The derivative is a major tool for investigating the behavior of a function. Since functions are ubiquitous, so are their derivatives. Velocity, growth rates, marginal costs, and material strain are all examples of derivatives. We motivate and define the derivative and compute a few examples, then discuss how features of a function are manifested in its derivative.

1. Section 2.1
The Derivative and Rates of Change
V63.0121.027, Calculus I
September 24, 2009
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OH.
. . . . . .

2. Regarding WebAssign
We feel your pain
. . . . . .

3. Explanations
From the syllabus:
Graders will be expecting you to express your ideas
clearly, legibly, and completely, often requiring
complete English sentences rather than merely just a
long string of equations or unconnected mathematical
expressions. This means you could lose points for
unexplained answers.
. . . . . .

4. Rubric
Points Description of Work
3 Work is completely accurate and essentially perfect.
Work is thoroughly developed, neat, and easy to read.
Complete sentences are used.
2 Work is good, but incompletely developed, hard to
read, unexplained, or jumbled. Answers which are
not explained, even if correct, will generally receive 2
points. Work contains “right idea” but is flawed.
1 Work is sketchy. There is some correct work, but most
of work is incorrect.
0 Work minimal or non-existent. Solution is completely
incorrect.
. . . . . .

5. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .

6. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
. . . . . .

7. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x2 at the point
(2, 4).
. . . . . .

8. Graphically and numerically
y
.
x m
. .
4 .
. . x
.
2
.
. . . . . .

9. Graphically and numerically
y
.
x m
3 5
. .
9 .
. .
4 .
. . . x
.
2
. 3
.
. . . . . .

10. Graphically and numerically
y
.
x m
3 5
2.5 4.25
. .25 .
6 .
. .
4 .
. . . x
.
22
. . .5
. . . . . .

11. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
. .41 .
4 .
. .
4 .
. .. x
.
.. .1
22
. . . . . .

12. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
. .0401 .
4 4
. .
. . x
.
22
. ..01
. . . . . .

13. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
. .
4 .
1 3
. .
1 .
. . . x
.
1
. 2
.
. . . . . .

14. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
. .
4 .
1.5 3.5
. .25 .
2 . 1 3
. . . x
.
1 2
. .5 .
. . . . . .

15. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
. .
4 . 1.9 3.9
. .61 .
3 .
1.5 3.5
1 3
. .. x
.
12
. ..9
. . . . . .

16. Graphically and numerically
y
.
x m
3 5
2.5 4.25
2.1 4.1
2.01 4.01
1.99 3.99
. .9601 .
3 4
. . 1.9 3.9
1.5 3.5
1 3
. . x
.
12
. ..99
. . . . . .

17. Graphically and numerically
y
.
x m
3 5
. .
9 . 2.5 4.25
2.1 4.1
2.01 4.01
. .25 .
6 . limit 4
1.99 3.99
. .41 .
4 .
. .0401 .
4 9601 .
3 . .61
3
4
. .. 1.9 3.9
1.5 3.5
. .25 .
2 . 1 3
. .
1 .
. . . ... . . x
.
1 1 . .2.1 3
. . .5 ..99 .5 .
12 .
2 9201
. . . . . .

18. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x2 at the point
(2, 4).
Upshot
If the curve is given by y = f(x), and the point on the curve is
(a, f(a)), then the slope of the tangent line is given by
f(x) − f(a)
mtangent = lim
x→a x−a
. . . . . .

19. Velocity
Problem
Given the position function of a moving object, find the velocity
of the object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can
be described by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the
ground. How fast is it falling one second after we drop it?
. . . . . .

20. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
. . . . . .

21. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5
. . . . . .

22. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
. . . . . .

23. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1
. . . . . .

24. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
. . . . . .

25. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01
. . . . . .

26. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
. . . . . .

27. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001
. . . . . .

28. Numerical evidence
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
. . . . . .

29. Velocity
Problem
Given the position function of a moving object, find the velocity
of the object at a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can
be described by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the
ground. How fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
= (−5) lim(1 + t) = −5 · 2 = −10
t→1
. . . . . .

30. y
. = h (t )
Upshot .
If the height function is given
by h(t), the instantaneous
velocity at time t0 is given by .
h(t) − h(t0 )
v = lim
t→t0 t − t0
h(t0 + ∆t) − h(t0 )
= lim
∆t→0 ∆t
. . . t .
∆
t
.
t
.0 t
.
. . . . . .

31. Population growth
Problem
Given the population function of a group of organisms, find the
rate of growth of the population at a particular instant.
. . . . . .

32. Population growth
Problem
Given the population function of a group of organisms, find the
rate of growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the
function
3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the
fish population growing fastest in 1990, 2000, or 2010? (Estimate
numerically)?
. . . . . .

33. Derivation
Let ∆t be an increment in time and ∆P the corresponding change
in population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et
. . . . . .

34. Derivation
Let ∆t be an increment in time and ∆P the corresponding change
in population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so we want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et
Too hard! Try a small ∆t to approximate.
. . . . . .

35. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈
0.1
. . . . . .

36. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈ 0.000136
0.1
. . . . . .

37. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈ 0.000136
0.1
P(0.1) − P(0)
r2000 ≈ ≈
0.1
. . . . . .

38. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈ 0.000136
0.1
P(0.1) − P(0)
r2000 ≈ ≈ 0.75
0.1
. . . . . .

39. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈ 0.000136
0.1
P(0.1) − P(0)
r2000 ≈ ≈ 0.75
0.1
P(10 + 0.1) − P(10)
r2010 ≈ ≈
0.1
. . . . . .

40. Numerical evidence
P(−10 + 0.1) − P(−10)
r1990 ≈ ≈ 0.000136
0.1
P(0.1) − P(0)
r2000 ≈ ≈ 0.75
0.1
P(10 + 0.1) − P(10)
r2010 ≈ ≈ 0.000136
0.1
. . . . . .

41. Population growth
Problem
Given the population function of a group of organisms, find the
rate of growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the
function
3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the
fish population growing fastest in 1990, 2000, or 2010? (Estimate
numerically)?
Solution
The estimated rates of growth are 0.000136, 0.75, and 0.000136.
. . . . . .

42. Upshot
The instantaneous population growth is given by
P(t + ∆t) − P(t)
lim
∆t→0 ∆t
. . . . . .

43. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
. . . . . .

44. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a
year is
C(q) = q3 − 12q2 + 60q
We are currently producing 5 tons a year. Should we change that?
. . . . . .

45. Comparisons
q C (q )
4
5
6
. . . . . .

46. Comparisons
q C (q )
4 112
5
6
. . . . . .

47. Comparisons
q C (q )
4 112
5 125
6
. . . . . .

48. Comparisons
q C (q )
4 112
5 125
6 144
. . . . . .

49. Comparisons
q C(q) AC(q) = C(q)/q
4 112
5 125
6 144
. . . . . .

50. Comparisons
q C(q) AC(q) = C(q)/q
4 112 28
5 125
6 144
. . . . . .

51. Comparisons
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144
. . . . . .

52. Comparisons
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144 24
. . . . . .

53. Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28
5 125 25
6 144 24
. . . . . .

54. Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25
6 144 24
. . . . . .

55. Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25 19
6 144 24
. . . . . .

56. Comparisons
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25 19
6 144 24 31
. . . . . .

57. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a
year is
C(q) = q3 − 12q2 + 60q
We are currently producing 5 tons a year. Should we change that?
Example
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should
produce more to lower average costs.
. . . . . .

58. Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but depends on units.
. . . . . .

59. Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but depends on units.
The marginal cost after producing q given by
C(q + ∆q) − C(q)
MC = lim
∆q→0 ∆q
is more useful since it’s unit-independent.
. . . . . .

60. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .

61. The definition
All of these rates of change are found the same way!
. . . . . .

62. The definition
All of these rates of change are found the same way!
Definition
Let f be a function and a a point in the domain of f. If the limit
f(a + h) − f(a)
f′ (a) = lim
h→0 h
exists, the function is said to be differentiable at a and f′ (a) is the
derivative of f at a.
. . . . . .

63. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
. . . . . .

64. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
(a 2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h) = 2a.
h→0
. . . . . .

65. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2).
. . . . . .

66. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x x
.
definition of the derivative to
find f′ (2).
Solution
1/x − 1/2 2−x .
f′ (2) = lim = lim . x
.
x→2 x−2 x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .

67. The Sure-Fire Sally Rule (SFSR) for adding Fractions
In anticipation of the question, “How did you get that?”
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x−2 x−2
2−x
=
2x(x − 2)
. . . . . .

68. The Sure-Fire Sally Rule (SFSR) for adding Fractions
In anticipation of the question, “How did you get that?”
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x−2 x−2
2−x
=
2x(x − 2)
. . . . . .

69. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′ ?
. . . . . .

70. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′ ?
If f is decreasing on an interval, f′ is negative (well,
nonpositive) on that interval
. . . . . .

71. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x x
.
definition of the derivative to
find f′ (2).
Solution
1/x − 1/2 2−x .
f′ (2) = lim = lim . x
.
x→2 x−2 x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .

72. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each
point x where f is differentiable, and come up with another
function, the derivative function.
What can we say about this function f′ ?
If f is decreasing on an interval, f′ is negative (well,
nonpositive) on that interval
If f is increasing on an interval, f′ is positive (well,
nonnegative) on that interval
. . . . . .

73. Graphically and numerically
y
.
x m
3 5
. .
9 . 2.5 4.25
2.1 4.1
2.01 4.01
. .25 .
6 . limit 4
1.99 3.99
. .41 .
4 .
. .0401 .
4 9601 .
3 . .61
3
4
. .. 1.9 3.9
1.5 3.5
. .25 .
2 . 1 3
. .
1 .
. . . ... . . x
.
1 1 . .2.1 3
. . .5 ..99 .5 .
12 .
2 9201
. . . . . .

74. What does f tell you about f′ ?
Fact
If f is decreasing on (a, b), then f′ ≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
. . . . . .

75. What does f tell you about f′ ?
Fact
If f is decreasing on (a, b), then f′ ≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
still!
. . . . . .

76. What does f tell you about f′ ?
Fact
If f is decreasing on (a, b), then f′ ≤ 0 on (a, b).
Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
f(x + ∆x) − f(x)
still! Either way, < 0, so
∆x
f(x + ∆x) − f(x)
f′ (x) = lim ≤0
∆x→0 ∆x
. . . . . .

77. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .

78. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
. . . . . .

79. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0
. . . . . .

80. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0
Note the proper use of the limit law: if the factors each have a
limit at a, the limit of the product is the product of the limits.
. . . . . .

81. How can a function fail to be differentiable?
Kinks
f
.(x)
. x
.
. . . . . .

82. How can a function fail to be differentiable?
Kinks
f
.(x) . ′ (x )
f
. x
. . x
.
. . . . . .

83. How can a function fail to be differentiable?
Kinks
f
.(x) . ′ (x )
f
.
. x
. . x
.
. . . . . .

84. How can a function fail to be differentiable?
Cusps
f
.(x)
. x
.
. . . . . .

85. How can a function fail to be differentiable?
Cusps
f
.(x) . ′ (x )
f
. x
. . x
.
. . . . . .

86. How can a function fail to be differentiable?
Cusps
f
.(x) . ′ (x )
f
. x
. . x
.
. . . . . .

87. How can a function fail to be differentiable?
Vertical Tangents
f
.(x)
. x
.
. . . . . .

88. How can a function fail to be differentiable?
Vertical Tangents
f
.(x) . ′ (x )
f
. x
. . x
.
. . . . . .

89. How can a function fail to be differentiable?
Vertical Tangents
f
.(x) . ′ (x )
f
. x
. . x
.
. . . . . .

90. How can a function fail to be differentiable?
Weird, Wild, Stuff
f
.(x)
. x
.
This function is differentiable
at 0.
. . . . . .

91. How can a function fail to be differentiable?
Weird, Wild, Stuff
f
.(x) . ′ (x )
f
. x
. . x
.
This function is differentiable But the derivative is not
at 0. continuous at 0!
. . . . . .

92. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .

93. Notation
Newtonian notation
f ′ (x ) y′ (x) y′
Leibnizian notation
dy d df
f(x)
dx dx dx
These all mean the same thing.
. . . . . .

94. Meet the Mathematician: Isaac Newton
English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687
. . . . . .

95. Meet the Mathematician: Gottfried Leibniz
German, 1646–1716
Eminent philosopher as
well as mathematician
Contemporarily
disgraced by the
calculus priority dispute
. . . . . .

96. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .

97. The second derivative
If f is a function, so is f′ , and we can seek its derivative.
f′′ = (f′ )′
It measures the rate of change of the rate of change!
. . . . . .

98. The second derivative
If f is a function, so is f′ , and we can seek its derivative.
f′′ = (f′ )′
It measures the rate of change of the rate of change! Leibnizian
notation:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2
. . . . . .

99. function, derivative, second derivative
y
.
. (x ) = x 2
f
.′ (x) = 2x
f
.′′ (x) = 2
f
. x
.
. . . . . .