My Students
1Fall 2010
1It is based mostly on the textbook, Smith and Minton’s Calculus Early Transcendental Functions, 3rd Ed and it has been reorganized and retyped by Jae Lee.
1 Limits and Continuity 7
1.1 A Brief Preview of Calculus: Tangent Lines and the Length of a Curve . . . 7
1.2 The Concept of Limit . . . 7
1.3 Computation of Limits . . . 10
1.4 Continuity and Its Consequences . . . 11
1.5 Limits Involving Infinity: Asymptotes . . . 13
1.5.1 Vertical Asymptote . . . 13
1.5.2 Horizontal Asymptote . . . 14
1.5.3 Slant Asymptote . . . 15
1.6 Formal Definition of the Limit . . . 15
1.7 Limits and Loss–of–Significance Errors . . . 15
2 Differentiation 17 2.1 Tangent Lines and Velocity . . . 17
2.2 The Derivative. . . 17
2.2.1 Definition . . . 17
2.2.2 Differentiability. . . 18
2.2.3 Meaning: Geometrical . . . 20
2.2.4 Alternative Derivative Notations . . . 21
2.3 Computation of Derivatives: The Power Rule . . . 22
2.3.1 Power Rule . . . 22
2.3.2 General Derivative Rules: Linearity . . . 22
2.3.3 Higher Order Derivatives . . . 23
2.3.4 Physical Meaning of Derivative: Rate of Change . . . 23
2.4 The Product and Quotient Rules . . . 24
2.4.1 Product Rule . . . 24
2.4.2 Quotient Rule . . . 24
2.4.3 Applications . . . 24
2.5 The Chain Rule . . . 25
2.5.1 Prerequisite . . . 25
2.5.2 Chain Rule . . . 25
2.5.3 Derivative of Inverse Function . . . 26
2.6 Derivatives of Trigonometric Functions . . . 27
2.6.1 Prerequisites: Basic Formulas . . . 27
2.6.2 Derivatives of Trigonometric Functions . . . 27
2.6.3 Applications . . . 28
2.7 Derivatives of Exponential and Logarithmic Functions . . . 29
2.7.1 Prerequisites: Basic Formulas . . . 29
2.7.3 Derivative of the Natural Logarithm . . . 30
2.7.4 Logarithmic Differentiation . . . 31
2.8 Implicit Differentiation and Inverse Trigonometric Functions . . . 32
2.8.1 Implicit Differentiation . . . 32
2.8.2 Derivatives of the Inverse Trigonometric Functions . . . 33
2.9 The Mean Value Theorem . . . 35
3 Applications of Differentiation 37 3.1 Linear Approximations and Newton’s Method . . . 37
3.2 Indeterminate Forms and L’Hˆopital’s Rule . . . 37
3.2.1 Introduction. . . 37
3.2.2 Indeterminate Forms: 0/0 and Infinity/Infinity . . . 38
3.2.3 L’Hˆopital’s Rule . . . 38
3.2.4 Other Indeterminate Forms . . . 39
3.3 Maximum and Minimum Values . . . 40
3.3.1 Absolute Extrema. . . 40
3.3.2 Local Extrema . . . 40
3.3.3 Critical Number . . . 41
3.4 Increasing and Decreasing Functions . . . 43
3.4.1 Increasing and Decreasing Functions . . . 43
3.4.2 Critical Point Classification . . . 43
3.5 Concavity and the Second Derivative Test . . . 45
3.5.1 Concavity . . . 45
3.5.2 Second Derivative Test . . . 46
3.6 Overview of Curve Sketching. . . 47
3.7 Optimization . . . 48
3.7.1 Guideline . . . 48
3.7.2 Area. . . 48
3.7.3 Volume . . . 49
3.7.4 Closest Point to Curve . . . 49
3.7.5 Soda Can & Highway . . . 50
3.8 Related Rates . . . 52 3.8.1 Spreading Oil . . . 52 3.8.2 Ladder . . . 53 3.8.3 Car Speed . . . 54 3.8.4 Economics . . . 55 3.8.5 Flying Jet . . . 55
3.9 Rates of Change in Economics and the Sciences . . . 56
4 Integration 57 4.1 Antiderivatives . . . 57
4.1.1 Antiderivative. . . 57
4.1.2 Indefinite Integral. . . 57
4.2 Sums and Sigma Notation . . . 60
4.3 Area . . . 60
4.4 The Definite Integral . . . 61
4.5 The Fundamental Theorem of Calculus . . . 62
4.5.1 The Fundamental Theorem of Calculus . . . 62
4.5.2 Application of Fundamental Theorem of Calculus. . . 63
4.6 Integration by Substitution . . . 65
4.6.2 Integration By Substitution . . . 66
4.6.3 Integration By Substitution in Definite Integral . . . 67
4.7 Numerical Integration . . . 67
4.8 The Natural Logarithm as an Integral . . . 68
5 Applications of the Definite Integral 69 5.1 Area Between Curves . . . 69
5.1.1 Region Bounded by Upper and Lower Curves . . . 69
5.1.2 Region Bounded by Right and Left Curves . . . 71
5.2 Volume: Slicing, Disks and Washers . . . 72
5.2.1 Volume by Slicing . . . 72
5.2.2 Method of Disks . . . 74
5.2.3 Method of Washers . . . 76
5.3 Volume by Cylindrical Shells . . . 77
5.4 Arc Length and Surface Area . . . 78
5.5 Projectile Motion . . . 78
5.6 Applications of Integration to Physics and Engineering . . . 78
5.7 Probability. . . 78
6 Integration Techniques 79 6.1 Review of Formulas and Techniques . . . 79
6.2 Integration by Parts . . . 82
6.2.1 Integration by Parts: No Repetition . . . 82
6.2.2 Integration by Parts: Repetition . . . 84
6.2.3 Integration by Parts Formula for Definite Integrals . . . 87
6.3 Trigonometric Techniques of Integration . . . 88
6.3.1 Integrals Involving Powers of Trigonometric Functions . . . 88
6.3.2 Trigonometric Substitution . . . 92
6.4 Integration of Rational Functions using Partial Fractions . . . 94
6.4.1 Division Algorithm . . . 94
6.4.2 Form I.(ax+b)(cx+d)S(x) , where the degree of the polynomial S(x) is less than 2 . . . 94
6.4.3 Form II.(ax+b)S(x) n, where the degree of the polynomial S(x) is less than n . . . 98
6.4.4 Form III. S(x) (ax2+bx+c)(dx2+ex+ f ), where the degree of S(x) is less than 4 . . . 99
6.4.5 Brief Summary of Integration Techniques . . . 101
6.5 Integration Tables and Computer Algebra Systems . . . 101
Limits and Continuity
§1.1 A Brief Preview of Calculus: Tangent Lines and the Length of a Curve. Skip. Please read the textbook.
§1.2 The Concept of Limit. Consider the function,
f (x) = x
2− 4
x− 2 ,
which has the domain all real numbers except 2, i.e.,R − {2}.
We raise a question: As x approaches 2, what value does f approach? We think of two ways: Graphical way and Computational/Analytical way.
Figure 1.1: Graph of f (x) = x
2− 4
x− 2
1. Graph (See the figure1.1):
(1) As x approaches 2 from the left, the graph shows that the values of f are closer to 4. (2) As x approaches 2 from the right, the graph shows that the values of f are closer to 4.
In simple notations, (1) as x→ 2−, f (x)→ 4 and (2) as x → 2+, f (x)→ 4. In much simpler way, they are expressed by
lim
x→2−f (x) = 4, x→2lim+f (x) = 4,
each of which is called the one–sided limit and reads that the limit of f as x approaches 2 from the left is 4 and the limit of f as x approaches 2 from the right is 4, respectively.
When both one–sided limits are equal, we say that the limit of f as x approaches 2 is 4 and it is expressed by lim
x→2f (x) = 4.
2. Computation:
The given function f can be simplified by
f (x) = x
2− 4
x− 2 =
(x− 2)(x + 2)
x− 2 = x + 2.
It is straightforward to see that as x→ 2, f (x) = x + 2 → 4. Thus we have lim
Let us consider another function,
g(x) = x
2− 5
x− 2 ,
which has the domainR − {2}. Question: as x approaches 2, what value does g approach? 1. Graph (See the figure1.2):
(1) As x approaches 2 from the left, the graph shows that the values of g get bigger and bigger.
(2) As x approaches 2 from the right, the graph shows that the values of g get negatively smaller and smaller.
Figure 1.2: Graph of f (x) = x
2− 5
x− 2
In simple notations, they are expressed by lim
x→2−g(x) =−∞, xlim→2+g(x) = +∞.
That is, each one–sided limit does not exist, which implies that as x approaches 2, the limit of g does not exist.
2. Computation:
As x goes to 2, the numerator of g, x2− 5, approaches −1. But, the denominator of g, x − 2, approaches 0. It allows us to expect that g cannot get closer to a finite number as x approaches 0. Thus, the limit of g does not exist as x goes to 2.
We summarize as follows: A limit exists if and only if both corresponding one-sided limits exist and are equal. That is,
lim
x→af (x) = L, for some number L, if and only if xlim→a−f (x) = limx→a+f (x) = L.
Figure 1.3: Graph of f Example 1.2.1. Use the graph in figure1.3to determine
lim
Figure 1.4: Graph of f (x) = sin x
x
ANSWER. We observe
lim
x→1−f (x) = 2, x→1lim+f (x) =−1, limx→1f (x) = does not exist, x→−1lim f (x) = 1. □
Example 1.2.2. Evaluate lim
x→0
sin(x)
x .
ANSWER. By the figure1.4, we get
lim x→0 sin(x) x = 1. □ Figure 1.5: Graph of f (x) = x |x|
Before we move to the next example, please remember the definition of the absolute function:
|x| =
{
x when x≥ 0
−x when x < 0.
Example 1.2.3. Evaluate lim
x→0
x |x|.
ANSWER. The function x/|x| has the domain R − {0}.
1. Graph: By the figure 1.12a, we have lim x→0− x |x| =−1, xlim→0+ x |x| = 1.
Since both one–sided limits are not equal, so lim
x→0
x
2. Computation: By the definition of the absolute function, we have x |x| = { x x = 1 when x > 0 x −x =−1 when x < 0. That is, lim x→0− x |x|= limx→0− x −x = limx→0−−1 = −1, x→0lim+ x |x| = limx→0+ x x= limx→0+1 = 1.
Since both one–sided limits are not equal, so lim
x→0
x
|x| does not exist. □
§1.3 Computation of Limits. Skip. Please read the textbook.
§1.4 Continuity and Its Consequences.
Definition 1.4.1. A function f is continuous at x = a when 1. f (a) is defined (i.e., a should be in the domain of f ), 2. lim
x→af (x) exists,
3. lim
x→af (x) = f (a).
Simply, f is continuous at x = a if and only if lim x→af (x) = f (a) = f ( lim x→ax ) .
If f is not continuous at x = a, then f is said to be discontinuous at x = a.
If f is continuous at any point in an interval I, then f is said to be continuous in the interval I. Comment: Graphically, “ f is continuous at x = a” means that the graph of f is connected at x = a.
Figure 1.6: Graph of f (x) = x
2+ 2x− 3
x− 1
Example 1.4.2. Determine where f (x) =x
2+ 2x− 3
x− 1 is continuous.
ANSWER. The domain of f isR − {1} and f is a rational function of polynomials. So we may expect that f
is continuous everywhere except x = 1.
1. Graph (See the figure1.6): By the figure, we observe f is continuous everywhere except x = 1. 2. Computation: A simple computation shows
f (x) = x
2+ 2x− 3
x− 1 =
(x− 1)(x + 3)
x− 1 = x + 3,
which is continuous everywhere, because its graph is a line. Hence, f is continuous everywhere except
Figure 1.7: Graph of f (x) = 1
x2 and h(x) = cos
1
x
Example 1.4.3. Find all discontinuities of f (x) = 1
x, g(x) = 1 x2, and h(x) = cos ( 1 x ) .
ANSWER. The functions f , g, and h have the same domainR − {0}. From the figure 1.7, we observe that
g and h are continuous everywhere except x = 0. It is easy to see that the function f is also continuous
everywhere except x = 0 by its graph. (See the first example in Section1.5Limit involving Infinity.) □ From your experience with the graphs of some common functions, the following result should come as no surprise.
Theorem 1.4.4.
1. All polynomials are continuous everywhere.
2. sin x, cos x, and the arctangent tan−1x are continuous everywhere.
3. exis continuous everywhere.
4. √nx is continuous for all x, when n is odd and for x > 0, when n is even.
5. ln x is continuous for x > 0.
Theorem 1.4.5. Suppose that f and g are continuous at x = a. Then all of the following are true: 1. ( f± g) is continuous at x = a,
2. f· g is continuous at x = a, 3. f
g is continuous at x = a if g(a)̸= 0.
Corollary 1.4.6. Suppose that g is continuous at x = a and f is continuous at g(a). Then, the composition
§1.5 Limits Involving Infinity: Asymptotes. □ 1.5.1 Vertical Asymptote.
Figure 1.8: Graph of f (x) = 1
x
Example 1.5.1. Examine lim
x→0
1
x.
ANSWER. By the figure1.8, we observe lim x→0− 1 x =−∞, x→0lim+ 1 x =∞. □
Comment: The graph, figure1.8, of f (x) = 1/x shows that as x→ 0, f (x) → ±∞. As x → 0, f (x) gets closer to the line x = 0. This line x = 0 is called the vertical asymptote of f .
Definition 1.5.2. The line x = a is called a vertical asymptote of the curve if at least one of the following statements is true:
lim
x→af (x) =∞, x→alim−f (x) =∞, x→alim+f (x) =∞,
lim
x→af (x) =−∞, x→alim−f (x) =−∞, x→alim+ f (x) =−∞.
For the various vertical asymptotes, see the figure below.
Example 1.5.3. Find all vertical asymptotes of f (x) = 1
x2.
ANSWER. By the graph of f (x) = 1/x2, we observe
lim
x→0−f (x) =∞ = limx→0+ f (x).
Example 1.5.4. Find all vertical asymptotes of f (x) = tan(x). ANSWER. By the graph of f , we observe
lim
x→−π/2−f (x) =x→limπ/2− f (x) =∞ = limx→3π/2−f (x) =··· =x→lim2n+1 2 π−
f (x),
lim
x→−π/2+f (x) =x→limπ/2+ f (x) =−∞ = limx→3π/2+f (x) =··· =x→lim2n+1 2 π+
f (x),
where n is any integer. Thus, the lines x = 2n + 1
2 π, n = 0,±1,±2,..., are all vertical asymptotes of f . □ □ 1.5.2 Horizontal Asymptote.
Let us examine lim
x→−∞
1
x and limx→∞
1
x. By the graph of y = 1/x, we observe
lim x→−∞ 1 x = 0, x→∞lim 1 x = 0.
That is, the graph of y = 1/x appears to approach the horizontal line y = 0, as x→ ±∞. In this case, we call
y = 0 a horizontal asymptote of y = 1/x.
Definition 1.5.5. The line y = L is called a horizontal asymptote of the curve y = f (x) if either lim
x→−∞f (x) = L or x→∞lim f (x) = L.
For the various horizontal asymptotes, see the figure below.
Figure 1.9: Graph of f (x) = 5x− 7 4x + 3
Example 1.5.6. Find all horizontal asymptotes of f (x) = 5x− 7 4x + 3.
ANSWER. It is not easy to sketch the graph of f . So we use the analytic method. f (x) = 5x− 7 4x + 3= 5x− 7 4x + 3· 1/x 1/x = 5− 7/x 4 + 3/x, xlim→∞f (x) = limx→∞ 5− 7/x 4 + 3/x = 5 4. Thus, the curve of f has the only one horizontal asymptote y = 5
4. See the figure1.9. □ □ 1.5.3 Slant Asymptote.
Skip. Please read the textbook. §1.6 Formal Definition of the Limit. Skip. Please read the textbook.
§1.7 Limits and Loss–of–Significance Errors. Skip. Please read the textbook.
Differentiation
§2.1 Tangent Lines and Velocity. Skip. Please read the textbook. §2.2 The Derivative.
□ 2.2.1 Definition.
Definition 2.2.1. The derivative of the function f at x = a is defined as
f′(a) = lim
h→0
f (a + h)− f (a)
h , (2.2.1)
provided the limit exists. If the limit exists, we say f is differentiable at x = a. An alternative form of (2.2.1) is
f′(a) = lim
b→a
f (b)− f (a)
b− a . (2.2.2)
Example 2.2.2. Use the definition of the derivative to find the derivative of f (x) = 3x2+ 2x− 1 at x = 1. ANSWER. By the definition, we have
f′(1) = lim h→0 f (1 + h)− f (1) h = lim h→0 3(1 + h)2+ 2(1 + h)− 1 − (3 + 2 − 1) h = limh→0(8 + 3h) = 8. □
Exercise 2.2.3. Use the definition of the derivative to find the derivative of f (x) = x2+ 2x at x = 2. Exercise 2.2.4. Find the derivative of f (x) =√x at x = 1.
Definition 2.2.5. The derivative of f is a function f′given by
f′(x) = lim
h→0
f (x + h)− f (x)
h . (2.2.3)
The process of computing a derivative is called differentiation. f is differentiable on an interval I if it is differentiable at every point in I.
Example 2.2.6. Find the derivative of f (x) = 3x2+ 2x− 1. ANSWER. By the definition, we have
f′(x) = lim h→0 f (x + h)− f (x) h = lim h→0 3(x + h)2+ 2(x + h)− 1 − (3x2+ 2x− 1) h = limh→0(6x + 2 + 3h) = 6x + 2. □
Exercise 2.2.7. Find the derivatives: f (x) = x2+ 2x.
f (x) =2
x, where x̸= 0.
Figure 2.1: Graphical Relations between f and f′
Example 2.2.8. Sketch the graph of f′, when the graph of f is given as the one in the left–hand side of the figure2.1.
ANSWER. The graph of f′is given in the right–hand side of the figure2.1with the graph of f itself.
Expla-nation in class. □
Figure 2.2: Graphical Relations between f and f′
Example 2.2.9. Sketch the graph of f (x), when the graph of f′(x) is given as the one in the left–hand side of the figure2.2.
ANSWER. The graph of f is given in the right–hand side of the figure2.2with the graph of f′itself.
Expla-nation in class. □
□ 2.2.2 Differentiability.
• Graphical Interpretation: We recall that the continuity of f at x = a graphically corresponds to the con-nectedness of the graph of y = f (x) at x = a. Then what is the graphical interpretation of the differentiability?
For a differentiable function f at x = a, its graph is smoothly connected at x = a.
• Relationship between Continuity and Differentiability:
1. If a function is differentiable at x = a, then is it continuous at x = a? The answer is YES. It’s because a smoothly connected graph at x = a is obviously connected.
2. If a function is continuous at x = a, then is it differentiable at x = a? The answer is NO. For instance, the function f (x) =|x − 3| is continuous at x = 3, but it is not differentiable at x = 3, because graphically its graph is not smooth at x = 3.
Theorem 2.2.10. If f (x) is differentiable at x = a, then f (x) is continuous at x = a.
(It is same as saying, if a function is not continuous at x = a, then it is not differentiable at x = a and so it
cannot have a derivative at x = a.)
Figure 2.3: Non–Differentiability There are four cases of non–differentiability. See the figure2.3.
1. Discontinuity: if a graph of f (x) is not connected at x = a, then f (x) is discontinuous at x = a and thus
f (x) is not differentiable at x = a. That is, f′(a) does not exist.
2. Corner Point: if a graph of f (x) has a corner point at x = a, then f (x) is not differentiable at x = a. That is, f′(a) does not exist.
3. Vertical Tangent Line: if a graph of f (x) has a vertical tangent line at x = a, then f (x) is not differen-tiable at x = a. That is, f′(a) does not exist.
4. Cusp: if a graph of f (x) has the shape of cusp at x = a, then f (x) is not differentiable at x = a. That is,
□ 2.2.3 Meaning: Geometrical.
Consider the graph of a function y = f (x). Let us choose a point P on the curve at x = a. Then the point
P has the coordinate (a, f (a)). Clearly we can draw so many straight lines passing through this point P.
However, when we give the condition “the line should touch (not cross) only the point P in the neighborhood
of the point P on the curve”, we can find only one line. We refer the line as the tangent line to the curve of y = f (x) at x = a.
• Equation of Line (from High School):
1. Point–Slope: if a line has a slope m and passes through a point (a, b), then the equation of the line is
y− b = m(x − a). (“Point–Slope” Equation of Line) (2.2.4) 2. Point–Point: if a line passes through (a, b) and (c, d), then the line has the slope b− d
a− c and so by the
formula above, the equation of the line is
y− b = b− d
a− c(x− a) or y− d = b− d
a− c(x− c). (“Point–Point” Equation of Line)
Applying the formula (2.2.4) to the tangent line to the curve of y = f (x) at x = a, we can get the equation of the tangent line:
y− f (a) = (slope of tangent line)(x − a), (2.2.5) where we don’t know the slope of the tangent line yet.
Here we raise two questions:
(a) If the graph of y = f (x) is a straight line, then what is the tangent line to the graph of y = f (x) at x = a? In this case, the tangent line is the line itself. So the equation of the tangent line at any point is exactly same as y = f (x).
(b) How can we find the slope of the tangent line to the curve of y = f (x) at x = a? The answer comes from the derivative of f (x) at x = a.
The derivative of f at x = a geometrically means the slope of tangent line to curve of y = f (x) at x = a. Now, from the formula (2.2.5), we deduce one of the most important formulas in this course: the equation of the tangent line to the curve of y = f (x) at x = a is
y− f (a) = f′(a)(x− a), (2.2.6) which you must memorize.
Example 2.2.11. Find the slope of the tangent line to the curve of f (x) =√x at x = 4 and write the equation
of the tangent line at x = 4.
ANSWER. We compute f (4 + h)− f (4) h = √ 4 + h−√4 h = (√ 4 + h−√4 h )(√ 4 + h +√4 √ 4 + h +√4 ) = 4 + h− 4 h(√4 + h +√4 ) = 1 √ 4 + h +√4 f′(4) = lim h→0 f (4 + h)− f (4) h = limh→0 1 √ 4 + h +√4 = 1 √ 4 + 0 +√4 = 1 4,
which is the slope of the tangent line to the curve of f (x) =√x at x = 4. By the formula (2.2.6), the equation of the tangent line at x = 4 is
y− f (4) = f′(4)(x− 4), i.e., y−√4 = 1
4(x− 4), i.e., y =
x
□ 2.2.4 Alternative Derivative Notations.
For a function y = f (x), its derivative is expressed by
y′, f′(x),
which are called Newton’s notations for the derivative. We also have Leibniz’ notations:
dy dx, d dxy, d dxf (x), d f (x) dx .
When we compute the values of the derivative function of y = f (x) at x = a, we use the following notations:
y′ x=a, f ′(a), dy dx x=a , d dxy x=a , d dxf (x) x=a , d f (x) dx x=a .
§2.3 Computation of Derivatives: The Power Rule. □ 2.3.1 Power Rule.
Recalling that the derivative means the slope of the tangent line, it is easy to understand the following two facts.
1. Any constant function f (x) = c has the derivative f′(x) = 0. 2. The identity function f (x) = x has the derivative f′(x) = 1. Theorem 2.3.1(Power Rule). For any natural number n,
f (x) = xnhas the derivative f′(x) = nxn−1. Example 2.3.2. Compute the derivative of f (x) = x7and g(t) = t28.
Theorem 2.3.3(General Power Rule). For any real number r,
f (x) = xrhas the derivative f′(x) = rxr−1.
Example 2.3.4. Compute the derivative of f (x) =√3x5and g(t) = 1/t and h(s) = 1/√3s2. □ 2.3.2 General Derivative Rules: Linearity.
Theorem 2.3.5. If f and g are differentiable and c is any constant, then (1) [ f (x) + g(x)]′= f′(x) + g′(x)
(2) [ f (x)− g(x)]′= f′(x)− g′(x) (3) [c f (x)]′= c f′(x)
Those three rules can be expressed as one rule: for differentiable functions f and g and any constant a and b, [a f (x) + bg(x)]′= a f′(x) + bg′(x).
Example 2.3.6. Find the derivatives: f (x) = x2 + x3. g(t) = 3t4 . h(s) = 2s6+ 3√ s . f (x) =4x2− 3x + 2 √ x x . f (x) = 3x2+ 5x− 2. g(s) = 3 s2− s 4+√s . h(t) = 2 3t+ 2t 4− 3.
Exercise 2.3.7. Let f (x) = 2x3− 3x2− 12x + 5. Find all x where f′(x) > 0 and all x where f′(x) < 0. Exercise 2.3.8. Find the equation of the tangent line to the graph of the given function and point: f (x) = 4 − 4x +2 x at x = 1. y = x3− 6x2+ 5 at x = 4. y = 6√3 x2−√4 x at x = 1.
□ 2.3.3 Higher Order Derivatives.
Given a function f , we have computed the derivative f′. In fact, this derivative is called the first derivative of f . If we can also compute the derivative of f′ (,i.e., the derivative of the derivative), then it is written by
f′′ and called the second derivative of f . If we can compute the derivative of f′′, then it is written by f′′′and called the third derivative of f . Below, we show common notations for the first five derivatives of f , where we assume that y = f (x).
Order Prime Notation Leibniz Notation 0 y = f (x) f 1 y′= f′(x) d f dx 2 y′′= f′′(x) d 2f dx2 3 y′′′= f′′′(x) d 3f dx3 4 y(4)= f(4)(x) d 4f dx4 5 y(5)= f(5)(x) d 5f dx5 .. . ... ...
Example 2.3.9. For f (x) = 3x4− 2x2+ 1, compute as many derivatives as possible. Exercise 2.3.10. Find f′′′(x) of f (x) = 3x4− 3x2+ 13.
Exercise 2.3.11. For f (x) = x4, find f′(x), f′′(x), f′′′(x), f(4)(x) and f(5)(x).
□ 2.3.4 Physical Meaning of Derivative: Rate of Change.
The geometrical meaning of the derivative of y = f (x) is the slope of the tangent line to the curve of y = f (x). The physical meaning of the derivative is the (instantaneous) rate of change. From physics, the velocity is the instantaneous rate of change of the distance and the acceleration is the instantaneous rate of change of the velocity. Hence, in a motion of an object, the velocity v(t) is the derivative of the distance function and the acceleration a(t) is the derivatives of the velocity v(t). That is,
a(t) = v′(t) = dv(t) dt .
Example 2.3.12. The motion of a particle is described by the function s(t) = 2t3−5t2+ 3t + 4, where s(t) is measured in centimeters and t in seconds. Find the acceleration as a function of time. What is the acceleration after 2 seconds?
§2.4 The Product and Quotient Rules. □ 2.4.1 Product Rule.
Theorem 2.4.1. Suppose f and g are differentiable, i.e., f′and g′exist. Then
[ f (x)g(x)]′= f′(x)g(x) + f (x)g′(x)
Example 2.4.2. Find the derivatives: f (x) = (x2− 1)√x . f (x) = (x3 + x− 3)2. f (x) = (2x4− 3x + 5) ( x2−√x +2 x ) .
Exercise 2.4.3. For f (x) = (x2− x)(x3+ x2− x + 1), find f′(1).
Exercise 2.4.4. Let f (x) = (g(x))2. Find f′(x) in terms of g(x) and g′(x).
Exercise 2.4.5. Find an equation of tangent line to y = (x4− 3x2+ 2x)(x3− 2x + 3) at x = 0. □ 2.4.2 Quotient Rule.
Theorem 2.4.6. Suppose f and g are differentiable and g(x)̸= 0. Then d dx [ f (x) g(x) ] = f ′(x)g(x)− f (x)g′(x) g2(x)
Example 2.4.7. Find the derivatives: f (x) =2x− 1 x + 1 . g(t) = t3 t2+ 1. h(x) = x2− 2 x2+ 1.
Example 2.4.8(Reciprocal Rule). Let f (x) = 1
g(x), where g(x)̸= 0. Find f
′(x) in terms of g(x) and g′(x).
□ 2.4.3 Applications.
Example 2.4.9. Suppose that a product currently sells for $25, with the price increasing at the rate of $2 per year. At this price, consumers will buy 150 thousand items, but the number sold is decreasing at the rate of 8 thousand per year. At what rate is the total revenue changing? Is the total revenue increasing or decreasing? Example 2.4.10. A golf ball of mass 0.05 kg struck by a golf club of mass m kg with speed 50 m/s will have an initial speed of u(m) = 83m
m + 0.05 m/s. Show that u
′(m) > 0 and interpret this result in golf terms. Compare
§2.5 The Chain Rule. □ 2.5.1 Prerequisite.
We recall that the derivative of y = f (x) is
y′= f′(x) = dy dx = d dxy = d f (x) dx = d dx f (x). In fact, we may fully express them as the derivative of y = f (x) with respect to x. Let us discuss why “with respect to” is important.
Example 2.5.1. (1) Find the derivative of f (x) = x2with respect to x. (2) Find the derivative of g(x) = x2with respect to t.
ANSWER. (1) By the power rule, we have
f′(x) = d
dx f (x) = d dxx
2= 2x.
(2) However, when we differentiate the function g(x) of x with respect to t, the function g(x) can be regarded as a constant function in the viewpoint of t, because there is no t in g(x) = x2. Now that g(x) is a constant in the viewpoint of t, so its derivative with respect to t should be zero. That is,
d dtg(x) = d dtx 2= 0. □ □ 2.5.2 Chain Rule.
Thanks to the power rule and the linearity, we can compute
(1) the derivative of an elementary function such as a polynomial and (2) the derivative of sum and difference of functions.
Moreover, thanks to the Product and Quotient Rules, we can compute (3) the derivative of the product of functions and
(4) the derivative of the quotient/fraction of functions.
However, we have one more operation on the functions: composition. The Chain Rule is the derivative rule for the composite function.
Let us discuss an example.
Example 2.5.2. Find the derivative of y = f (x) where f (x) =(x3+ 2x)7with respect to x. ANSWER. Let u = g(x) where g(x) = x3+ 2x. Then we observe
du dx = d dxg(x) = 3x 2 + 2, and y = f (x) =(x3+ 2x)7= (g(x))7= u7, i.e., y = u7.
So when we differentiate y with respect to u, we have
dy du=
d duu
7= 7u6.
But since we are looking for the derivative of y = f (x) with respect to x, we need to multiply the whole equation by du dx = 3x 2+ 2. That is, dy dx= dy du· du dx = 7u 6·(3x2+ 2)= 7(x3+ 2x)6(3x2+ 2). □
Theorem 2.5.3(CHAIN RULE). If g is differentiable at x and f is differentiable at g(x), then the composite function h = f◦ g defined by h(x) = f (g(x)) is differentiable at x and h′is given by the product
h′(x) = f′(g(x))· g′(x).
In Leibniz notation, if y = f (u) and u = g(x) are both differentiable functions, then dy dx = dy du du dx.
In words, the chain rule says that the derivative of f◦ g is the derivative of the outside function f multiplied by the derivative of the inside function g.
Example 2.5.4. Use the Chain Rule to find the derivative of the function h(x) = (x2+ 1)5.
ANSWER. The outside function f (x) = x5 has the derivative f′(x) = 5x4, while the inside function g(x) =
x2+ 1 has the derivative g′(x) = 2x. Thus, by the Chain Rule, h(x) has the derivative
h′(x) = f′(g(x))· g′(x) = 5 (g(x))4· 2x = 10x(g(x))4= 10x(x2+ 1)4. □
As you can see, when a composite function is given, it is very important to find out the outside and inside function to get the derivative of the composite function.
Example 2.5.5. Differentiate the functions: y =(x3+ x− 1)5. f (x) =√x2+ 1 . g(s) = s3√4s + 1 . h(x) = 8x (x3+ 1)2. k(x) = 8 (x3+ 1)2. Example 2.5.6. Compute d dt √ 100 + 8t .
Remark 2.5.7(Generalized Chain Rule). For a composite function k = ( f◦g)◦h of three functions f , g and
h, i.e., k(x) = f (g(h(x))), we have the derivative:
k′(x) = f′(g(h(x)))· g′(h(x))· h′(x).
It is true by the exactly same argument as the one developed for the composite function of two functions above.
□ 2.5.3 Derivative of Inverse Function.
We recall from High School that g is the inverse function of f if ( f◦ g)(x) = f (g(x)) = x and (g ◦ f )(x) =
g( f (x)) = x. The inverse function of f is denoted by f−1. So when g is the inverse function of f , we get
g = f−1.
Let g(x) be the inverse function of f (x), i.e., f (g(x)) = x and g( f (x)) = x. Differentiating both sides of
f (g(x)) = x with respect to x and applying the Chain Rule, we get f′(g(x))g′(x) = 1, i.e., g′(x) = 1
f′(g(x)) provided f′(g(x))̸= 0. Thus, we have the following theorem.
Theorem 2.5.8. If f is differentiable at all x and has an inverse function g(x) = f−1(x), then
g′(x) = 1
f′(g(x)) provided f′(g(x))̸= 0.
§2.6 Derivatives of Trigonometric Functions. □ 2.6.1 Prerequisites: Basic Formulas.
It is suggested that you should memorize the followings. Theorem 2.6.1.
sin (A + B) = sin A cos B + cos A sin B, sin (A− B) = sinAcosB − cosAsinB cos (A + B) = cos A cos B− sinAsinB, cos (A− B) = cosAcosB + sinAsinB
tan (A + B) = tan A + tan B
1− tanAtanB, tan (A− B) =
tan A− tanB 1 + tan A tan B.
sin (2A) = 2 sin A cos A, cos (2A) = cos2A− sin2A = 1− 2sin2A = 2 cos2A− 1
sin2A = 1− cos(2A)
2 , cos
2A = 1 + cos (2A)
2 1 = sin2A + cos2A, 1 + tan2A = sec2A.
□ 2.6.2 Derivatives of Trigonometric Functions.
The following lemma is used to prove the theorems on the derivatives of trigonometric functions. Lemma 2.6.2.
lim
θ→0sinθ = 0 θ→0limcosθ = 1
lim θ→0 sinθ θ = 1 θ→0lim 1− cosθ θ = 0.
The results above can be proved by using the graphs or L’Hˆopital’s Rule (Section3.2). The definition of the derivative and the lemma above imply the Theorem.
Theorem 2.6.3. The trigonometric functions have the following derivatives: d dxsin x = cos x d dxcos x =−sinx d dxtan x = sec 2x d dxcot x =−csc 2x d
dxsec x = sec x tan x
d
dxcsc x =−cscxcotx, where sec x = 1
cos x, csc x = 1
sin x and cot x = 1 tan x. Example 2.6.4. Find the derivatives:
f (x) = x5cos x. g(x) = sin2x. h(t) = 4tant − 5csct. k(x) = sinxcosx. f (x) = cos(x3). g(x) = cos3x. h(x) = cos(3x). f (θ) = cosθ 1 + sinθ. g(x) = secxtanx. h(x) = sec x 1 + tan x. k(x) = sin ( 2x x + 1 ) .
Example 2.6.5. For f (x) = sin x + cos x + 149x74, find f(75)(x) and f(150)(x), where f(75)(x) means the 75th derivative of f (x) and f(150)(x) means the 150th derivative of f (x).
□ 2.6.3 Applications.
Example 2.6.6. Find an equation of the tangent line to y = 3 tan x− 2cscx at x =π/3.
Look at the figure2.4Spring–mass system. The vertical displacement of a weight suspended from a spring, in the absence of damping (i.e., when resistance to the motion, such as air resistance, is negligible), is given by
u(t) = a cos(ωt) + b sin(ωt),
whereω is the frequency, t is time and a and b are constants.
Example 2.6.7. Suppose that u(t) measures the displacement (measured in inches) of a weight suspended from a spring t seconds after it is released and that
u(t) = 4 cost.
Find the velocity at any time t and determine the maximum velocity.
Figure 2.4: Spring–Mass System and Electric Circuit
Example 2.6.8. Look at the figure2.4 A simple circuit. If the capacitance is 1 (farad), the inductance is 1 (henry) and the impressed voltage is 2t2(volts) at time t, then a model for the total charge Q(t) in the circuit at time t is
Q(t) = 2 sint + 2t2− 4 (coulombs).
The current is defined to be the rate of change of the charge with respect to time and so is given by
I(t) = dQ(t)
dt (amperes). Compare the current at times t = 0 and t = 1.
§2.7 Derivatives of Exponential and Logarithmic Functions. □ 2.7.1 Prerequisites: Basic Formulas.
From the High School, we recall:
xaxb= xa+b, xaya= (xy)a, xab= x(ab), (xa)b= x(ab),
x−a= 1
xa, x
b−a= xb
xa, x
abc = x(a(bc)), x0= 1.
On the exponential function y = ax, let us recall:
1. The function y = ax is defined only when a > 0. The function y = ax with a < 0 is discussed in the Complex Analysis (Math 315).
2. The range of the function y = axis always positive and its graph passes through the point (0, 1).
3. It is differentiable everywhere and does not have any vertical asymptote. But its horizontal asymptote is
y = 0.
We also recall the laws on the natural logarithmic function: ln (ab) = ln a + ln b, lna b = ln a− lnb, lna b= b ln a, log ab = ln b ln a, ln 1 = 0, a = e ln a= ln ea.
On the natural logarithmic function y = ln x, let us recall:
1. The function y = ln x has the domain of all positive real numbers and its graph passes through the point (1, 0).
2. It is differentiable on the domain and has the vertical asymptotes x = 0. □ 2.7.2 Derivatives of the Exponential Functions.
Theorem 2.7.1. For any constant a > 0,
d dxa
x= axln a,
where ln a = logea and e is the Euler constant e≈ 2.71828.
Example 2.7.2. Compute the derivative of y = 101−x2.
Example 2.7.3. If the value of a 100-dollar investment doubles every year, its value after t years is given by
v(t) = 1002t. Find the instantaneous percentage rate of change of the worth.
Since ln e = logee = 1, the derivative of f (x) = exis
d dxe
x= exln e = ex.
We now have the following result. Theorem 2.7.4.
d dxe
Figure 2.5: Spring–Mass System
Example 2.7.5. If we build damping (i.e., resistance to the motion due to friction, for instance) into our model spring-mass system (see the figure2.5 Spring–mass system), the vertical displacement at time t of a weight hanging from a spring can be described by
u(t) = Aeαtcos(ωt) + Beαtsin(ωt),
where A, B,α andω are constants. For each of
(1) u(t) = e−tcost and (2) v(t) = e−t/6cos 4t, sketch a graph of the motion of the weight and find its velocity at any time t. Example 2.7.6. Find the derivative:
f (x) = 3ex2. g(x) = xe2/x . h(x) = 32x2 . k(t) = esint. y = ex− e−x ex+ e−x. y = e2x sin(3x).
Example 2.7.7. Find the 1000th derivative of f (x) = xe−x.
□ 2.7.3 Derivative of the Natural Logarithm. Theorem 2.7.8. For x > 0,
d
dx(ln x) = 1
x.
Example 2.7.9. Find the derivative: f (x) = xlnx.
g(x) = lnx3.
h(x) = ln(x2
+ 1). k(x) = lnlnx.
Example 2.7.10. Find an equation of the tangent line to the curve y = ln (
xex2
)
at x = 1.
Example 2.7.11. The concentration x of a certain chemical after t seconds of an autocatalytic reaction is given by x(t) = 10
9e−20t+ 1. Show that x
′(t) > 0 and use this information to determine that the concentration
□ 2.7.4 Logarithmic Differentiation.
A clever technique called logarithmic differentiation uses the rules of logarithms to help find derivatives of certain functions for which we don’t presently have derivative formulas. For instance, note that the function
f (x) = xxis neither a power function because the exponent is not a constant and nor an exponential function because the base is not constant.
Example 2.7.12. Find the derivative: f (x) = xx , x > 0. g(x) = xln x. h(x) = xsin x. k(x) = (cosx)x .
§2.8 Implicit Differentiation and Inverse Trigonometric Functions. □ 2.8.1 Implicit Differentiation.
Suppose that x and y satisfy the equation
x2+ y2= 4
whose graph is a circle centered at the origin with radius 2. Our goal is to describe the slope of the tangent line to the curve at each point (x, y). Just as in the case where y is a function of x, we can define dy/ dx to be the slope of the tangent line to the graph at a point (x, y). To find dy/ dx, we proceed as follows:
Step 1.Differentiate each side of the equation with respect to x: d dx ( x2+ y2)= d dx4, d dxx 2+ d dxy 2= 0, 2x + 2ydy dx = 0. The equation of the circle implies
x2+ y2= 4 ⇐⇒ y2= 4− x2 ⇐⇒ y = ±√4− x2,
i.e., y is a function of x and so y2 is a composite function having the variable x and thus the Chain Rule implies d
dxy
2= 2ydy
dx. Be careful! y is not a constant with respect to x. Step 2.Solve for dy/ dx:
2x + 2ydy dx = 0, 2y dy dx =−2x, dy dx=− 2x 2y =− x y.
This is the process known as implicit differentiation. As you can see, it is based on the Chain Rule. Example 2.8.1. Suppose that x and y satisfy the equation
x2− xy + y2= 1
whose graph is an ellipse. (1) Find dy/ dx and (2) find the points on the ellipse where the tangent line is horizontal or vertical.
Example 2.8.2. Find the slope of the tangent line to the graph of x3y2= xy3+ 6 at the point (2, 1).
Example 2.8.3. Find y′(x) = dy/ dx for x2+ y3− 2y = 3. Then, find the slope of the tangent line at the point (2, 1).
Example 2.8.4. Find y′(x) = dy/ dx for x2y2− 2x = 4 − 4y. Then, find an equation of the tangent line at x = 2.
Example 2.8.5. Suppose that van der Waals’ equation for a specific gas is (
P + 5 V2
)
(V− 0.03) = 9.7.
Thinking of the volume V as a function of pressure P, use implicit differentiation to find the derivative dV dP at the point (P,V ) = (5, 1).
Remark 2.8.6(Aside: General Version of van der Waals’ equation). ( P +an 2 V2 ) (V− nb) = nRT.
where P is the pressure of the fluid, V is the total volume of the container containing the fluid, a is a measure of the attraction between the particles, b is the volume excluded by a mole of particles, n is the number of moles, R is the gas constant and T is the absolute temperature.
If you are interested, find dP dV, dV dT and dT dP and check dP dV dV dT dT dP =−1.
Example 2.8.7. Find y′′(x) = d
2y
dx2 implicitly for y
2+ 2e−xy= 6. Then find the value of y′′ at the point (0, 2).
Example 2.8.8. Use implicit differentiation to find an equation of the tangent line to the curve at the given point:
(1) 2(x2+ y2)2= 25(x2− y2), (3, 1); (2) y2= x3+ 3x2, (1,−2).
The graphs of (1) and (2) are called the lemniscate and the Tschirnhausen cubic, respectively. See the figure2.6.
Figure 2.6: Lemniscate 2(x2+ y2)2= 25(x2− y2) (Left) and Tschirnhausen Cubic y2= x3+ 3x2(Right) □ 2.8.2 Derivatives of the Inverse Trigonometric Functions.
Let us recall from the Section 2.5 The Chain Rule: a function g is called the inverse function of f if ( f◦ g)(x) = f (g(x)) = x and (g ◦ f )(x) = g( f (x)) = x. The inverse function of f is denoted by f−1 and so g = f−1.
The trigonometric functions have the inverse functions when we restrict the domains of sin x, cos x and tan x by [−π/2,π/2], [0,π] and [−π/2,π/2], respectively. The inverse functions of sin x, cos x and tan x are denoted
by either sin−1x, cos−1x, and tan−1x or arcsin x, arccos x and arctan x.
But be careful!
arcsin x = sin−1x̸= (sinx)−1= 1 sin x, which is also same to other trigonometric functions.
• Domain and Range of Inverse Trigonometric Functions
1. Considering the restricted domain [−π/2,π/2], sin x has the range [−1,1]. It implies that the inverse
function sin−1x has the domain [−1,1] and the range [−π/2,π/2].
2. Considering the restricted domain [0,π], cos x has the range [−1,1]. It implies that the inverse function cos−1x has the domain [−1,1] and the range [0,π].
3. Considering the whole domain (−π/2,π/2), tan x has and rangeR. Its inverse function tan−1x has the
domainR and range (−π/2,π/2).
See the figures2.7,2.8and2.9below on the graphs of the trigonometric and inverse trigonometric functions. Using the implicit differentiation or derivatives of inverse functions in the Section2.5 The Chain Rule, we deduce the following derivatives of the inverse trigonometric functions.
Theorem 2.8.9. d dxsin −1x =√ 1 1− x2, (−1 < x < 1), d dxcos −1x =−√ 1 1− x2, (−1 < x < 1), d dxtan −1x = 1 1 + x2, d dxcot −1x =− 1 1 + x2,
Figure 2.7: Graphs of sin x (Left) on [−π/2,π/2] and sin−1x (Right) on [−1,1]
Figure 2.8: Graphs of cos x (Left) on [0,π] and cos−1x (Right) on [−1,1]
Figure 2.9: Graphs of tan x (Left) on (−π/2,π/2) and tan−1x (Right) onR
d dxsec −1x = 1 |x|√x2− 1, (|x| > 1), d dxcsc −1x =− 1 |x|√x2− 1, (|x| > 1).
(1) Implicit Differentiation Technique: Let y = sin−1x. Then,
y = sin−1x ⇐⇒ siny = x −π
2 ≤ y ≤
π
2. Implicitly differentiating the second equation, we get
cos ydy dx = dx dx = 1 ⇐⇒ dy dx = 1 cos y.
The equation sin y = x implies cos y =√1− x2, because sin2y + cos2y = 1, i.e., cos2y = 1− sin2y = 1− x2
and−π/2≤ y ≤π/2. Hence, the result becomes
dy dx = 1 cos y= 1 √ 1− x2 ⇐⇒ d dxsin −1x =√ 1 1− x2 − 1 < x < 1.
(2) Formula on Derivative of Inverse Function (Section2.5): Since g(x) = sin−1x is the inverse of the sine
function f (x) = sin x with−π/2≤ x ≤π/2, the formula implies g′(x) = 1 f′(g(x)) ⇐⇒ g ′(x) = 1 cos g(x) ⇐⇒ d dxsin −1x = 1 cos(sin−1x).
Let sin−1x =θ ∈ [−π/2,π/2]. Then x = sinθ and again sin2θ+ cos2θ = 1 implies cos2θ = 1− sin2θ = 1− x2and cosθ =√1− x2 (, becauseθ ∈ [−π/2,π/2]). Hence, the result becomes
dy dx = 1 cos(sin−1x) = 1 cosθ = 1 √ 1− x2 ⇐⇒ d dxsin −1x =√ 1 1− x2 − 1 < x < 1. □
Example 2.8.10. Compute the derivative: cos−1(3x2).
(sec−1x)2. tan−1(x3).
Example 2.8.11. One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of 130 ft/s (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? See the figure2.10below.
Figure 2.10: Baseball
§2.9 The Mean Value Theorem. Skip. Please read the textbook.
Applications of Differentiation
§3.1 Linear Approximations and Newton’s Method. Skip. Please read the textbook.
§3.2 Indeterminate Forms and L’Hˆopital’s Rule. □ 3.2.1 Introduction.
For the given two polynomials,
P(x) = anxn+ an−1xn−1+··· + a2x2+ a1x + a0, and Q(x) = bmxm+ bm−1xm−1+··· + b2x2+ b1x + b0,
where a0, . . . , anand b0, . . . , bmare all constants, let us discuss the limit of the rational function P(x)/Q(x) as
x→ a.
(1) a is a number:
(i) If Q(a)̸= 0 and P(a) ̸= 0, then
lim
x→a
P(x)
Q(x) = some number.
(ii) If Q(a) = 0 but P(a)̸= 0, then
lim
x→a
P(x)
Q(x)=±∞.
(iii) If Q(a)̸= 0 but P(a) = 0, then
lim
x→a
P(x) Q(x) = 0.
(iv) If Q(a) = 0 and P(a) = 0, then we use L’Hˆopital’s Rule (Subsection3.2.3). (2) a =±∞:
(i) If deg Q = m > n = deg P, then
lim
x→a
P(x) Q(x) = 0.
(ii) If deg Q = m < n = deg P, then
lim
x→a
P(x)
Q(x)=±∞.
(iii) If deg Q = m = n = deg P, then
lim x→a P(x) Q(x) = an bm .
Example 3.2.1. Evaluate the limit: lim x→1 x2+ 5 x + 1 , x→1lim x2+ 5 x− 1 , x→1lim x− 1 x2+ 5, x→∞lim x2+ 1 x3+ 5, x→∞lim x3+ 5 x2+ 1, x→∞lim 2x2+ 3x− 5 x2+ 4x− 11.
We recall a useful theorem: Theorem 3.1 on page 88 (textbook) in Section 3.1 Computation of Limits. Theorem 3.2.2. Suppose that lim
x→af (x) and limx→ag(x) both exist and let c be any constant. The following then
(i) lim
x→a[c· f (x)] = c · limx→af (x),
(ii) lim
x→a[ f (x)± g(x)] = limx→af (x)± limx→ag(x),
(iii) lim x→a[ f (x)· g(x)] = [ lim x→af (x) ][ lim x→ag(x) ] , (iv) lim x→a f (x) g(x) = limx→af (x) limx→ag(x)
(if limx→ag(x)̸= 0).
□ 3.2.2 Indeterminate Forms: 0/0 and Infinity/Infinity.
In the previous subsection, we have discussed the limit of the rational function of polynomials. Now, we consider a particular case with general functions f (x) and g(x),
lim
x→a
f (x) g(x).
If limx→a f (x) = 0 = limx→ag(x), then the limit is in the form of
0 0. If limx→a f (x) =±∞ = limx→ag(x), then the limit is in the form of ∞
∞.
Those two forms are called the indeterminate forms. We use L’Hˆopital’s Rule to compute those indetermi-nate forms.
□ 3.2.3 L’Hˆopital’s Rule.
Theorem 3.2.3. Suppose that f and g are differentiable on the interval (a, b), except possibly at some fixed point c∈ (a,b) and that g′(x) = 0 on (a, b), except possibly at c.
Suppose further that lim
x→c
f (x)
g(x) has the indeterminate form
0 0 or
∞
∞ and that limx→c
f′(x) g′(x) = L (or±∞). Then, lim x→c f (x) g(x) = limx→c f′(x) g′(x).
The conclusion of Theorem 3.2.3 also holds if limx→cg(x)f (x) is replaced with any of the limits limx→c+ f (x)
g(x),
limx→c− g(x)f (x), limx→∞g(x)f (x) or limx→−∞g(x)f (x). (In each case, we must make appropriate adjustments to the hypotheses.)
Example 3.2.4. Evaluate the limit. lim θ→0 sinθ θ , θ→0lim 1− cosθ θ , θ→0lim 1− cosθ sinθ . Example 3.2.5. Evaluate the limit.
lim x→∞ ex x, xlim→∞ x2 ex, xlim→∞ ln x ex , xlim→∞ ex ln x. Example 3.2.6. Evaluate the limit.
lim x→0 x2 ex− 1, xlim→0+ ln x csc x.
□ 3.2.4 Other Indeterminate Forms.
We study other indeterminate forms via examples. Example 3.2.7(∞ − ∞ form). Evaluate the limit.
lim x→0 ( 1 x2− 1 x4 ) , lim x→0 ( 1 ln(x + 1)− 1 x ) .
Example 3.2.8(0· ∞ form). Evaluate lim
x→∞
1
xln x.
Example 3.2.9(1∞form). Evaluate lim
x→1+x 1 x−1.
Example 3.2.10(00form). Evaluate lim
x→0+(sin x)
x
.
Example 3.2.11(∞0form). Evaluate lim
x→∞(x + 1)
2 x.
§3.3 Maximum and Minimum Values. □ 3.3.1 Absolute Extrema.
Definition 3.3.1. For a function f defined on a set S of real numbers and a number c∈ S, (1) f (c) is the absolute maximum of f on S if f (c)≥ f (x) for all x ∈ S and
(2) f (c) is the absolute minimum of f on S if f (c)≤ f (x) for all x ∈ S.
An absolute maximum or an absolute minimum is referred to as an absolute extremum. If a function has more than one extremum, we refer to these as extrema (the plural form of extremum).
Example 3.3.2. Locate any absolute extrema of the given function on the given interval: f (x) = x2− 9 on (−∞,∞).
f (x) = x2− 9 on (−3,3).
f (x) = x2− 9 on [−3,3].
We have seen that a function may or may not have absolute extrema, depending on the interval on which we are looking.
Example 3.3.3. Locate all absolute extrema of the given function on the given interval: f (x) =1
x on [−3,3].
f (x) = cosx on (−∞,∞).
Example 3.3.4. From the graph of the function f (x) = x2, we see that this function has the absolute minimum value of f (0) = 0 at x = 0, but no absolute maximum value.
Example 3.3.5. From the graph of the function f (x) = x3, we see that this function has neither an absolute maximum value nor an absolute minimum value. In fact, it has no local extreme values either.
Theorem 3.3.6(Extreme Value Theorem). A continuous function f defined on a closed and bounded interval [a, b] attains both an absolute maximum and an absolute minimum on that interval. That is, if f is continuous
on a closed and bounded interval [a, b], then f attains an absolute maximum value f (c) and an absolute
minimum value f (d) at some numbers c and d in [a, b].
Example 3.3.7. Find the absolute extrema of f (x) = 1/x on the interval [1, 3]. □ 3.3.2 Local Extrema.
Definition 3.3.8.
(1) f (c) is a local maximum of f if f (c)≥ f (x) for all x in some open interval containing c. (2) f (c) is a local minimum of f if f (c)≤ f (x) for all x in some open interval containing c. In either case, we call f (c) a local extremum of f .
Local maxima and minima (the plural forms of maximum and minimum, respectively) are sometimes referred to as relative maxima and minima, respectively.
Notice from the figure above that each local extremum seems to occur either (i) at a point where the tangent line is horizontal (i.e., where f′(x) = 0), or (ii) at a point where the tangent line is vertical (i.e., where f′(x) is undefined), or (iii) at a corner (again, where f′(x) is undefined).
Example 3.3.9. Locate any local extrema for the given function and describe the behavior of the derivative at the local extremum: f (x) = 9− x2and f (x) =|x|.
Figure 3.1: Various Local Extrema □ 3.3.3 Critical Number.
Definition 3.3.10. A number c is called a critical number of f if (1) c is in the domain of f and
(2) either f′(c) = 0 or f′(c) is undefined.
We recall the observation that local extrema occur only at points where the derivative is zero or undefined. We state this formally in a theorem.
Theorem 3.3.11(Fermat’s Theorem). Suppose that f (c) is a local extremum (local maximum or local mini-mum). Then c must be a critical number of f , i.e., either f′(c) = 0 or f′(c) is undefined.
Example 3.3.12. Find the critical numbers and local extrema of the given function: f (x) = 2x3− 3x2− 12x + 5.
f (x) = (3x + 1)2/3.
f (x) = x3/5(4− x).
Remark 3.3.13 (Very Important Remark). Fermat’s Theorem says that local extrema can occur only at critical numbers. This does not say that there is a local extremum at every critical number.
Example 3.3.14. Find the critical numbers and local extrema of the given function: f (x) = x3and f (x) = x1/3.
You should not forget the first condition on the critical number. The critical number must be in the domain of
the function.
Example 3.3.15. Find all the critical numbers of f (x) = 2x
2
x + 2.
We have observed that local extrema occur only at critical numbers and that continuous functions must have an absolute maximum and an absolute minimum on a closed, bounded interval. However, we haven’t yet really been able to say how to find these extrema. The following Theorem is particularly useful.
Theorem 3.3.16(CANDIDATE OF ABSOLUTE EXTREMA). Suppose that f is continuous on the closed
in-terval [a, b]. Then, the absolute extrema of f must occur
(i) either at an endpoint (a or b) of the interval, or (ii) at a critical number.
When we use the terms maximum, minimum or extremum without specifying absolute or local, we will
always be referring to absolute extrema.
Remark 3.3.17(CLOSEDINTERVALMETHOD). The Theorem above gives us a simple procedure for finding the absolute extrema of a continuous function on a closed, bounded interval:
Step 1. Find all critical numbers in the interval and compute function values at these points. Step 2. Compute function values at the endpoints.
Step 3. The largest function value is the absolute maximum and the smallest function value is the absolute minimum.
Example 3.3.18. Find the absolute extrema of the given function on the given interval: f (x) = 2x3− 3x2− 12x + 5 on [−2,4].
f (x) = 4x3− 8x2+ 5x on [0, 1].
f (x) = x3− 3x2+ 1 on [−1/2,4].
f (x) = x + 2cosx on [0,2π].
Sometimes we need to use a calculator or a computer for the computation.
Example 3.3.19. Find the absolute extrema of the given function on the given interval: f (x) = 4x5/4− 8x1/4 on [0, 4].
§3.4 Increasing and Decreasing Functions.
In this section, we see how to determine which critical numbers correspond to local extrema. At the same time, we will learn more about the connection between the derivative and graphing.
□ 3.4.1 Increasing and Decreasing Functions. Definition 3.4.1. For every x1, x2∈ I with x1< x2,
1. f (x1) < f (x2) −→ f is (strictly) increasing on an interval I
(i.e., f (x) gets larger as x gets larger).
2. f (x1) > f (x2) −→ f is (strictly) decreasing on an interval I
(i.e., f (x) gets smaller as x gets larger).
Theorem 3.4.2. Suppose f is differentiable on an interval I. 1. f′(x) > 0 for all x∈ I −→ f is increasing on I.
2. f′(x) < 0 for all x∈ I −→ f is decreasing on I.
Example 3.4.3. Find the intervals where the function is increasing or decreasing. (Hint: Sign Chart.) f (x) = 3x4− 4x3− 12x2+ 5.
y = x3− 3x2− 9x + 1.
y = sin2
x.
Example 3.4.4. Draw a graph of the given function showing all local extrema: f (x) = 2x3
+ 9x2− 24x − 10.
f (x) = 3x4+ 40x3− 0.06x2− 1.2x. (Hint: f′(x) = 12(x− 0.1)(x + 0.1)(x + 10).)
□ 3.4.2 Critical Point Classification.
Given a critical number c of a function f , how can we tell whether f (c) is a local extreme value? And if it is, how can we tell whether it’s a local maximum or a local minimum?
Theorem 3.4.5(First Derivative Test). Suppose that f is continuous on the interval [a, b] and c∈ (a,b) is a critical number.
1. If f′(x) > 0 for all x∈ (a,c) and f′(x) < 0 for all x∈ (c,b) (i.e., f changes from increasing to decreasing at c), then f (c) is a local maximum.
2. If f′(x) < 0 for all x∈ (a,c) and f′(x) > 0 for all x∈ (c,b) (i.e., f changes from decreasing to increasing at c), then f (c) is a local minimum.
3. If f′(x) has the same sign on (a, c) and (c, b), then f (c) is not a local extremum.
a c b Classification
f′ + 0 − f (c) is a local maximum f′ − 0 + f (c) is a local minimum f′ + 0 + f (c) is not a local extremum f′ − 0 − f (c) is not a local extremum
Example 3.4.6. Find the local extrema of the given function. (Hint: Sign Chart.) f (x) = 3x4− 4x3− 12x2+ 5.
f (x) = x + 2sinx on the interval [0,2π]. f (x) = 3x5
+ 5x3.
f (x) = x5/3− 3x2/3
. f (x) =1− x
x2 .
f (x) = x2/3(15x2− 72x + 120).
Example 3.4.7. (1) Find all critical numbers and (2) use the First Derivative Test to classify each as the location of a local maximum, local minimum or neither.
y = x4+ 4x3− 2.
y = xe−2x. y = x
§3.5 Concavity and the Second Derivative Test. □ 3.5.1 Concavity.
When we say the graph of a function is increasing, it is not clear because there are two types of increasing graphs: concave up and concave down. Similarly, we have two types of decreasing graphs. So we have total 4 types: increasing concave up, increasing concave down and decreasing concave up, decreasing concave down. In this section, we use the Second Derivative Test to determine the concavity.
Definition 3.5.1. For a function f that is differentiable on an interval I, the graph of f is 1. concave up on I if f′is increasing on I or
2. concave down on I if f′is decreasing on I.
Be Careful: The condition is about the increment of the derivative of f , not f itself.
Theorem 3.5.2. Suppose that f′′exists on an interval I.
1. f′′(x) > 0 on I −→ the graph of f is concave up on I. 2. f′′(x) < 0 on I −→ the graph of f is concave down on I.
Example 3.5.3. Determine where the graph of f (x) = 2x3+ 9x2−24x−10 is concave up and concave down and draw a graph showing all significant features of the function.
ANSWER. We find the critical numbers and the roots of f′′= 0 and form the sign chart.
f′(x) = 6x2+ 18x− 24 = 6(x + 4)(x − 1) = 0 at x =−4 or x = 1. So f has the critical numbers −4 and 1.
f′′(x) = 12x + 18 = 6(2x + 3) = 0 at x =−3
2.
x −4 −3
2 1
Sign of f′ + 0 − − − 0 +
Increment of f Inc. Local Max. Dec. Dec. Dec. Local Min. Inc.
Sign of f′′ − − − 0 + + +
Concavity of f Con. Down Con. Down Con. Down No Concavity Con. Up Con. Up Con. Up
From the table, we conclude f is concave downward on (−∞,−3/2) and upward on (−3/2,∞). Significant Features:
1. The graph of f increases on (−∞,−4) ∪ (1,∞), while it decreases on (−4,1).
2. The graph of f has the local maximum f (−4) = 102 at x = −4 and local minimum f (1) = −23 at
x = 1.
3. The graph of f is concave upward on ( −3 2,∞ ) and downward on ( −∞,−3 2 ) . 4. The concavity of the graph of f changes at the point
( −3 2, 79 2 ) on the curve. □
Example 3.5.4. Determine the intervals where the graph of the given function is concave up and concave down:
f (x) = x4− 6x2+ 2x + 3.
g(x) = x + 3(1 − x)1/3
