∞) = R.
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A function f is differentiable on the closed interval [a, b] if f is differentiable on the open interval
(a, b) and the one-sided limits f+′ (a) and f−′ (b) both exist.
Differentiability on an interval of the form [a, b), [a, ∞), (a, b] or (−∞, b] is defined in the obvious way,
using the one-sided limit at the endpoint which belongs to the interval.
A function f is called a differentiable function if f is differentiable at each point of its domain.
Tutorial 3.1.1. 1. [C1] §3.1, p. 228: 22, 24–27, 29, 30. You may use any of the two equivalent
definitions of the derivative.
2. Find f ′ (a) if it exists, where
(
3x2
if x ≤ 1
(a)
f (x) =
and a = 1.
2x3 + 1 if x > 1
(
x3
if x ≤ 1
(b)
f (x) =
and a = 1.
cx + b if x > 1
3.2
Tangents
The derivative of a function at a point also has a geometric interpretation.
3.2.1
Non-vertical tangents
Worked Example 3.2.1. Find the equation of the tangent line to the graph of f (x) = x2 at x = −3.
Tutorial 3.2.1. 1. §3.1, p. 228: 11, 12, 14, 16, 18.
2. For the cases in Tutorial 3.1.1.2 where the derivative exists, find the corresponding tangent line.
3.2.2
Vertical tangents
The function f has a vertical tangent at a provided that all three of the following conditions hold:
• f is continuous at a,
• f is differentiable near a, i. e., f ′ (x) exists for x , a with x close to a, and
• lim | f ′ (x)| = ∞.
x→a
Vertical tangent
Vertical tangent
Vertical tangents at endpoints
cusp
a
a
a
b
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The definition for vertical tangents may be modified to include vertical tangent lines at an endpoint of
the domain of the function. Thus, if f is continuous on [a, b] but is undefined outside [a, b], then f has
a vertical tangent at x = a if lim+ | f ′ (x)| = ∞ and f has a vertical tangent at x = b if lim− | f ′ (x)| = ∞.
x→a
x→b
√
Worked Example 3.2.2. Let f (x) = 3 x. Show that f has a vertical tangent at x = 0.
Tutorial 3.2.2. Which of the following
functions has a vertical tangent? Justify you answer.

−1 if x < 0,


√
a. f (x) = 3 + x, b. sgn(x) = 0
if x = 0,



−1 if x > 0.
3.3
Differentiability vs. Continuity
Since both continuity and differentiability involve limits, we can investigate the relationship between
these two concepts.
1. A function can be continuous at a given point without being differentiable there.
Worked Example 3.3.1. Let
(
f (x) =
x2
if x ≤ 1,
x
if x > 1.
Investigate the continuity and differentiability of f at x = 1.
2. Every function that is differentiable at a point is continuous there.
Theorem 3.1. If a function f is differentiable at x = a, then f is continuous at x = a.
Proof. Suppose that f is differentiable at x = a. We must show that f is continuous at x = a, i .e., we
need to show that lim f (x) = f (a).
x→a
If x is in the domain of f and x , a, then f (x) may be written as
Å
ã
f (x) − f (a)
f (x) = f (a) +
(x − a).
x−a
Then
ï
Å
ã
ò
f (x) − f (a)
(x − a)
lim f (x) = lim f (a) +
x→a
x→a
x−a
ïÅ
ã
ò
f (x) − f (a)
= lim f (a) + lim
(x − a)
x→a
x→a
x−a
Å
ã
f (x) − f (a)
= f (a) + lim
lim (x − a)
x→a
x→a
x−a
= f (a) + f ′ (a) lim (x − a)
x→a
= f (a) + f (a) · 0
′
= f (a).
Hence f is continuous at x = a.
□
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The above theorem shows that
DIFFERENTIABLE ⇒ CONTINUOUS
or contrapositively,
NOT CONTINUOUS ⇒ NOT DIFFERENTIABLE
Worked Example 3.3.2. Consider the greatest integer function f (x) = ⌊x⌋ at x = 0. Discuss the
continuity and differentiability of f at x = 0.
Cases in which a function f fails to be differentiable at x = a are:
1. f is discontinuous at a. Graphically, if we observe a removable, jump or an infinite discontinuity
at a.
2. f is continuous at a and the graph of f is pointed (e. g. |x|) or makes a sharp turn or corner (e.g.
Worked Example 3.3.1),
3. f has a vertical tangent at a (e.g. Worked Example 3.2.2).
Tutorial 3.3.1. [C1] §3.2, p. 244: 78, 79.
3.4
The derivative as a function
We observe that if for a given function f we let X = {a ∈ R : f ′ (a) exists}, then we can associate with
each x ∈ X the number f ′ (x) and we obtain a function f ′ with domain X:
f ′ : X→R
x 7→ f ′ (x).
The function f ′ is called the derivative of f and, by definition, the values of f ′ at each x are given by
the following (or an appropriate one-sided limit)
f ′ (x) = lim
h→0
f (x + h) − f (x)
h
(if it exists).
Worked Example 3.4.1. Let f (x) = |x|. Find the derivative function f ′ of f .
3.5
Derivative Rules
It is quite cumbersome on each occasion to calculate the derivative of a differentiable function at a given
point using first principles. We will now develop formulas that give the derivatives of known functions
that we will encounter. The proofs of these results use the definition of the derivative and are examinable.
The expressions
d
d f (x) d f
f (x),
,
, D x f, f x or f ′
dx
dx dx
are commonly used to denote the first derivative of the function f (x).
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Theorem 3.2 (Derivative of constants). If c ∈ R and f (x) = c for each x, then f ′ (x) = 0.
Proof.
c−c
f (x + h) − f (x)
= lim
= lim 0 = 0.
h→0 h
h→0
h→0
h
f ′ (x) = lim
□
The above theorem says that the derivative of constant functions are zero, i. e.,
d
(c) = 0.
dx
∀c ∈ R
Theorem 3.3 (Constant Multiple Rule, factorization of constants). If c ∈ R and f is differentiable, then
c f is differentiable and (c f )′ (x) = c f ′ (x).
Proof.
(c f )′ (x) = lim
h→0
c( f (x + h)) − c( f (x))
f (x + h) − f (x)
(c f )(x + h) − (c f )(x)
= lim
= c lim
= c f ′ (x).
h→0
h→0
h
h
h
□
Thus
∀c ∈ R
d
d
[c f (x)] = c f (x).
dx
dx
Theorem 3.4 (Power Rule for positive integer power functions). If f (x) = xn where n ∈ Z+ , then
f ′ (x) = nxn−1 .
Proof. We must show that if f (x) = xn where n is a positive integer, then f ′ (a) = nan−1 for every a. We
will use the factorization formula
xn − an = (x − a)(xn−1 + axn−2 + a2 xn−3 + a3 xn−4 + · · · + an−2 x + an−1 ).
We then have
f (x) − f (a)
x−a
x n − an
= lim
x→a x − a
(x − a)(xn−1 + axn−2 + a2 xn−3 + a3 xn−4 + · · · + an−2 x + an−1 )
= lim
x→a
x−a
n−1
n−2
2 n−3
3 n−4
= lim (x + ax + a x + a x + · · · + an−2 x + an−1 ) (polynomial with n terms)
f ′ (a) = lim
x→a
x→a
n−1
=a
+ aan−2 + a2 an−3 + a3 an−4 + · · · + an−2 a + an−1
=a
+a
n−1
= na
n−1
n−1
+a
n−1
+a
n−1
+ ··· + a
n−1
+a
n−1
(polynomials are continuous)
(n times)
.
□
Theorem 3.4 says
∀n ∈ N
d n
x = nxn−1 .
dx
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Worked Example 3.5.1. 1.
2.
29
d
d
d
d 1
d √
π = 0, e = 0, (−4) = 0,
= 0 and
2 = 0.
dx
dx
dx
dx 2
dx
d
d
(3x15 ) = 3 x15 = 3(15)x15−1 = 45x14 .
dx
dx
Theorem 3.5 (Sum Rule). If f and g are differentiable functions, then f + g is differentiable and
( f + g)′ (x) = f ′ (x) + g′ (x).
Proof. In class.
Hence
d
d
d
[ f (x) + g(x)] =
f (x) + g(x).
dx
dx
dx
Theorem 3.6 (Difference Rule). If f and g are differentiable functions, then f − g is differentiable and
( f − g)′ (x) = f ′ (x) − g′ (x).
Proof. We can apply the Sum Rule and the Constant Multiple Rule.
( f − g)′ (x) = ( f + [−g])′ (x) = f ′ (x) + (−g)′ (x) = f ′ (x) − g′ (x).
Hence
d
d
d
[ f (x) − g(x)] =
f (x) − g(x).
dx
dx
dx
Worked Example 3.5.2. Find the first derivatives of the following functions.
1. x3 − 4x2 + 7.
2. 3t2 (t2 + 2).
w3 − 7w
3.
.
2w
4. Let g(x) = |x2 − 4|. Sketch the graph of g. Use the graph of g to sketch the graph of g′ .
Theorem 3.7 (Product Rule). If f and g are differentiable functions then f g is differentiable and
( f g)′ (x) = f ′ (x)g(x) + f (x)g′ (x).
Proof. In class.
Hence
d
[ f (x)g(x)] =
dx
Å
ã
d
d
f (x) g(x) + f (x) g(x).
dx
dx
or
(Derivative of product)= (Derivative of first factor)× (Second factor)+ (First factor) × (Derivative of second factor)
It is sometimes convenient to write the Product Rule in the form
(uv)′ = uv′ + vu′
□
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Worked Example 3.5.3. Use the product rule to find the first derivatives of the following functions.
1. (x3 + 1)(2x2 + 8x − 5).
2. (8w − 1)(w2 + 4w + 7)(w3 − 5).
Theorem 3.8 (Quotient Rule). If f and g are differentiable functions and g(x) , 0, then
tiable and
f
is differeng
Å ã′
f
f ′ (x)g(x) − f (x)g′ (x)
(x) =
.
g
[g(x)]2
In class.
Hence
Å
d
dx
Å
f (x)
g(x)
ã
=
ã
d
d
f (x) g(x) − f (x) g(x)
dx
dx
[g(x)]2
Worked Example 3.5.4. Use the Quotient Rule to evaluate the first derivative of the following functions.
2x − 7
x2 + 9
v3 − 1
2. 3
v +1
1.
We can now extend the Power Rule to the case in which the exponent is a negative integer using the
Quotient Rule.
Theorem 3.9 (Power Rule for negative integer power functions). If f (x) = xn and n ∈ Z− , then f ′ (x) =
nxn−1 .
Proof. Since n ∈ Z− , n = −m for some positive integer m. Then
Å ã
d n
d −m
d
1
x =
x =
f ′ (x) =
dx
dx
dx xm
Å
ã
d
d
1 xm − (1) xm
dx
dx
=
2
m
(x )
(0)xm − (1)(mxm−1 )
=
x2m
m−1
−mx
=
x2m
= −mxm−1−2m
= −mx−m−1
= nxn−1
(∵ n = −m)
∀ n ∈ Z+
d −n
x = −nx−n−1
dx
Hence
□
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31
The Power Rule is also extended to include rational exponents, the proof of which will be done as a
consequence of section 3.7 on Implicit Differentiation of these notes. The Power Rule holds in fact for
all reals.
Theorem 3.10 (General Power Rule). If r is any real number, then
d r
x = rxr−1 .
dx
Worked Example 3.5.5. Use the generalized Power Rule to find the derivatives of the following functions.
1
1. f (x) = .
x
√
2. g(x) = x.
3. h(x) = xe+π .
√
4. v(t) = t 2 +
√
√
t + 2.
Tutorial 3.5.1. [C1] §3.3, p. 263: 106–112, 116, 117, 122, 125.
The next critically important rule is for composite functions.
Theorem 3.11 (Chain Rule). If f and g are both differentiable functions, then so is the composite
function f ◦ g and
( f ◦ g)′ (x) = f ′ (g(x))g′ (x).
The Chain Rule may also be expressed in the following form. If y = f (u) and u = g(x) are both
differentiable functions of x, then
Å ãÅ ã
dy
dy
du
=
dx
du
dx
Worked Example 3.5.6. Use the Chain Rule to find the first derivatives of the following.
1. (x2 − 1)100 .
1
.
(4w2 + 6w − 7)3
√
3. v2 + v + 4.
√
4. x x2 + 1.
x 23
5.
.
x2 + 16
2.
It is apparent that the Chain Rule gives the following neat calculations.
∀r ∈ R
d
[ f (x)]r = r[ f (x)]r−1 f ′ (x)
dx
32
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and
f ′ (x)
d p
f (x) = p
dx
2 f (x)
The following is a summary of the theorems and pertinent results of this section.
Function y
Its derivative
constant c, c ∈ R
0
f +g
f ′ + g′
f −g
f ′ − g′
cf, c ∈ R
cf′
fg
f ′ g + f g′
f
g
f ′ g − f g′
g2
xr , r ∈ R
rxr−1
√
1
√
x
f ◦g
dy
dx
2 x
( f ′ ◦ g)g′
Worked Example 3.5.7. Use the Derivative Rules in the following questions.
5
1. Find the equation of the normal line to the graph of f (x) =
at x = −2.
1 + x2
√
2. Let f (x) = x − 3 3 x. Find the points on the graph of f where f has horizontal and vertical tangents.
Tutorial 3.5.2. 1. [C1] §3.6, p. 297: 214, 215, 219, 220, 221, 223, 228, 229, 238.
2. [C1] §3.3, p. 264: 137, 138, 140, 141, 142.
3. For the following functions, find the points on the graphs where the tangents are horizontal or vertical.
1
(a) f (x) = x − ,
x
4
1
3
(b) g(x) = x + 4x 3 ,
3
(c) h(x) = x(x + 2) 5 ,
1
(d) d(x) = xr − x r on [0, ∞], where r > 1.
3.6
Trigonometric Derivatives
To find the derivatives of trigonometric functions we make use of Theorems 1.8 and 1.9 which respectively proves
sin θ
1 − cos θ
lim
= 1 and lim
= 0.
θ→0 θ
θ→0
θ
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33
The trigonometric functions are differentiable in their domains and their derivatives are now given using
the limit definition (incorporating the above limits in Theorems 1.8 and 1.9) together with the Derivative
Rules and trigonometric identities.
Theorem 3.12 (Derivative of sin x).
d
sin x = cos x
dx
Proof. In class.
Using a similar proof to that of Theorem 3.12 we have the following. We present here an alternate proof
using the aid of a trigonometric identity and the Chain Rule.
Theorem 3.13 (Derivative of cos x).
d
cos x = − sin x
dx
Proof. In class.
Theorem 3.14 (Derivative of tan x).
d
tan x = sec2 x
dx
Proof. In class.
Theorem 3.15 (Derivative of cosec x).
d
cosec x = − cosec x cot x
dx
Proof.
Å
ã
d
d
1
cosec x =
dx
dx sin x
d
(sin x)−1
=
dx
d
= − (sin x)−2
sin x
dx
1
= − 2 (cos x)
sin x
Å
ã
1
cos x =−
sin x
sin x
= − cosec x cot x.
Theorem 3.16 (Derivative of sec x).
d
sec x = sec x tan x
dx
(Chain Rule)
□
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Proof.
d
d
sec x =
dx
dx
Å
1
cos x
ã
d
cos x
dx
cos2 x
−(1)
=
(Quotient Rule)
sin x
cos2 x
ãÅ
ã
Å
sin x
1
=
cos x
cos x
= sec x tan x.
=
Theorem 3.17 (Derivative of cot x).
□
d
cot x = − cosec2 x
dx
Proof.
d
d cos x cot x =
dx
dx sin x
Å
ã
d
d
cos x sin x − cos x sin x
dx
dx
=
2
sin x
=
=
(Quotient Rule)
− sin x sin x − cos x cos x
sin2 x
− sin2 x − cos2 x
sin2 x
= − cosec2 x.
Worked Example 3.6.1. Find the derivatives of the following.
1. y = sin x cos x.
π
2. f ′
if f (x) = x cot x.
4
√
3. sec x
4. g(t) = sin3 (4t).
p
5. y = x2 + cos3 (sin(3x)).
Herewith a summary of the trigonometric derivatives.
□
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35
y
dy
dx
Using Chain Rule with u = f (x)
sin x
cos x
y = sin u ⇒ y′ = (cos u)
cos x
− sin x
y = cos u ⇒ y′ = −(sin u)
tan x
sec2 x
y = tan u ⇒ y′ = (sec2 u)
cosec x
− cosec x cot x
y = cosec u ⇒ y′ = −(cosec u cot u)
sec x
sec x tan x
y = sec u ⇒ y′ = (sec u tan u)
du
dx
cot x
− cosec2 x
y = cot u ⇒ y′ = −(cosec2 u)
du
dx
du
dx
du
dx
du
dx
du
dx
Tutorial 3.6.1. 1. [C1] §3.5, p. 285: 175, 177–183.
2. [C1] §3.6, p. 297: 216–218, 222, 224, 227, 230, 233–237.
3. Does the curve y = sin (x − sin x) have horizontal tangents at the x-axis? Give reasons.
3.7
Implicit Differentiation
√
Equations of the form y = f (x) give y explicitly as a function of x, e. g., y = x3 + 1 and y = sin x.
Thus far we have been differentiating functions of this type. In some cases y is not given explicitly as a
dy
function of x, of which we need to find
as illustrated in the next example.
dx
Worked Example 3.7.1. Suppose that y is a function of x that satisfies the equation
3x3 y − 4y − 2x + 1 = 0.
Find
dy
by first finding y explicitly.
dx
It is generally not so simple to write such equations explicitly as y = f (x). Such functions (like the above
example) are said to be given implicitly by the equation. When these functions are differentiable, it is
dy
possible to find
without solving the equation for y. This method is called implicit differentiation.
dx
dy
This technique to find
is given below.
dx
Implicit differentiation:
Differentiate both sides of the equation w.r.t x
taking into account the Chain Rule
dy
and then solve the resulting equation for .
dx
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We may now differentiate
3x3 y − 4y − 2x + 1 = 0
implicitly as shown in class.
When using the method of implicit differentiation it is important to remember that y is a function of x
and that we are differentiating with respect to x. Hence, powers of y are powers of functions and the
Chain Rule is required for their differentiation. Also be prepared to use combinations of the Chain Rule,
Product Rule and Quotient Rule.
Worked Example 3.7.2. 1. Find
dy
by implicit differentiation if x2 + y2 = 4.
dx
2. Find the equation of the tangent line at (x, y) = ( π6 , 5π
6 ) to the curve
π
.
2
x sin y + y sin x =
3. If p, q ∈ Z and q , 0 prove that
p p
d qp
x = x q −1 .
dx
q
(This proves the Power Rule for all rationals).
p
Proof. Let y = x q . Then yq = x p . Implicit differentiation then gives:
yq = x p
dy
⇒ qyq−1
= px p−1
dx
dy px p−1
⇒
=
dx qyq−1
px p−1
= Ä p äq−1
q xq
=
px p−1
p
qx p− q
p
p
= x p−1−p+ q
q
p p
= x q −1 .
q
p
Hence, since y = x q we have proved that
p
d qp
x = qp x q −1 .
dx
4. Find y′ if xy tan yx = x2 + y3 .
Tutorial 3.7.1. 1. Å
[C1]ã §3.8, p. 317: 300–302, 304, 305, 307–309, 316–318.
dy
1
2. Find
if y sin
= 1 − xy.
dx
y
□
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37
Derivatives of Exponential and Logarithmic Functions
Consider the exponential functions f (x) = a x , a > 0. All such functions have (0, 1) as their only
intercept. The derivative at x = 0 of each exponential function is of great significance. Using the limit
definition to calculate f ′ (0) we get
f (0 + h) − f (0)
h
0+h
a − a0
= lim
h→0
h
ah − 1
= lim
.
h→0
h
f ′ (0) = lim
h→0
Thus for each exponential function f (x) = a x , the slope of the tangent at x = 0 is given by the limit
ah − 1
.
h→0
h
f ′ (0) = lim
We will use the following fact, without proof: Of all possible exponential functions, the natural exponential function f (x) = e x is the only one whose tangent line at x = 0 has slope 1.
Thus for f (x) = e x , f ′ (0) = 1, i. e.,
eh − 1
=1
h→0
h
lim
Consequently, we have the following important derivative theorem.
Theorem 3.18 (Derivative of e x ).
d x
e = ex
dx
Proof. In class.
Note: Multiples of the natural exponential function are the only functions whose derivatives are themselves, i. e.,
d
Ce x = Ce x
dx
Worked Example 3.8.1. Find y′ in the following.
√
1. y = e x +1 .
2
2. y = e2x cos(πx).
3. xey − √ye x =√2.
4. y = e x + e x .
To find the derivative of y = a x for any exponential function we resort to the properties of exponentials
and logarithms, in particular
eln x = x and ln xr = r ln x.
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38
Theorem 3.19 (Derivative of a x ).
d x
a = a x ln a.
dx
Proof. In class.
Worked Example 3.8.2. Find y′ if
1. y = 5 x .
2
2. y = 35x 23x .
Tutorial 3.8.1. 1. [C1] §3.9, p. 331: 331, 332, 334, 337–339.
dy
2
2. Find
if e x y = 2x + y2 .
dx
3. Using the Chain Rule we have
d f (x)
e
=
dx
4. Using the Chain Rule we have
d f (x)
a
=
dx
Theorem 3.20 (Derivative of ln x).
1
d
ln x =
dx
x
Proof. In class.
Together with the Chain Rule we have
d
f ′ (x)
ln[ f (x)] =
dx
f (x)
Worked Example 3.8.3. Find y′ if
1. y = xr for r ∈ R.
2. y = ln(ln x).
√
3. y = ln(x + x2 + 1).
4. y3 + x2 ln y = 5x + 3.
x2 (2x − 1)3
5. y = ln
.
(x + 5)2
Theorem 3.21 (Derivative of loga x).
d
1
loga x =
.
dx
x ln a
Proof. In class.
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39
Worked Example 3.8.4. Find f ′ (x) if
6x + 4
.
1. f (x) = log5
2x − 3
2. f (x) = ln(log10 x).
Herewith a summary of exponential and logarithmic derivatives.
y
dy
dx
Using Chain Rule with u = f (x)
ex
ex
y = eu ⇒ y′ = eu
ax
a x ln a
y = au ⇒ y′ = au ln a
ln x
1
x
y = ln u ⇒ y′ =
loga x
1
x ln a
y = loga u ⇒ y′ =
du
dx
du
dx
1 du
u dx
1 du
u ln a dx
Tutorial 3.8.2. 1. [C1] §3.9, p. 331: 333, 340–342, 344, 345.
2. Using the Chain Rule we have
d
loga f (x) =
dx
3.9
Linear Approximation and Differentials
Using the derivative, we will illustrate one of the simplest types of approximation methods, namely
linear approximation. The idea is to use the value of the function and its derivative at a single point to
find approximate values of the function at nearby points.
Assume that f is a differentiable function at x = a and that we are interested in approximate values of
f (x) for x near a. We use the tangent line of f at (a, f (a)) as an approximation of f (x) when x is near a.
The tangent line is a graph of a linear function. The approximation
f (x) ≈ f (a) + f ′ (a)(x − a) for x ≈ a
is the linear approximation of f at a. The linear function
L(x) = f (a) + f ′ (a)(x − a)
whose graph is the tangent line is called the linearization of f at a.
Worked Example 3.9.1. Find the linearization
√ (linear approximation) of the function f (x) =
near 1 and use it to approximate the value of 1.02.
√
x at x
In many problems we are interested in the change, or approximate change, of values y = f (x) that
correspond to a change in x. If x changes from x1 to x2 , this amount of change is denoted by ∆x.
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Similarly ∆y (∆ f ) will denote the change in the dependent variable y that corresponds to the change ∆x.
Thus
∆y = f (x2 ) − f (x1 ) = f (x1 + ∆x) − f (x1 ).
In the linear approximation, if ∆x = x − a represents the change in x, the corresponding change in
y = f (x) is ∆y = f (x) − f (a). If f is differentiable at a, then we know that
f (x) ≈ f (a) + f ′ (a)(x − a)
so that
f (x) − f (a) ≈ f ′ (a)(x − a),
i. e.,
∆y ≈ f ′ (a) ∆x.
We use a notation that suggests the derivative in the linear approximation. We write dy for ∆y and dx
for ∆x. The differential of y = f (x) at x = a is defined as dy where
dy = f ′ (a) dx.
The symbols dy and dx represent variables with dy dependent on dx. dy is called the differential
approximation of the change in y = f (x).
Worked Example 3.9.2. Find the differentials dy of the following functions.
√
1. y = 1 + x2 .
2. y = (1 + 2r)−4 .
3. y = t2 et .
The following are rules for working with differentials. If c ∈ R and u and v are functions of x, then
differential of a constant
dc = 0
factorization
d(cu) = c du
differential sum
d(u + v) = du + dv
differential product
d(uv) = v du + u dv
differential quotient
d
differential power
d(xn ) = nxn−1 dx
u
v
=
v du − u dv
v2
Tutorial 3.9.1. [C1] §4.2, pp. 364–365: 46, 49–56, 58–60, 68–71, 87–89.
Chapter 4
Introduction to Integration
NOTE: Integration is particularly important to those students studying statistics as it is used widely in
continuous distributions.
LEARNING OUTCOMES:
On completion of this chapter you should (tick the checkbox when you have mastered the skill)
□ 1. know the definition of an antiderivative,
□ 2. be able to verify that a function is an antiderivative of a given function,
□ 3. understand the indefinite integral as an antiderivative,
□ 4. know the antiderivatives or integrals of all the known functions,
□ 5. know the properties of the definite integral,
□ 6. know how to apply the Fundamental Theorem of Calculus (FTC) in computing a definite integral,
□ 7. know and understand how to use the Substitution Rule in evaluating a definite or indefinite
integral,
□ 8. know and understand how to use the technique of Integration by Parts (IBP) in evaluating a
definite or indefinite integral,
□ 9. know how to solve the examinable tutorial problems and the worked out problems from these
notes on this chapter.
For a given function f we have been able to compute the (unique) derivative f ′ . In this chapter we
consider the following question.
For a given function f , can we find a function (or functions) F such that F ′ (x) = f (x) for each x in
an interval?
4.1
Antidifferentiation and the Indefinite Integral
A function F is called an antiderivative of the function f on an interval I if F ′ (x) = f (x) for all x ∈ I.
Worked Example 4.1.1. Verify that F(x) is an antiderivative of f (x) in the following.
1. F(x) = −x cos x + sin x and f (x) = x sin x.
√
3
2. F(x) = 31 (x2 + 1) 2 and f (x) = x x2 + 1.
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3. F(x) = x ln x − x and f (x) = ln x.
4. F(x) = e x and f (x) = e x .
ax
5. F(x) =
and f (x) = a x .
ln a
Antiderivatives are never unique. Since the derivative of a constant is zero, it follows that if F is an
antiderivative of f , then so are the functions G(x) = F(x) + C for every number C. Functions of this
type, F(x) + C, are the only possible antiderivatives of f .
If F is an antiderivative of f , then the most general antiderivative of f on an interval I is F(x) + C,
where C is an arbitrary constant. The most general antiderivative is denoted by the symbol
Z
f (x) dx
and called theRindefinite integral of f . The function f (x) is called the integrand of the indefinite
integral. Here and dx are merely parts of a symbol and have no individual meaning. Thus if we know
that F is an antidervative of f , then the most general antiderivative or indefinite integral of f is
Z
f (x) dx = F(x) + C.
Based on the previous Worked Example we may conclude the following results.
R
Worked Example 4.1.2. 1. x sin x dx = −x cos x + sin x + C.
R √
3
2. x x2 + 1 dx = 13 (x2 + 1) 2 + C.
3.
R
ln x dx = x ln x − x + C.
4.
R
e x dx = e x + C.
5.
R
a x dx =
ax
+ C.
ln a
The process of indefinite integration is thus the procedure of finding the most general antiderivative of
a given function.
Z
f (x) dx is all solutions to the question “What must I differentiate to get the integrand f (x)?”
Theorem 4.1 (Integral of power functions). Let f (x) = xr where −1 , r ∈ R. Then
Z
Proof. Let F(x) =
Thus
xr dx =
1 r+1
x + C.
r+1
1 r+1
x . Then
r+1
F ′ (x) =
1
(r + 1)xr+1−1 = xr .
r+1
Z
1 r+1
x + C.
r+1
xr dx =
□
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43
Theorem 4.2 (Integral of 1x ).
Z
1
dx = ln |x| + C.
x
Proof. In class.
Theorems 4.1 and 4.2 give the integral of the power functions for each r ∈ R:
Z
Note.

 1 xr+1 + C
r
x dx = r + 1

ln |x| + C
if r , −1,
if r = −1.
1
d
f ′ (x)
d
ln |x| = and the Chain Rule give
ln | f (x)| =
. Thus
dx
x
dx
f (x)
Z
f ′ (x)
dx = ln | f (x)| + C.
f (x)
Worked Example 4.1.3. Find y′ if
1. y = ln | sin x|.
2. y = ln | sec x|.
3. y = ln | sec x + tan x|.
4. y = ln | cosec x − cot x|.
The derivative knowledge acquired in the previous chapter and the examples above provide a table of
antiderivatives or indefinite integrals.
1st Semester Calculus Study Guide
44
e x dx = e x + C
Z
a x dx =
ax
+C
ln a
Z
ln x dx = x ln x − x + C
Z
sin x dx = − cos x + C
Z
cos x dx = sin x + C
Z
Z
Z
tan x dx = ln | sec x| + C
cosec x dx = ln | cosec x − cot x| + C
Z
sec x dx = ln | sec x + tan x| + C
Z
cot x dx = ln | sin x| + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
Z
cosec2 x dx = − cot x + C
Theorem
4.3 (Properties
integral).
R
R of the indefinite
R
1. [ f (x) + g(x)]dx = f (x) dx + g(x) dx
2.
R
3.
R
[ f (x) − g(x)]dx =
k f (x) dx = k
R
R
f (x) dx −
R
g(x) dx
f (x) dx for any k ∈ R.
Worked Example 4.1.4. Find the indefinite integrals of the following functions.
1. f (x) = 10x4 − 6x3 + 5.
2. g(u) = 2eu − 3 sin u + 2 cos u.
x2 + 3x + 1
.
x
4. t(x) = x(3x2 + 1)2 .
3. h(x) =
5. s(t) = 2t + 3 sec t − ln t3 .
It is important to note that
Z
f ′ (x) dx = f (x) + C.
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45
Worked Example 4.1.5. 1. Find f if f ′ (t) = 2 cos t − sec2 t, − π2 < t < π2 .
2. Find g if g′ (x) = 12x2 − 6x + 1 and g(1) = 5.
Tutorial 4.1.1. 1. [C1] §4.10, p. 497: 470–481, 490, 492, 494, 496, 498.
2. [C1] §4.10, p. 497: 499–502.
4.2
The Definite Integral & FTC
A definite integral is a particular limit involving continuous functions that we will encounter and evaluate in more detail in Chapter 8, a limit of Riemann Sums.
Let f be a continuous function that is defined on a closed interval [a, b]. A partition of [a, b] is any
decomposition of [a, b] into subintervals of the form
[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], . . . , [xn−1 , xn ]
where n is a positive integer and xi are numbers such that
a = x0 < x1 < x2 < x3 < · · · < xn−1 < xn = b.
The length of each subinterval [xi−1 , xi ] is denoted by ∆xi , i. e,
∆xi = xi − xi−1 .
We will mostly partition the interval [a, b] into n subintervals of equal length. Their common length is
then
b−a
∆x =
n
and the points of subdivision are given by the formula
xi = a + i∆x.
The symbol for a definite integral and the limit that defines it are
Zb
f (x) dx = lim
n→∞
a
n
X
f (xi∗ )∆x
i=1
where xi∗ is any sample point in the subinterval [xi−1 , xi ]. a and b are called the limits of integration,
particularly a is the lower limit and b the upper limit. Note that the limit exists and so
Rb
f (x) dx ∈ R.
a
The process of finding the number represented by the limit defining
Rb
f (x) dx is called integration or
a
evaluating the integral. In the notation of the definite integral, the letter x is a dummy variable; the
symbol used for the independent variable is immaterial. Thus
Zb
a
f (x) dx =
Zb
a
f (t) dt =
Zb
f (r) dr = . . .
etc.
a
The definite integral and the indefinite integral (an antiderivative) are related by the Fundamental Theorem of Calculus, part 2 (FTC2). The FTC2 provides a means of calculating the definite integral without
evaluating the limit that defines it. The FTC1 and FTC2 will be encountered in more detail in the second
semester.
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Theorem 4.4 (FTC2). If f is continuous on [a, b], then
Zb
f (x) dx = F(x)
òb
= F(b) − F(a),
a
a
where F is any antiderivative of f .
Note. If f is non-negative and continuous on [a, b], then the definite integral corresponds to the area
under the graph of f .
Worked Example 4.2.1. Evaluate the integrals using FTC2.
1.
R4 1
1
2
2.
√ dx.
x
R 4 + x2
1
x3
dx.
π
3.
R4
sec2 x dx.
− π4
The following are properties of the definite integral.
Zb
f (x) dx = −
Za
a
Za
f (x) dx
b
f (x) dx = 0
a
Zb
c f (x) dx = c
a
Zb
Zb
f (x) dx for constants c
a
( f (x) + g(x)) dx =
a
Zb
Zb
f (x) dx +
a
( f (x) − g(x)) dx =
a
Zb
g(x) dx
a
Zb
Zb
f (x) dx −
a
g(x) dx
a
Tutorial 4.2.1. [C1] §5.3, p. 563–564: 170–173, 176–180, 182–184.
4.3
The Substitution Rule
lt is clear from the FTC2 that the indefinite integral
ation of the definite integral
Rb
a
R
f (x) dx = F(x) + C greatly facilitates the evalu-
f (x) dx. When we differentiate a composite function we use the Chain
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47
Rule. In attempting to calculate an indefinite integral, we often use the Chain Rule in reverse called the
Substitution Rule or the Substitution Method.
Theorem 4.5 (Indefinite Integral Substitution Rule). If u = g(x) is a differentiable function whose range
is contained in an interval I and f is continuous on I, then
Z
Z
′
f (g(x))g (x) dx =
f (u) du.
Worked Example 4.3.1. Use the Substitution Rule to evaluate the following indefinite integrals.
1.
R
(3x + 2)7 dx.
R √3
2. x 7 − 6x2 dx.
R
x2 sin x3 dx.
R cos x
4.
dx.
√
R 1 + sin x
3.
tan 4x sec2 4x dx.
5.
Worked Example 4.3.2. Evaluate
1.
R
e2x dx.
2.
R
sin 5x dx.
Theorem 4.6 (Definite Integral Substitution Rule). If g is differentiable on [a, b], if g′ is continuous on
[a, b], and if f is continuous on the range of g, then
Zb
f (g(x))g (x) dx =
Zg(b)
′
f (u) du.
g(a)
a
Worked Example 4.3.3. Evaluate the following definite integrals.
1.
R1
x(x2 + 1)3 dx.
0
π
2.
R3 sin θ
0
π
3.
R
cos2 θ
dθ.
sin2 x dx.
−π
Tutorial 4.3.1. 1. [C1] §5.5, p. 592–593: 261, 263, 265–273, 275, 276, 278, 279, 282, 286.
2. [C1] §5.5, p. 593: 292, 293, 295–297.
3. Let a and b be positive numbers. Show that
Za
Za
i. if f is continuous on [0, a], then
f (x) dx =
f (a − x) dx;
0
Zab
ii.
a
1
dx =
x
Zb
1
1
dx.
x
0
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4. Let f be a continuous function with antiderivative F and let k be a nonzero real number. Show that
Z
1
f (kx) dx = F(kx) + C.
k
Also formulate and prove a corresponding formula for definite integrals.
Use the above formula to complete the following table:
Z
Z
4.4
ekx dx =
Z
sin kx dx =
Z
cos kx dx =
Z
sec2 kx dx =
sec kx tan kx dx =
Integration by Parts
Theorem 4.7 (Integration by parts, IBP). If u = f (x) and v = g(x) where f ′ and g′ are continuous, then
Z
Z
u dv = uv − v du.
Worked Example 4.4.1. Use IBP to solve the following integrals.
1.
R
xe x dx.
2.
R
ln x dx.
3.
R
e x cos x dx.
Evaluating a definite integral using IBP leads to the following
Zb
òb Zb
u dv = uv − v du.
a
a
a
Alternatively, the indefinite integral may first be solved and then using that solution we may evaluate the
definite integral.
Worked Example 4.4.2. Evaluate
π
Z2
x cos x dx.
0
Tutorial 4.4.1. 1. [C2] §3.1, pp. 270–271: 6, 10, 13, 15–17, 23, 53–55.
2. In questions 6, 10, 13 from part 1, evaluate the definite integrals over the interval [0, 2].
Chapter 5
Differentiation Techniques
LEARNING OUTCOMES:
On completion of this chapter you should (tick the checkbox when you have mastered the skill)
□ 1. be able to find the higher order derivatives of a given function,
□ 2. be able to find the n -degree Taylor poynomial of a given function at a given point,
□ 3. be able to find the derivative of the inverse f of a given invertible function f at a given point
without deriving the formula for f ,
□ 4. be able to find and know the proofs of the derivatives (and integrals) of the inverse trigonometric
functions,
□ 5. be able to use the technique of logarithmic differentiation to evaluate derivatives involving complicated products, quotients or powers,
□ 6. know how to solve the examinable tutorial problems and the worked out problems from these
th
−1
−1
notes on this chapter.
5.1
Higher Order Derivatives and Taylor Polynomials
Since the derivative f ′ of a function f is itself a function, we may consider successive derivatives of
functions. The derivative of the first derivative is called the second derivative, the derivative of the
second derivative is the third derivative, etc. We use the following notations for y = f (x).
1st derivative
2nd derivative
3rd derivative
..
.
nth derivative
dy
dx
Å ã
d dy
d2 y
= 2
dx dx
dx
Ç
å
2
d d y
d3 y
=
dx dx2
dx3
..
.
Ç
å
d dn−1 y
dn y
=
dx dxn−1
dxn
49
y′ = f ′
y(1) = f (1)
y′′ = f ′′
y(2) = f (2)
y′′′ = f ′′′
y(3) = f (3)
..
.
y(n) = f (n)
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Worked Example 5.1.1. 1. Find the first four derivatives of
3
(a) y = 4x2 − 5x + 8 − .
x
(b) f (x) = sin x.
1
(c) h(x) = . Hence find a formula for the nth derivative h(n) (x).
x
2. Find y′′ if x2 − 3xy + y2 = 4.
Note. If f is any function that has n derivatives at x = a, we can form the nth -degree Taylor polynomial
of f about a (note that 0! = 1! = 1)
T n (x) = f (a) + f ′ (a)(x − a) +
=
n
X
f (k) (a)(x − a)k
k=0
k!
f ′′ (a)(x − a)2 f ′′′ (a)(x − a)3
f (n) (a)(x − a)n
+
+ ··· +
2!
3!
n!
.
Note that the first-order Taylor polynomial of f about a is
T 1 (x) = f (a) + f ′ (a)(x − a)
so that the graph of T 1 is the tangent line at a.
Worked Example 5.1.2. 1. If f (x) = ln x, find the fourth degree Taylor polynomial at x = 1.
2. Find the Taylor polynomial of degree n of f (x) = e x at x = 0.
Taylor polynomials will be studied in more detail in the last block of the second semester in Chapter 11
of these notes.
Tutorial 5.1.1. 1. Find the first and second derivatives of
x2 − 5
a. y = x5 − 3x2 − 2x, b. y =
, c. y = e x cos x, d. y = sec x.
x
2. a. Find the Taylor polynomial of order (degree) 2 about 2 in part 1b and about 0 in part 1d.
b. Find th e Taylor polynomial of order 3 about 0 in parts 1a and 1c.
3. [C1] §3.8, p. 317: 322.
4. [C2] §6.3, p. 578: 117, 118, 120. Also find the Taylor polynomials of order 3 and 4.
5.2
Continuity and Differentiation of Inverse Functions
Recall that a function f is invertible if and only if f is a one-to-one function, i. e.,
f (x) = f (y) ⇒ x = y,
equivalently,
x , y ⇒ f (x) , f (y).
For such functions the inverse f −1 is defined by
f −1 (y) = x ⇔ y = f (x).
In particular, e x and ln x are invertible and are inverses of each other.
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Theorem 5.1. If f is a one-to-one continuous function defined on an interval, then its inverse function
f −1 is also continuous.
Given a one-to-one differentiable function f whose derivative is never zero, the inverse function is also
differentiable in its domain and its derivative may be found at given points without actually deriving the
formula for f −1 and then finding its derivative ( f −1 )′ , as illustrated by the following theorem.
Theorem 5.2. If f is a one-to-one differentiable function with inverse function f −1 and f ′ ( f −1 (a)) , 0,
then the inverse function is differentiable at a and
( f −1 )′ (a) =
Worked Example 5.2.1. 1. Let f (x) =
√
1
f ′ ( f −1 (a))
.
x − 2. Find ( f −1 )′ (2) by
(a) first finding a formula for f −1 , then differentiating to get a formula for ( f −1 )′ (x) and finally evaluating
( f −1 )′ (2),
(b) using Theorem 5.2 and confirm that (a) and (b) agree.
2. If g(x) = 2x + e x , show that g is invertible and then find (g−1 )′ (1) without finding the formula for g−1 .
Tutorial 5.2.1. [C1] §3.7, p. 306–307: 265, 266, 269–271, 273, 278, 289.
5.3
Inverse Trigonometric Derivatives and Integrals
Recall the definitions and properties of the inverse trigonometric functions done in Algebra. In this
section we derive the derivative formulas for the inverse trigonometric functions arcsin x, arccos x and
arctan x. We will also incorporate the indefinite integrals of these functions. All proofs are examinable.
Theorem 5.3 (Derivative of arcsin x).
d
1
arcsin x = √
,
dx
1 − x2
−1 < x < 1.
Proof. In class.
1
The above theorem informs that √
is an antiderivative of arcsin x so that the associated indefinite
1 − x2
integral is
Z
1
dx = arcsin x + C.
√
1 − x2
Worked Example 5.3.1. √Find y′ if
1. (a) y = x arcsin 2x + 12 1 − 4x2 .
Å
ã
x
(b) y = arcsin √
.
1 + x2
2. y = arcsin(cos x).
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Solution. For | cos x| < 1, i. e., x , kπ, k ∈ Z, we have
y = arcsin(cos x) ⇒
1
d
dy
= √
cos x
dx
1 − cos2 x dx
1
= p
(− sin x)
sin2 x
sin x
=−
| sin x|
(
−1 if sin x > 0,
=
1
if sin x < 0.
This shows that the graph of arcsin(cos x) is saw-tooth like, i. e., it is composed of line segments with
slopes alternating between 1 and −1.
A direct computation of the graph of y = arcsin(cos x) is given below.
y = arcsin(cos x) ⇒ sin y = cos x
π
⇔ cos y −
= cos x
2
π
⇔ y − = ±x + 2kπ, k ∈ Z.
2
Since y −
π
∈ [−π, 0], we therefore have
2
π

 − x + 2kπ
2
y= π

 + x − 2kπ
2
if x ∈ [2kπ, (2k + 1)π), k ∈ Z,
if x ∈ [(2k − 1)π, 2kπ), k ∈ Z.
Below is the graph of y = arcsin(cos x).
−3π
π
−π
−2π
3. Evaluate
1
Z4
0
1
dx.
√
1 − 4x2
4. Prove that
Z
x
1
dx = arcsin
+ C, a > 0
√
a
a2 − x2
2π
3π
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and then evaluate
Z
53
1
dx.
√
3 − x2
5. Use integration by parts to prove
Z
arcsin x dx = x arcsin x +
p
1 − x2 + C.
Theorem 5.4 (Derivative of arccos x).
1
d
arccos x = − √
.
dx
1 − x2
Proof. In class.
Worked Example 5.3.2. Find y′ if x2 + x arccos y = ye x .
The integral of arccos x may be omitted as integrals of that type can be realised in terms of arcsin x. The
more preferred derivative and integral is that of arcsin x.
Theorem 5.5 (Derivative of arctan x).
d
1
.
arctan x =
dx
1 + x2
Proof. In Class.
The corresponding indefinite integral is
Z
1
dx = arctan x + C.
1 + x2
Worked Example 5.3.3. 1. Find y′ if
x+1
(a) y = arctan
.
x−1
(b) y = ln arctan x2 .
2. Evaluate
Z1
0
ex
dx.
1 + e2x
3. Prove that
Z
x
1
1
dx
=
arctan
+ C.
a
a
a2 + x 2
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54
x
Proof. Let u = . Then au = x so that a du = dx. Then
a
Z
Z
1
1
dx = a
du
2
2
2
2 2
a +x
Z a +a u
1
=a
du
2
a (1 + u2 )
Z
1
1
du
=
a
1 + u2
1
= arctan u + C
a
x
1
+ C.
= arctan
a
a
4. Find
Z
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□
x+9
dx.
x2 + 9
5. Use integration by parts to prove that
Z
arctan x dx = x arctan x −
1
ln(1 + x2 ) + C.
2
Tutorial 5.3.1. 1. [C1] §3.7, p. 307: 279, 283–285.
2. [C1] §5.7, pp. 613–614: 391, 394, 397, 400, 401, 411, 413, 428.
3. Using the Chain Rule,
d
arcsin f (x) =
dx
5.4
d
arctan f (x) =
dx
Logarithmic Differentiation
Given y = f (x), y′ can also be found by the process called logarithmic differentiation. This method
is especially useful if f (x) involves complicated products, quotients, or powers. The properties of the
(natural) logarithm allows simplification of f (x).
The following outlines the technique of logarithmic differentiation.
Process to follow in logarithmic differentiation
1. Given:
y = f (x)
2. Take the natural logarithms on both sides and simplify using properties of the natural logarithm:
ln y = ln f (x)
3. Differentiate implicitly with respect to x:
d
d
y′
d
ln y =
ln f (x) ⇒
=
ln f (x)
dx
dx
y
dx
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4. Solve for y′ by cross-multiplication:
Å
d
y =y
ln f (x)
dx
′
ã
Worked Example 5.4.1. Use the method of logarithmic differentiation to find y′ if
x2 (x − 1)3
.
1. (Complicated quotient): y =
x+1
√ 2
2. (Complicated product): y = xe x (x2 + 1)10 .
x
3. (Complicated power): y = 2 x .
Theorem 5.6 (General Power Rule).
∀r ∈ R
d r
x = rxr−1
dx
Proof. Let y = xr . Using logarithmic differentiation we have
ln |y| = ln |xr |
= ln |x|r
= r ln |x|
y′ r
=
∴
y
x
ry
⇒ y′ =
x
rxr
=
x
Hence
d r
x = rxr−1 .
dx
□
Tutorial 5.4.1. [C1] §3.9, p. 331: 346–353.
5.5
Partial Differentiation
In the previous sections and chapters we have had a lot of experience differentiating functions of one
variable (for example, f (x)). But what about a function of two variables, say, f (x, y)? In this section we
will consider functions of more than one variable and discuss how to differentiate them.
Here we introduce the notion of a partial derivative of a function of more than one variable. A more
comprehensive reading into partial derivatives (Section 4.3 in [C3] or Chapter 14 in Thomas or Chapter 15 in Stewart) will facilitate your understanding of this section. Partial derivatives are used in solving
a particular type of first order differential equations as will be seen in Chapter 12 in the second semester.
They are also used in Statistics for regression and correlation.
We will be concerned mainly with functions of two variables f (x, y). The graph of a function f (x, y) of
two variables x and y is the set of all points (x, y, z) in three-dimensional space such that z = f (x, y).
Generally, f (x, y) represents a surface in three-dimensional space R3 .
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Worked Example 5.5.1. Let f (x, y) =
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sin(x + y)
. Find f (2, −2) and f (π, π2 ).
x
If f is a function of two variables, then the first partial derivative of f (x, y) with respect to x is
f (x + h, y) − f (x, y)
h→0
h
f x (x, y) = lim
(y is fixed)
and the first partial derivative of f (x, y) with respect to y is
fy (x, y) = lim
h→0
f (x, y + h) − f (x, y)
h
(x is fixed)
provided that these limits exist.
By the above limit definitions,
to find f x (x, y) we may regard y as a constant and differentiate f (x, y) with respect to x in the usual
way
and
to find fy (x, y) we may regard x as a constant and differentiate f (x, y) with respect to y in the usual
way.
Other common notations for the two first partial derivatives of f (x, y) are
fx =
∂f
∂x
and
fy =
∂f
.
∂y
For the geometric interpretation of the first partial derivatives, consult a textbook.
Worked Example 5.5.2. If f (x, y) = x2 y3 , evaluate f x (3, 5) and fy (3, 5).
Worked Example 5.5.3. Let f (x, y) = x2 y + sin(xy2 ). Find f x and fy .
Worked Example 5.5.4. Let z = xe xy . Find z x and zy .
If f is a function of two variables x and y, then f x and fy are also functions of two variables and we
may consider their first partial derivatives. These are called the second partial derivatives of f and are
denoted as follows:
Å ã
∂ ∂f
∂2 f
1. f xx = ( f x ) x =
=
. To find f xx we differentiate f partially with respect to x twice
∂x ∂x
∂x2
whilst keeping y constant.
Å ã
∂2 f
∂ ∂f
=
. To find f xy we differentiate f partially with respect to x whilst
2. f xy = ( f x )y =
∂y ∂x
∂y∂x
keeping y constant, then differentiate again partially with respect to y keeping x constant.
Å ã
∂ ∂f
∂2 f
3. fyx = ( fy ) x =
=
. To find fyx we differentiate f partially with respect to y whilst
∂x ∂y
∂x∂y
keeping x constant, then differentiate again partially with respect to x keeping y constant.
Å ã
∂ ∂f
∂2 f
4. fyy = ( fy )y =
=
. To find fyy we differentiate f partially with respect to y twice
∂y ∂y
∂y2
whilst keeping x constant.
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Worked Example 5.5.5. Find the second partial derivatives of f if f (x, y) = x3 y2 − 2x2 y + 3x.
Third and higher partial derivatives are defined in a similar fashion, e. g.,
Ç
å
∂ ∂2 f
∂3 f
f xxx =
=
∂x ∂x2
∂x3
and
∂
f xyx =
∂x
Ç
∂2 f
∂y∂x
å
=
∂3 f
.
∂x∂y∂x
For functions of three or more variables, the technique for calculating the partial derivatives works in
the same way as we have done above, with all variables being kept constant except the one with respect
to which we want to differentiate.
Tutorial 5.5.1. [C3] §4.3, pp. 386–387: 120, 121, 123, 124, 127, 135–141.
Chapter 6
Applications of Differentiation
ln this chapter we will consider some of the applications of the knowledge of the various differentation
techniques that you have acquired in the previous chapters.
6.1
Indeterminate Forms and l’Hôpital’s Rule
LEARNING OUTCOMES:
On completion of this section you should be able to (tick the checkbox when you have mastered the
skill)
□ 1. identify an indeterminate form of a limit,
□ 2. use l’Hôpital’s Rule in evaluating limits of indeterminate forms , , 1 , 0 · ∞, 0 , ∞ , ∞ − ∞,
□ 3. know how to solve the examinable tutorial problems and the worked out problems from the
0 ∞
0 ∞
∞
0
0
notes on this section.
Theorem 6.1 (l’Hôpital’s Rule). Suppose that f and g are differentiable functions and g′ , 0 near x = a
(except possibly at a). Suppose that
lim f (x) = 0
and
lim f (x) = ±∞
and
x→a
lim g(x) = 0
x→a
or that
x→a
lim g(x) = ±∞.
x→a
(In other words, we have an indeterminate form of the type 00 or ∞
∞ ). Then
f (x)
f ′ (x)
= lim ′
x→a g(x)
x→a g (x)
lim
if the limit on the right hand side exists (or is ∞ or −∞).
The theorem also applies to one-sided limits.
2x2 − 2
.
x→−1 x + 1
Worked Example 6.1.1 (Type 00 ). 1. lim
e2x − 1
.
x→0 e x − 1
2. lim
58
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1st Semester Calculus Study Guide
59
e x + e−x − 2
.
x→0 1 − cos 2x
2x − 1
4. lim
.
x→0
x
3. lim
2x2 − 2x + 3
.
x→∞
3x2 + 1
Worked Example 6.1.2 (Type ∞
∞ ). 1. lim
ln x
2. lim √ .
x→∞
x
Worked Example 6.1.3 (Type 0 · ∞). Evaluate lim x cosec x.
x→0
ã
Å
1 x
= e, using l’Hôpital’s Rule.
1+
x→∞
x
Worked Example 6.1.4 (Type 1∞ ). 1. Prove that lim
Å
ã
1 x
. Then y is an indeterminate of the form 1∞ . So we can rewrite this form
Proof. Let y = lim 1 +
x→∞
x
into the form 00 and use l’Hôpita1’s rule as follows.
ã
Å
1 x
“1∞ ”
y = lim 1 +
x→∞
x
Å
ã
1 x
⇒ ln y = ln lim 1 +
x→∞
x
Å
ã
1 x
= lim ln 1 +
since the function ln is continuous
x→∞
x
Å
ã
1
= lim x ln 1 +
x→∞
x
ã
Å
1
ln 1 +
“ 0”
x
= lim
x→∞
1
0
x
Ö
è
Å
ã
1
1
− 2
1
x
1+
l’H
Åx ã
= lim
x→∞
1
− 2
x
1
= lim
x→∞
1
1+
x
= 1.
Thus we have shown that ln y = 1. By the property of inverse functions, y = eln y = e1 = e. Since y is
the original limit we have proved that
Å
ã
1 x
lim 1 +
= e.
□
x→∞
x
Note. When using l’Hôpital’s rule, as illustrated above, indicate the indeterminate form (in the above
l’H
case 00 ) adjacent to the limit and then followed by the = notation indicating that l’Hôpital’s rule is used.
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1
2. lim (1 + 3x) 2x .
x→0
Worked Example 6.1.5 (Type 00 ). lim+ (sin x) x .
x→0
1
Worked Example 6.1.6 (Type ∞0 ). Evaluate lim x x .
x→∞
ã
1
1
.
Worked Example 6.1.7 (Type ∞ − ∞). Evaluate lim x
−
x→0 e − 1
x
Å
Note. ∞ · ∞ and ∞ + ∞ are not indeterminate forms. Do not misuse l’Hôpital’s Rule as the following
example illustrates.
cos 2x
.
x→0
x2
1
Solution. This is a limit of the form which is not an indeterminate form. The use of l’Hôpital’s Rule
0
produces the incorrect solution
Worked Example 6.1.8 (Misuse of l’Hôpital’s Rule). Evaluate lim
cos 2x
−2 sin 2x
−4 cos 2x
= lim
= lim
= −2
2
x→0
x→0
x→0
2x
2
x
lim
WRONG!
cos 2x
= ∞.
x→0
x2
However, the correct analysis yields lim
Note. Do not get influenced by quotients and immediately use l’Hôpital’s Rule. You must first identify
that we have one of the indeterminate forms 1∞ , 00 , 0 · ∞, 00 , ∞
∞ or ∞ − ∞ in order to use l’Hôpital’s
Rule.
Tutorial 6.1.1. [C1] §4.8, p. 470-471: 356, 358, 359, 361–365, 369, 371, 378, 380, 382–384, 388, 392
6.2
Related Rates
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. understand the idea of a related rate,
□ 2. be able to identify, model and solve a related rate problem using differential calculus,
□ 3. know how to solve the examinable tutorial problems and the worked out problems from these
notes in this section.
Let f (t) be a differentiable function where t represents time. The
average rate of change of f in the interval [t, t + h] is
f (t + h) − f (t)
h
and the
rate of change of f with respect to t is f ′ (t) =
df
f (t + h) − f (t)
= lim
.
dt h→0
h
f ′ (t) is often called the instantaneous rate of change of f with respect to t.
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61
Related Rate Strategy:
1. Read the problem very carefully.
2. Draw a diagram, where relevant, and introduce notation on the diagram assigning symbols
to all quantities that are functions of time.
3. Express the given information and indicate the quantities that vary on the diagram.
4. Specify, in derivative form, the unknown rate of change that you are looking for.
5. Write an equation that relates the various quantitites of the problem.
6. Use the geometry of the situation, if necessary, to eliminate one of the variables by substitution.
7. Use implicit differentiation, with the Chain Rule, to differentiate the simplified equation
w.r.t t.
8. Substitute the given information into the resulting equation and solve for the unknown
rate of change.
9. State the final answer, in coherent form, specifying the units that you are using.
Worked Example 6.2.1 (Sliding ladder, Pythagoras). A ladder 2m long is leaning against the side of a
building. If the bottom of the ladder slides away from the building horizontally at a rate of 0.2m/s, how
fast is the ladder sliding down the building when the top of the ladder is 1.2m above the ground?
Worked Example 6.2.2 (Cone, Similar Triangles). A conical tank 8m across the top and 6m deep is full
of water. The tank springs a leak at the bottom and loses water at a rate of 2 cubic meters per minute.
How fast is the water level dropping at the instance when the water is exactly 3 meters deep?
4
r
6
h
Worked Example 6.2.3 (Inclination, Trigonometry). A balloon leaves the ground 50m away from an
observer and rises vertically at a rate of 14m/min. At what rate is the angle of inclination of the observer’s
line of sight increasing at the instant when the balloon is exactly 50m above the ground?
Tutorial 6.2.1. [C1] §4.1, p. 350: 1–3, 9–11, 17, 29, 30.
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6.3
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Extrema and the Extreme Value Theorem
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. know and be able to explain the definition of an absolute (or global) maximum or minimum of
a function,
□ 2. know and be able to explain the definition of a local (or relative) maximum or minimum of a
funcion,
□ 3. know how to determine extrema graphically,
□ 4. know under which condition a given function will attain absolute extreme values on a closed
interval (the Closed Interval Method),
□ 5. know the statements and applications of the Extreme Value Theorem (EVT) and Fermat’s Theorem,
□ 6. know how to solve the examinable tutorial problems and worked out problems from these notes
on this section.
Let f be a function and c ∈ dom( f ).
f has an absolute (or global) maximum at c if f (c) ≥ f (x) for all x ∈ dom( f ).
f (c) is then the maximum of f in dom( f ). (c, f (c)) is the highest point of the graph of f .
f has an absolute (or global) minimum at c if f (c) ≤ f (x) for all x ∈ dom( f ).
f (c) is then the minimum of f in dom( f ). (c, f (c)) is the lowest point of the graph of f .
The minimum and maximum of a function f on an interval are called the extreme values of f or
extrema of f .
Worked Example 6.3.1. Find the absolute maximum and absolute minimum of the following functions
graphically.
1. f (x) = x2 + 1, 1 ≤ x ≤ 2.
2. g(x) = x2 + 1, −1 < x < 2.
x2 + 1
if x , 0,
2
if x = 0,
x2 + 1
if x , 0,
2
if x = 0,
(
3. h(x) =
(
4. t(x) =
− 1 ≤ x ≤ 2.
− 1 < x < 2.
Let c ∈ dom( f ). Then f has a local maximum at c if there exists an open interval (a, b) such that
c ∈ (a, b) and f (c) ≥ f (x) for all x ∈ (a, b) ∩ dom( f ). f (c) is also called a relative maximum.
Also, f has a local minimum at c if there exists an open interval (a, b) such that c ∈ (a, b) and f (c) ≤ f (x)
for all x ∈ (a, b) ∩ dom( f ). f (c) is also called a relative minimum.
The local maxima and minima are called local extrema.
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Worked Example 6.3.2. Find the local extrema of the following functions graphically.
1. f (x) = x2 + 1, −1 ≤ x ≤ 2.
2. g(x) = x2 + 1, −1 < x < 2.
(
x2 + 1 if x , 0,
3. h(x) =
− 1 ≤ x ≤ 2.
2
if x = 0,
(
x2 + 1 if x , 0,
4. t(x) =
− 1 < x < 2.
2
if x = 0,
5. f (x) = |x|.
6. g(x) = sin x.
Theorem 6.2 (Extreme Value Theorem – EVT). If f is continuous on a closed interval [a, b], then f
attains an absolute maximum value f (c) at some c ∈ [a, b] and an absolute minimum value f (d) at some
d ∈ [a, b].
Note. In the EVT, we cannot omit continuous or closed as Worked Example 6.3.1.3 (not continuous)
and 6.3.1.2 (open interval, no absolute maximum) illustrates. Continuity on a closed interval guarantees
the existence of BOTH absolute maximum AND absolute minimum.
Theorem 6.3 (Fermat’s Theorem). If f has a local maximum or minimum at an interior point c of
dom( f ) and if f is differentiable at c, then f ′ (c) = 0.
Proof. In class.
Note that the converse of Fermat’s Theorem is false, i. e., f ′ (c) = 0 ⇏ f (c) is a local maximum or a
local minimum.
Worked Example 6.3.3 (Converse of Fermat’s Theorem is false). Prove that f (x) = x3 has no local
extremum at x = 0.
A critical number of a function f is a number c ∈ dom( f ) such that
either f ′ (c) = 0 or f ′ (c) does not exist.
f ′ (c) does not exist
f ′ (c) = 0
Horizontal
tangent
c
Fermat’s Theorem may be restated as
lf f has a local maximum or minimum at c in the interior of dom( f ), then c is a critical number of f .
Thus the local extrema occur only at the critical numbers or at the endpoints of the domain. However,
the critical numbers of a function need not produce relative extrema.
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√3
Worked Example 6.3.4. Find the critical number(s) of f (x) = (x + 5)2 x − 4.
To find an absolute maximum or minimum of a continuous function on a closed interval, we note that
either it is local (in the interior of the domain of the function, in which case it occurs at a critical number
by Fermat’s Theorem) or it occurs at an endpoint of the closed interval. We thus have the following
method (The Closed Interval Method) to find absolute extrema of continuous functions on closed
intervals which we will exploit in Section 6.6.
The Closed Interval Method:
To find the absolute maximum and minimum values of a continuous function f on a closed interval
[a, b], use the following steps:
1. Find the critical numbers c ∈ (a, b) of f by finding x where f ′ (x) = 0 or where f ′ (x) does
not exist.
2. Evaluate f at each critical number c ∈ (a, b), i. e., evaluate f (c) for critical numbers c ∈
(a, b).
3. Evaluate f at each endpoint of [a, b], i. e., evaluate f (a) and f (b).
4. Compare the values of all f (c), f (a) and f (b).
The smallest of f (c), f (a) or f (b) is the global minimum. The greatest of these is the global
maximum.
Worked Example 6.3.5. Find the absolute maximum and minimum values of f (x) = 2 sin x − cos 2x
on [0, 2π].
2
Worked Example 6.3.6. Find the absolute maximum and minimum of g(x) = 2x − 3x 3 on [−1, 3].
Solution. We have
g (x) = 2 − 2x
′
− 31
=2−
2
1
x3
Ç
=2
1
x3 − 1
å
1
.
x3
The critical values are
1
g′ (x) = 0 ⇔ x 3 − 1 = 0 ⇔ x = 1
1
and g′ (x) does not exist ⇔ x 3 = 0 ⇔ x = 0.
Then comparing
2
g(−1) = 2(−1) − 3(−1) 3 = −2 − 3 = −5,
g(0) = 0,
g(1) = 2 − 3 = −1,
√3
g(3) = 6 − 3 9 ≈ −0.24
we see that we have an absolute maximum at x = 0 (critical point) and an absolute minimum at x = −1
(interval endpoint).
Tutorial 6.3.1. [C1] §4.3, pp. 376-377: 100–103, 108–112, 114, 116, 119, 121, 125.
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6.4
65
Rolle’s Theorem and the Mean Value Theorem
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. know Rolle’s Theorem and its proof,
□ 2. be able to use the IVT and apply Rolle’s Theorem,
□ 3. know the Mean Value Theorem (MVT) and its proof,
□ 4. be able to apply the MVT,
□ 5. know how to solve the examinable tutorial problems and the worked out problems from these
notes on this section.
Theorem 6.4 (Rolle’s Theorem). Let f be a function that satisfies the following three conditions:
1. f is continuous on the closed interval [a, b],
2. f is differentiable on the open interval (a, b),
3. f (a) = f (b).
Then there is a number c ∈ (a, b) such that f ′ (c) = 0.
Proof. There are three possible cases for f .
1. f is a constant function. Suppose that f (x) = k where k is some constant. Then clearly f ′ (x) = 0 for
each x. Thus for each c ∈ (a, b), f ′ (c) = 0.
2. f (x) > f (a) for some x ∈ (a, b).
f ′ (c) = 0
a
c
f ′ (c) = 0
a
b
c
b
By the first condition, f is continuous on [a, b] and thus we may apply the Extreme Value Theorem. By
the EVT f has a maximum value somewhere in [a, b]. Since f (a) = f (b) (by Condition 3), f must attain
its maximum value at some c ∈ (a, b). Then f has a local maximum at c. Since f is differentiable in
(a, b) by Condition 2, and c ∈ (a, b), f ′ (c) exists. Consequently, by Fermat’s Theorem, f ′ (c) = 0.
3. f (x) < f (a) for some x ∈ (a, b).
f ′ (c) = 0
a
c
f ′ (c) = 0
b
a
c
b
This case is similar to case 2, by the EVT f has a minimum value somewhere in [a, b]. Since f (a) = f (b),
f attains its minimum value at some c ∈ (a, b). Again, f ′ (c) = 0 by Fermat’s Theorem.
□
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We can see that in particular Rolle’s Theorem implies that if a differentiable curve crosses the x-axis
twice ( f (a) = f (b) = 0), there must be a point between successive crossings where the tangent to the
curve is parallel to the x-axis.
Rolle’s Theorem requires that f is continuous on [a, b] and differentiable on (a, b) with f (a) = f (b), i. e.,
all three conditions MUST be satisfied to guarantee at least one c ∈ (a, b) for which f has a horizontal
tangent (at c). Of course, there could be more than one point c in (a, b) for which the tangent lines are
horizontal at these points c.
Worked Example 6.4.1. Prove that the equation 2x − 1 − sin x = 0 has exactly one real root.
Worked Example 6.4.2. Let f (x) = |x − 2|. Then f satisfies f (1) = 1 = f (3), but there is no c ∈ (1, 3)
such that f ′ (c) = 0. Why does this not contradict Rolle’s Theorem?
Theorem 6.5 (Mean Value Theorem). Let f be a function that satisfies the following conditions:
1. f is continuous on the closed interval [a, b],
2. f is differentiable on the open interval (a, b).
Then there is a number c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
,
b−a
or equivalently,
f (b) − f (a) = f ′ (c)(b − a).
Proof. Consider the equation of the secant line AB.
f (x)
h(x)
A
y = f (x)
B
yAB
a
x
b
The slope of the secant line AB is
mAB =
f (b) − f (a)
.
b−a
Thus the equation of the secant line AB is
yAB − f (a) = mAB (x − a)
ã
Å
f (b) − f (a)
=
(x − a)
b−a
Å
ã
f (b) − f (a)
∴ yAB = f (a) +
(x − a).
b−a
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We now apply Rolle’s Theorem to the function
h(x) = f (x) − yAB = f (x) − f (a) −
Å
ã
f (b) − f (a)
(x − a)
b−a
defined as the difference between f and the function whose graph is the secant line AB.
We are going to show that h satisfies the conditions of Rolle’s Theorem.
1. Since yAB is a first degree polynomial, yAB is continuous on [a, b]. Thus h is a difference of continuous
functions, and hence h is also continuous on [a, b].
2. Since yAB is a first degree polynomial, yAB is differentiable on (a, b). Thus h is a difference of
differentiable functions and soÅh is also differentiable
on (a, b).
ã
f (b) − f (a)
3. Since h(a) = f (a) − f (a) −
(a − a) = 0 and
b−a
Å
ã
f (b) − f (a)
h(b) = f (b) − f (a) −
(b − a)
b−a
= f (b) − f (a) − ( f (b) − f (a))
= 0,
it follows that h(a) = h(b).
Thus h indeed satisfies the condition of Rolle’s Theorem. Hence we may conclude that there is c ∈ (a, b)
such that h′ (c) = 0. Thus
0 = h′ (c)
f (b) − f (a)
b−a
f (b) − f (a)
′
⇒ f (c) =
.
b−a
= f ′ (c) −
□
Worked Example 6.4.3. Prove that the function f (x) = x3 − 8x − 5 satisfies the hypothesis of the MVT
on the interval [1, 4] and find a number c ∈ (1, 4) that satisfies the conclusion of the MVT.
Worked Example 6.4.4. Given the function f (x) =
of the MVT.
x2 − 4x + 3
in the interval [1, 3] discuss the validity
x−2
Theorem 6.6. If f ′ (x) = 0 for all x in an interval (a, b), then f is a constant function on (a, b).
Proof. Let x1 , x2 ∈ (a, b) be distinct. We may suppose that x1 < x2 . Since f ′ (x) = 0 for all x ∈ (a, b),
f is differentiable on (x1 , x2 ) and continuous on [x1 , x2 ]. Applying the MVT to f on [x1 , x2 ] we find
c ∈ (x1 , x2 ) such that
f (x2 ) − f (x1 )
x2 − x1
f (x2 ) − f (x1 )
⇒0=
x2 − x1
⇒ 0 = f (x2 ) − f (x1 )
f ′ (c) =
∵ f ′ (x) = 0 for all x ∈ (a, b)
⇒ f (x2 ) = f (x1 ).
Thus f has the same value for any two numbers in (a, b). Consequently, f is constant on (a, b).
□
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Theorem 6.7. If f ′ (x) = g′ (x) for all x in an interval (a, b), then f − g is a constant on (a, b), that is,
f (x) = g(x) + C, where C is a constant.
Proof. Let F(x) = f (x) − g(x). Then F ′ (x) = f ′ (x) − g′ (x) = 0 for all x ∈ (a, b). By the above theorem,
F is constant on (a, b), i. e., f − g is constant on (a, b). Thus there is C ∈ R such that ( f − g)(x) = C for
all x ∈ (a, b), i. e., f (x) − g(x) = C or f (x) = g(x) + C for all x ∈ (a, b).
□
Tutorial 6.4.1. 1. [C1] §4.4, pp. 388-389: 152–156, 161, 162, 166–168, 174, 175, 177, 182, 184, 192.
2. Show that the functions y(x) = Ce x , C ∈ R, are the only functions which satisfy y′ = y. Hint:
Consider the function y(x)e−x .
6.5
Curve Sketching
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. be able to use the first derivative to find the interval(s) on which a given function is increasing
or decreasing,
□ 2. be able to use the First and Second Derivative Test to determine whether a function has a local
maximum or minimum at a critical number c or not,
□ 3. be able to explain the concepts of concave up and concave down,
□ 4. be able to use the second derivative to find the interval(s) on which a function is concave up and
concave down,
□ 5. know the definition and understand the notion of a point of inflection,
□ 6. be able to use the second derivative to find the points of inflection of a function,
□ 7. be able to draw any given curve using the curve sketching guidelines,
□ 8. know how to solve the examinable tutorial problems and worked out problems from these notes
on this section.
6.5.1
Increasing and Decreasing Functions and The First Derivative Test
A function f is increasing on an interval I if: ∀ x1 , x2 ∈ I, x1 < x2 ⇒ f (x1 ) < f (x2 )
and
a function f is decreasing on an interval I if: ∀ x1 , x2 ∈ I, x1 < x2 ⇒ f (x1 ) > f (x2 ).
The following theorem indicates how the derivative may be used to determine intervals on which a
function is increasing and decreasing. This theorem is called the I/D Test.
Theorem 6.8 (Increasing and Decreasing Test). Let f be a function that is continuous on a closed
interval [a, b] and differentiable on the open interval (a, b).
1. If f ′ (x) > 0 for each x ∈ (a, b), then f is increasing on [a, b].
2. If f ′ (x) < 0 for each x ∈ (a, b), then f is decreasing on [a, b].
3. If f ′ (x) = 0 for each x ∈ (a, b), then f is constant on [a, b].
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Proof. 1. Suppose that f ′ (x) > 0 for all x ∈ (a, b) and consider any two numbers x1 , x2 ∈ [a, b] such
that x1 < x2 . We wish to show that f (x1 ) < f (x2 ).
Since f is continuous on [a, b] and differentiable on (a, b), f is also continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). We can therefore apply the Mean Value Theorem to f on [x1 , x2 ]. Consequently, by
the MVT there is c ∈ (x1 , x2 ) such that
f ′ (c) =
f (x2 ) − f (x1 )
.
x2 − x1
Hence
f (x2 ) − f (x1 ) = f ′ (c)(x2 − x1 ).
Since x2 − x1 > 0 and since by the hypothesis, f ′ (c) > 0, the product f ′ (c)(x2 − x1 ) > 0, i. e., f (x2 ) −
f (x1 ) > 0. Hence f (x2 ) > f (x1 ).
2. The proof is similar to that of 1. above.
3. Theorem 6.6.
□
Guidelines for finding the intervals of INCrease and DECrease:
Let f be continuous on (a, b). To find the open intervals on which f is increasing and decreasing,
use the following steps:
1. Find the critical numbers c ∈ (a, b) of f by solving f ′ (x) = 0 AND finding x such that f ′ (x)
does not exist.
2. Use the critical numbers to create a first derivative table.
3. Determine and place the sign of f ′ at one test value in each open interval in the table.
4. Write down the sign of f ′ in the table.
5. On each interval, use the I/D Test to determine whether f is INCreasing or DECreasing.
Worked Example 6.5.1. Find the open intervals where the function f (x) = 4x5 − 15x4 − 20x3 + 110x2 −
120x + 40 is increasing and decreasing.
The next theorem is a consequence of the I/D Test, which makes locating the local extrema of a function
more explicit.
Theorem 6.9 (The First Derivative Test). Let c ∈ (a, b) be a critical number of a function f that is
continuous on [a, b] and differentiable on (a, b), except possibly at c.
1. If f ′ (x) > 0 for a < x < c and f ′ (x) < 0 for c < x < b, i. e., f ′ (x) changes from positive to negative at
c, then f (c) is a local maximum of f .
2. If f ′ (x) < 0 for a < x < c and f ′ (x) > 0 for c < x < b, i. e., f ′ (x) changes from negative to positive at
c, then f (c) is a local minimum of f .
3. If f ′ (x) > 0 or f ′ (x) < 0 for each x ∈ (a, b) except x = c, i. e., f ′ (x) does not change sign at c, then
f (c) is not a local extremum of f .
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70
y
MATH1036
y
f ′ (x) < 0
f ′ (x) > 0
c
Local maximum
f ′ (x) < 0
x
f ′ (x) > 0
c
Local minimum
y
x
y
f ′ (x) > 0
f ′ (x) < 0
f ′ (x) < 0
f ′ (x) > 0
c
No max or min
x
c
No max or min
x
First Derivative Test Guidelines for finding local maxima and local minima
1. Complete the I/D Test producing the first derivative table.
2. Based on the change of the sign of f ′ around the critical values, determine the type of local
extremum.
Note The following symbolic tables can help to memorize the First Derivative Test.
(a) f ′ changes sign from positive to negative around c ⇒ f (c) is a local maximum.
c
f ′ (c)
+
0
−
INC
LOC
MAX
DEC
Slope
f
(b) f ′ changes sign from negative to positive around c ⇒ f (c) is a local minimum.
c
f ′ (c)
−
0
+
DEC
LOC
MIN
INC
Slope
f
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71
(c) f ′ > 0 around c ⇒ f is increasing around c and c is neither a local maximum nor a local minimum.
c
f ′ (c)
+
0
+
INC
not LOC MAX
not LOC MIN
INC
Slope
f
(d) f ′ < 0 around c ⇒ f is decreasing around c and c is neither a local maximum nor a local minimum.
c
f ′ (c)
−
0
−
DEC
not LOC MAX
not LOC MIN
DEC
Slope
f
x
using the First Derivative
Worked Example 6.5.2. Find local maxima and minima of f (x) = 2
x +4
Test.
1
Worked Example 6.5.3. Use the First Derivative Test to find the local extrema of g(x) = (x4 − 4x3 ).
9
Tutorial 6.5.1. 1. [C1] §3.4, p. 273: 150–152, 158.
2. [C1] §4.5, pp. 403: 201, 202, 204.
3. [C1] §4.3, pp. 377–378: 123, 126, 144.
6.5.2
Concavity and The Second Derivative Test
You have seen that the first derivative locates the intervals of increase and decrease of a function f by
providing the slope of f . This gives a rough idea of the straight line shape of f . The bending or curving
of the graph of f , giving its true shape, is determined by the second derivative of f . The nature of the
bending or curving of f is called the concavity of f . More precisely,
If the graph of f lies ABOVE all of its tangents on an interval I, then f is CONCAVE UPWARD on I.
If the graph of f lies BELOW all of its tangents on an interval I, then f is CONCAVE DOWNWARD on I.
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The following theorem shows us how to use the second derivative of a function to determine the intervals
on which the function is concave up (CU) and concave down (CD).
Theorem 6.10 (Concavity Test). Let f be a function whose second derivative exists on an open interval I.
1. If f ′′ (x) > 0 for all x ∈ I, then f is concave upward on I.
2. If f ′′ (x) < 0 for all x ∈ I, then f is concave downward on I.
Note. In more advanced mathematics, one considers convex instead of concave up. A function f is
called convex if the straight line segment connecting two points on the graph of f lies on or above the
graph of f . Every concave up functions is convex, but the converse is not true. For example, straight
lines are convex but not concave up.
Guidelines for finding the intervals on which f is CU and CD:
To find the open intervals on which f is concave up and concave down, use the following steps:
1. Identify intervals (a, b) on which f is continuous and twice differentiable.
2. Find the numbers c ∈ (a, b) such that f ′′ (c) = 0.
3. Use the above results to determine open intervals and a second derivative table.
4. Determine and place the sign of f ′′ at one test value in each open interval in the table.
5. Generate the sign of f ′′ in the table.
6. Use the Concavity Test to determine whether f is CU or CD on each interval.
1
3
Worked Example 6.5.4. Determine the open intervals on which the graph of f (x) = x4 − x2 is
4
2
concave up and concave down.
x2 + 1
Worked Example 6.5.5. Determine the open intervals on which the graph of g(x) = 2
is concave
x −4
up and concave down.
We notice that in the above examples there are points in the domain of the functions at which the
concavity changes. If the function is continuous at such points of concavity change, then that point is
called a point of inflection. More precisely,
A point P = (c, f (c)) on the graph of a function f is a point of inflection if there exists an open
interval (a, b) with c ∈ (a, b) with f continuous at c such that one of the following statements hold:
(i) f ′′ (x) > 0 if a < x < c and f ′′ (x) < 0 if c < x < b, i. e., f changes from concave up to concave
down at c or
(ii) f ′′ (x) < 0 if a < x < c and f ′′ (x) > 0 if c < x < b, i. e., f changes from concave down to concave
up at c.
Three types of points of inflection are shown below. Note that the graph crosses its tangent line at a
point of inflection if the tangent exists at that point.
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y
73
y
Concave
Up
Concave
Down
Concave
Up
Point of inflection
Point of inflection
Concave
Down
x
x
y
Concave
Down
Point of inflection
Concave
Up
x
Worked Example 6.5.6. Find the points of inflection of f (x) = sin x + cos x on [0, 2π].
In addition to testing for concavity, the second derivative can be used to perform a simple test for local
maxima and minima. The test is based on the fact that if the graph of a function f is concave up on an
open interval containing c, and f ′ (c) = 0, then f (c) must be a local minimum. Similarly, if the graph of
a function f is concave downward on an open interval containing c, and f ′ (c) = 0, then f (c) must be a
local maximum.
Theorem 6.11 (The Second Derivative Test). Suppose that a function f is differentiable on an open
interval (a, b) containing c and the second derivative of f on (a, b) exists.
1. If f ′ (c) = 0 and f ′′ (c) > 0, then f (c) is a local minimum.
2. If f ′ (c) = 0 and f ′′ (c) < 0, then f (c) is a local maximum.
Proof. 1. If f ′ (c) = 0 and f ′′ (c) > 0, then
f ′ (x) − f ′ (c)
f ′ (x)
= lim
.
x→c
x→c x − c
x−c
0 < f ′′ (c) = lim
Hence
f ′ (x)
>0
x−c
for x , c close to c. If x < c, then x − c < 0 and f ′ (x) < 0 for x close to c. Also, if x > c, then x − c > 0
and f ′ (x) > 0 for x close to c. Thus, f ′ changes sign from negative to positive at c. The First Derivative
Test then implies that f (c) is a local minimum.
2. Similar to 1.
□
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We have derived two tests for local maxima and local minima: The First Derivative Test and The
Second Derivative Test. Either one can be used at your preference but note that if f ′′ (c) = 0 then
The Second Derivative Test is not applicable. In such cases you have no choice but to resort to
The First Derivative Test.
Worked Example 6.5.7. If f (x) = 5x3 − 3x5 , use The Second Derivative Test (where possible) to find
the local maxima and minima of f . Discuss concavity, find the points of inflection and sketch the graph
of f .
The graph above does not depict all features and is only a rough sketch. The intercepts are missing.
This leads to the culmination of the applications of the derivative into the procedure to sketch a curve
showing all important features.
Tutorial 6.5.2. [C1] §4.5, pp. 404–406: 211, 212, 214, 215, 221–223, 225, 230.
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6.5.3
1st Semester Calculus Study Guide
75
A summary of Curve Sketching
Curve sketching guidelines:
1. Find the DOMAIN of f .
2. Determine the INTERCEPTS:
x-intercept: let y = 0 and solve for x,
y-intercept: let x = 0 and evaluate y.
3. Check for SYMMETRY:
f (x) = f (−x) ⇒ even function ⇒ y-axis symmetry,
f (x) = − f (−x) ⇒ odd function ⇒ symmetry about origin.
4. Find the ASYMPTOTES:
Horizontal: Evaluate lim f (x) and lim f (x). If lim f (x) = L or lim f (x) = L then y = L
x→∞
x→−∞
x→∞
x→−∞
is HA.
Vertical: Look at all a < dom( f ) with points x near a in the domain of f . If lim+ f (x) = ±∞
x→a
or lim− f (x) = ±∞, then x = a is VA.
x→a
p(x)
, where p and q are polynomials
q(x)
r(x)
with deg p = deg q + 1, do long division to get f (x) = (mx + b) +
with deg r < deg q. Then
q(x)
y = mx + b is SA.
Slant or Oblique (for rational functions): If f (x) =
5. Find the INTERVALS OF INCREASE and DECREASE: Use the Inc/Dec Test
Regions of increase: f ′ (x) > 0 on interval I ⇒ f is increasing on I,
Regions of decrease: f ′ (x) < 0 on interval I ⇒ f is decreasing on I.
6. Determine the LOCAL MAXIMA and LOCAL MINIMA:
Find the critical numbers of f : Solve for x where f ′ (x) = 0 AND f ′ (x) does not exist.
Use the 1st Derivative Test: If c is a critical number, then
f ′ changes sign from positive to negative around c ⇒ f (c) is a local maximum,
f ′ changes sign from negative to positive around c ⇒ f (c) is a local minimum.
or use the 2nd Derivative Test: If c is a critical number and f ′′ (c) , 0, then
f ′′ (c) > 0 ⇒ f (c) is a local minimum,
f ′′ (c) < 0 ⇒ f (c) is a local maximum.
7. Find the CONCAVITY and POINTS OF INFLECTION: Use Concavity Test. Find f ′′ (x).
f ′′ (x) > 0 ⇒ CU,
f ′′ (x) < 0 ⇒ CD.
P is a point of inflection if f is continuous at P and concavity changes around P.
8. Use the above information to SKETCH the curve giving all points of interest.
Worked Example 6.5.8 (No Asymptotes). Draw the graph of the function f (x) = x5 − 5x3 .
5
2
Worked Example 6.5.9 (Vertical tangent). Sketch the graph of f (x) = x 3 + 5x 3 .
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Worked Example 6.5.10 (Vertical and Horizontal Asymptotes). Sketch the graph of the function f (x) =
5 − x2
.
1 − x2
Worked Example 6.5.11 (Slant Asymptote). Sketch the graph of g(x) =
x2 − 9
.
2x − 4
Tutorial 6.5.3. 1. [C1] §4.5, p. 406: 231, 232, 234, 235, 237, 239, 240. The use of a calculator, or
better some kind of computer algebra on an electronic device, is optional. Also investigate if there are
intercepts and asymptotes, that is, follow the curve sketching guidelines given in this subsection.
2. [C1] Section 4.6, pp. 437–438: 296, 298.
6.6
Optimization
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. be able to identify and know what is meant by an optimization problem,
□ 2. be able to set up the function that has to be maximized or minimized in a given word problem
about optimization,
□ 3. be able to find the absolute extreme values of the optimization function,
□ 4. be able to solve a given optimization problem,
□ 5. know how to solve the examinable tutorial problems and the worked out problems from these
notes on this section,
Strategy for Optimization:
1. Carefully read and understand the problem.
2. Draw a diagram, where relevant, and introduce notation on the diagram assigning symbols
to all given and required quantities.
3. Assign a symbol (the optimization function) to the quantity that is to be maximized or
minimized.
4. Express the optimization function in terms of one variable, using the geometry of the situation.
5. Find the critical values of the optimization function.
6. Determine the absolute maximum or minimum value of the optimization function.
7. State the final answer, in coherent form, specifying the units that you are using.
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Worked Example 6.6.1. You need to fence in a rectangular vegetable patch so that it fits into a righttriangular plot whose sides measure 4m and 12m. What is the maximum area for the vegetable patch?
Solution. Let x be the length of the rectangle and y be the width of the rectangle.
P
4
Q
x
R
y
S
12
T
Then the area of the rectangle is A = xy. We need A as a function of one variable. We use similar
triangles. Since ∆PQR ∼ ∆PS T ,
4−y
4
= ,
x
12
which gives 12(4 − y) = 4x, i. e., x = 3(4 − y). Hence
A = xy = 3(4 − y)y = 12y − 3y2 .
The critical numbers of A occur only where
dA
= 0. Thus
dy
dA
= 0 ⇔ 12 − 6y = 0.
dy
Hence y = 2 is the only critical number.
Since the domain of A is 0 ≤ y ≤ 4, we compare A(0), A(2) and A(4) to find the maximum area using
the Closed Interval Method. Now
A(0) = 0,
A(2) = 12(2) − 3(2)2 = 12,
A(4) = 3(4 − 4)4 = 0.
Thus the maximum area occurs when y = 2. In this case, x = 3(4 − y) = 3(4 − 2) = 6. Hence the largest
rectangle has area 12m2 .
Worked Example 6.6.2. Two posts, one 6m high and the other 9m high, stand 20m apart. They are to
be stayed by two wires, attached to a single stake, running from the ground level to the top of each post.
Where should the stake be placed to use the least wire?
Worked Example 6.6.3. Four meters of wire is to be used to form a square and a circle. How much
of the wire should be used for the square and how much should be used for the circle to enclose the
maximum total area?
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Worked Example 6.6.4. A manufacturer wishes to make a closed box with a square base and rectangular sides. The base of the box and all four sides are to be made of metal costing R10/m2 . The lid is to
be constructed of a thick glass costing R50/m2 . What are the dimensions of the box of greatest volume
that can be constructed for R720?
Worked Example 6.6.5. A right circular cylinder is inscribed in a right circular cone so that the centrelines of the cylinder and the cone coincide. The cone has height 6m and radius of base 3m.
(a) Express the volume of the cylinder as a function of its radius.
(b) Find the dimensions and the volume of the cylinder with maximal volume.
Tutorial 6.6.1. 1. [C1] §4.7, p. 451–453: 315, 316, 320, 324–326, 332–334, 342, 351.
2. a. For a cylinder of given volume, what are its dimensions which give the smallest surface area?
Check with some cans how close they are to these optimal dimensions.
b. Can you give some reasons why cans may not be of optimal dimensions? Can you devise a cost
function which is more general than the surface area?
Chapter 7
Hyperbolic Functions
This chapter is for self-study during the break between Semester One and Semester Two. You will not
be given any formal lectures on this examinable chapter, but you are welcome to ask your lecturers and
tutors for help. The reference for the section on hyperbolic functions is [C1], Sections 1.5 and 6.9.
LEARNING OUTCOMES:
On completion of this section you should (tick the checkbox when you have mastered the skill)
□ 1. know the definitions of the hyperbolic functions sinh x, cosh x, tanh x, sech x, coth x and cosech x
□
□
□
□
□
□
and know the sketches of the graphs of sinh x, cosh x and tanh x given in [C1], pages 108, 110,
745,
2. know the hyperbolic identities and their derivations given in [C1], pages 110 and 111, 1–9.
3. know the derivatives (and the proofs) of the hyperbolic functions given in [C1], page 746,
4. know the inverse hyperbolic functions sinh−1 x, cosh−1 x and tanh−1 x and their sketches given
in [C1], pages 748 and 749,
5. know how to express the inverse hyperbolic
terms of natural
√ functions in−1
√ logarithms, i. e.,−1know
−1
2 + 1), cosh x = ln(x +
how Å
to prove
and
use
sinh
x
=
ln(x
+
x
x2 − 1) and tanh x =
ã
1+x
1
ln
,
2
1−x
6. know the derivatives (and the proofs) of the inverse hyperbolic functions given in [C1], page
750, and know how to use the inverse hyperbolic functions in integration techniques as given
on that page,
7. know how to solve the examinable tutorial problems and the worked out problems from [C1]
on this section.
Tutorial 7.0.1. 1. [C1], §6.9, p. 754: 381–384.
2. Prove that cosh2 x − sinh2 x = 1. There are at least three ways to prove this identity.
3. Consider the hyperbolic functions sinh, cosh, tanh. Given the following values, find the values of the
other two hyperbolic functions.
4
4
3
a. sinh x = , b. cosh x = , c. tanh x = − .
3
3
4
Are these values unique?
Tutorial 7.0.2. 1. [C1], §6.9, p.754: 385, 388, 391, 392, 394. You do not need to graph the functions.
2. [C1], §6.9, p.754: 395–397, 399–401.
3. [C1], §6.9, p.754: 405, 407, 409, 411.
4. [C1], §6.9, p.755: 415, 417.
79
80
1st Semester Calculus Study Guide
Tutorial 7.0.3. Prove the following formulas:
ä
Ä
√
(a) sinh−1 x = ln x + x2 + 1 ,
Ä
ä
√
(b) cosh−1 x = ln x + x2 − 1 ,
ã
Å
1+x
1
.
(c) tanh−1 x = ln
2
1−x
MATH1036