1. 9 The Derivative Limits One-sided Limits and Continuity The Derivative
Basic Rules of Differentiation
The Product and Quotient Rules: Higher-Order Derivatives
The Chain Rule
Differentiation of Exponential and Logarithmic Functions
Marginal Functions in Economics

2. Basic Rules of Differentiation
9.4
Basic Rules of Differentiation

3. Four Basic Rules
We’ve learned that to find the rule for the derivative f ′of a function f, we first find the difference quotient
But this method is tedious and time consuming, even for relatively simple functions.
This chapter we will develop rules that will simplify the process of finding the derivative of a function.

4. Rule 1: Derivative of a Constant
We will use the notation
to mean “the derivative of f with respect to x at x.”
Rule 1: Derivative of a constant
The derivative of a constant function is equal to zero.

5. Rule 1: Derivative of a Constant
We can see geometrically why the derivative of a constant must be zero.
The graph of a constant function is a straight line parallel to the x axis.
Such a line has a slope that is constant with a value of zero.
Thus, the derivative of a constant must be zero as well. y
f(x) = c x

6. Rule 1: Derivative of a Constant
We can use the definition of the derivative to demonstrate this:

7. Rule 2: The Power Rule Rule 2: The Power Rule
If n is any real number, then

8. Rule 2: The Power Rule
Lets verify this rule for the special case of n = 2.
If f(x) = x2, then

9. Rule 2: The Power Rule Practice Examples: If f(x) = x, then

10. Rule 2: The Power Rule
Practice Examples:
Find the derivative of

11. Rule 2: The Power Rule
Practice Examples:
Find the derivative of

12. Rule 3: Derivative of a Constant Multiple Function
If c is any constant real number, then

13. Rule 3: Derivative of a Constant Multiple Function
Practice Examples:
Find the derivative of

14. Rule 3: Derivative of a Constant Multiple Function
Practice Examples:
Find the derivative of

15. Rule 4: The Sum Rule
Rule 4: The Sum Rule

16. Rule 4: The Sum Rule
Practice Examples:
Find the derivative of

17. Rule 4: The Sum Rule
Practice Examples:
Find the derivative of

18. Applied Example: Conservation of a Species
A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of conservation measures be carried out over the next decade to save a certain species of whale from extinction.
After implementing the conservation measure, the population of this species is expected to be
where N(t) denotes the population at the end of year t.
Find the rate of growth of the whale population when t = 2 and
t = 6.
How large will the whale population be 8 years after implementing the conservation measures?

19. Applied Example: Conservation of a Species
Solution
The rate of growth of the whale population at any time t is given by
In particular, for t = 2, we have
And for t = 6, we have
Thus, the whale population’s rate of growth will be 34 whales per year after 2 years and 338 per year after 6 years.

20. Applied Example: Conservation of a Species
Solution
The whale population at the end of the eighth year will be

21. The Product and Quotient Rules
9.5
The Product and Quotient Rules

22. Rule 5: The Product Rule
The derivative of the product of two differentiable functions is given by

23. Rule 5: The Product Rule
Practice Examples:
Find the derivative of

24. Rule 5: The Product Rule
Practice Examples:
Find the derivative of

25. Rule 6: The Quotient Rule
The derivative of the quotient of two differentiable functions is given by

26. Rule 6: The Quotient Rule
Practice Examples:
Find the derivative of

27. Rule 6: The Quotient Rule
Practice Examples:
Find the derivative of

28. Applied Example: Rate of Change of DVD Sales
The sales ( in millions of dollars) of DVDs of a hit movie t years from the date of release is given by
Find the rate at which the sales are changing at time t.
How fast are the sales changing at:
The time the DVDs are released (t = 0)?
And two years from the date of release (t = 2)?

29. Applied Example: Rate of Change of DVD Sales
Solution
The rate of change at which the sales are changing at time t is given by

30. Applied Example: Rate of Change of DVD Sales
Solution
The rate of change at which the sales are changing when the DVDs are released (t = 0) is
That is, sales are increasing by $5 million per year.

31. Applied Example: Rate of Change of DVD Sales
Solution
The rate of change two years after the DVDs are released (t = 2) is
That is, sales are decreasing by $600,000 per year.

32. Higher-Order Derivatives
The derivative f ′ of a function f is also a function.
As such, f ′ may also be differentiated.
Thus, the function f ′ has a derivative f ″ at a point x in the domain of f if the limit of the quotient
exists as h approaches zero.
The function f ″ obtained in this manner is called the second derivative of the function f, just as the derivative f ′ of f is often called the first derivative of f.
By the same token, you may consider the third, fourth, fifth, etc. derivatives of a function f.

33. Higher-Order Derivatives
Practice Examples:
Find the third derivative of the function f(x) = x2/3 and determine its domain.
Solution
We have and
So the required derivative is
The domain of the third derivative is the set of all real numbers except x = 0.

34. Higher-Order Derivatives
Practice Examples:
Find the second derivative of the function f(x) = (2x2 +3)3/2
Solution
Using the general power rule we get the first derivative:

35. Higher-Order Derivatives
Practice Examples:
Find the second derivative of the function f(x) = (2x2 +3)3/2
Solution
Using the product rule we get the second derivative:

36. Applied Example: Acceleration of a Maglev
The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function
s = 4t2 (0  t  10)
What is the maglev’s acceleration after 30 seconds?
Solution
The velocity of the maglev t seconds from rest is given by
The acceleration of the maglev t seconds from rest is given by the rate of change of the velocity of t, given by
or 8 feet per second per second (ft/sec2).

37. 9.6
The Chain Rule

38. Deriving Composite Functions
Consider the function
To compute h′(x), we can first expand h(x)
and then derive the resulting polynomial
But how should we derive a function like H(x)?

39. Deriving Composite Functions
Note that is a composite function:
H(x) is composed of two simpler functions
So that
We can use this to find the derivative of H(x).

40. Deriving Composite Functions
To find the derivative of the composite function H(x):
We let u = f(x) = x2 + x and y = g(u) = u100.
Then we find the derivatives of each of these functions
The ratios of these derivatives suggest that
Substituting x2 + x + 1 for u we get

41. Rule 7: The Chain Rule If h(x) = g[f(x)], then
Equivalently, if we write y = h(x) = g(u),
where u = f(x), then

42. The Chain Rule for Power Functions
Many composite functions have the special form
h(x) = g[f(x)]
where g is defined by the rule
g(x) = xn (n, a real number)
so that
h(x) = [f(x)]n
In other words, the function h is given by the power of a function f.
Examples:

43. The General Power Rule If the function f is differentiable and
h(x) = [f(x)]n (n, a real number),
then

44. The General Power Rule Practice Examples: Find the derivative of
Solution
Rewrite as a power function:
Apply the general power rule:

45. The General Power Rule Practice Examples: Find the derivative of
Solution
Apply the product rule and the general power rule:

46. The General Power Rule Practice Examples: Find the derivative of
Solution
Rewrite as a power function:
Apply the general power rule:

47. The General Power Rule Practice Examples: Find the derivative of
Solution
Apply the general power rule and the quotient rule:

48. Applied Problem: Arteriosclerosis
Arteriosclerosis begins during childhood when plaque forms in the arterial walls, blocking the flow of blood through the arteries and leading to heart attacks, stroke and gangrene.

49. Applied Problem: Arteriosclerosis
Suppose the idealized cross section of the aorta is circular with radius a cm and by year t the thickness of the plaque is
h = g(t) cm
then the area of the opening is given by
A = p (a – h)2 cm2
Further suppose the radius of an individual’s artery is 1 cm (a = 1) and the thickness of the plaque in year t is given by
h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm

50. Applied Problem: Arteriosclerosis
Then we can use these functions for h and A
h = g(t) = 1 – 0.01(10,000 – t2)1/ A = f(h) = p (1 – h)2
to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule:

51. Applied Problem: Arteriosclerosis
For example, at age 50 (t = 50),
So that
That is, the area of the arterial opening is decreasing at the rate of 0.03 cm2 per year for a typical 50 year old.

52. Differentiation of the Exponential and Logarithmic Functions
9.7
Differentiation of the Exponential and Logarithmic Functions

53. Rule 8 Derivative of the Exponential Function
The derivative of the exponential function with base e is equal to the function itself:

54. Examples Find the derivative of the function Solution
Using the product rule gives

55. Examples Find the derivative of the function Solution
Using the general power rule gives

56. Rule 9 Chain Rule for Exponential Functions
If f(x) is a differentiable function, then

57. Examples
Find the derivative of the function
Solution

58. Examples
Find the derivative of the function
Solution

59. Examples
Find the derivative of the function
Solution

60. Examples
Find the derivative of the function
Solution

61. Examples
Find the derivative of the function
Solution

62. Rule 10 Derivative of the Natural Logarithm
The derivative of ln x is

63. Examples
Find the derivative of the function
Solution

64. Examples
Find the derivative of the function
Solution

65. Rule 11 Chain Rule for Logarithmic Functions
If f(x) is a differentiable function, then

66. Examples
Find the derivative of the function
Solution

67. Examples
Find the derivative of the function
Solution

68. Marginal Functions in Economics
9.8
Marginal Functions in Economics
MC(251) = C(251) – C(250)
= [ (251) – 0.2(251)2]
– [ (250) – 0.2(250)2]
= 45,599.8 – 45,500
= 99.80

69. Marginal Analysis
Marginal analysis is the study of the rate of change of economic quantities.
These may have to do with the behavior of costs, revenues, profit, output, demand, etc.
In this section we will discuss the marginal analysis of various functions related to:
Cost
Average Cost
Revenue
Profit

70. Applied Example: Rate of Change of Cost Functions
Suppose the total cost in dollars incurred each week by Polaraire for manufacturing x refrigerators is given by the total cost function
C(x) = x – 0.2x (0  x  400)
What is the actual cost incurred for manufacturing the 251st refrigerator?
Find the rate of change of the total cost function with respect to x when x = 250.
Compare the results obtained in parts (a) and (b).

71. Applied Example: Rate of Change of Cost Functions
Solution
The cost incurred in producing the 251st refrigerator is
C(251) – C(250) = [ (251) – 0.2(251)2]
– [ (250) – 0.2(250)2]
= 45,599.8 – 45,500
= 99.80
or $99.80.

72. Applied Example: Rate of Change of Cost Functions
Solution
The rate of change of the total cost function
C(x) = x – 0.2x2
with respect to x is given by
C´(x) = 200 – 0.4x
So, when production is 250 refrigerators, the rate of change of the total cost with respect to x is
C´(x) = 200 – 0.4(250)
= 100
or $100.

73. Applied Example: Rate of Change of Cost Functions
Solution
Comparing the results from (a) and (b) we can see they are very similar: $99.80 versus $100.
This is because (a) measures the average rate of change over the interval [250, 251], while (b) measures the instantaneous rate of change at exactly x = 250.
The smaller the interval used, the closer the average rate of change becomes to the instantaneous rate of change.

74. Applied Example: Rate of Change of Cost Functions
Solution
The actual cost incurred in producing an additional unit of a good is called the marginal cost.
As we just saw, the marginal cost is approximated by the rate of change of the total cost function.
For this reason, economists define the marginal cost function as the derivative of the total cost function.

75. Applied Example: Marginal Cost Functions
A subsidiary of Elektra Electronics manufactures a portable music player.
Management determined that the daily total cost of producing these players (in dollars) is
C(x) = x3 – 0.08x2 + 40x
where x stands for the number of players produced.
Find the marginal cost function.
Find the marginal cost for x = 200, 300, 400, and 600.
Interpret your results.

76. Applied Example: Marginal Cost Functions
Solution
If the total cost function is:
C(x) = x3 – 0.08x2 + 40x
then, its derivative is the marginal cost function:
C´(x) = x2 – 0.16x + 40

77. Applied Example: Marginal Cost Functions
Solution
The marginal cost for x = 200, 300, 400, and 600 is:
C´(200) = (200)2 – 0.16(200) + 40 = 20
C´(300) = (300)2 – 0.16(300) + 40 = 19
C´(400) = (400)2 – 0.16(400) + 40 = 24
C´(600) = (600)2 – 0.16(600) + 40 = 52
or $20/unit, $19/unit, $24/unit, and $52/unit, respectively.

78. Applied Example: Marginal Cost Functions
Solution
From part (b) we learn that at first the marginal cost is decreasing, but as output increases, the marginal cost increases as well.
This is a common phenomenon that occurs because of several factors, such as excessive costs due to overtime and high maintenance costs for keeping the plant running at such a fast rate.

79. Applied Example: Marginal Revenue Functions
Suppose the relationship between the unit price p in dollars and the quantity demanded x of the Acrosonic model F loudspeaker system is given by the equation
p = – 0.02x (0  x  20,000)
Find the revenue function R.
Find the marginal revenue function R′.
Compute R′(2000) and interpret your result.

80. Applied Example: Marginal Revenue Functions
Solution
The revenue function is given by
R(x) = px
= (– 0.02x + 400)x
= – 0.02x x
(0  x  20,000)

81. Applied Example: Marginal Revenue Functions
Solution
Given the revenue function
R(x) = – 0.02x x
We find its derivative to obtain the marginal revenue function:
R′(x) = – 0.04x + 400

82. Applied Example: Marginal Revenue Functions
Solution
When quantity demanded is 2000, the marginal revenue will be:
R′(2000) = – 0.04(2000) + 400
= 320
Thus, the actual revenue realized from the sale of the 2001st loudspeaker system is approximately $320.

83. Applied Example: Marginal Profit Function
Continuing with the last example, suppose the total cost (in dollars) of producing x units of the Acrosonic model F loudspeaker system is
C(x) = 100x + 200,000
Find the profit function P.
Find the marginal profit function P′.
Compute P′ (2000) and interpret the result.

84. Applied Example: Marginal Profit Function
Solution
From last example we know that the revenue function is
R(x) = – 0.02x x
Profit is the difference between total revenue and total cost, so the profit function is
P(x) = R(x) – C(x)
= (– 0.02x x) – (100x + 200,000)
= – 0.02x x – 200,000

85. Applied Example: Marginal Profit Function
Solution
Given the profit function
P(x) = – 0.02x x – 200,000
we find its derivative to obtain the marginal profit function:
P′(x) = – 0.04x + 300

86. Applied Example: Marginal Profit Function
Solution
When producing x = 2000, the marginal profit is
P′(2000) = – 0.04(2000) + 300
= 220
Thus, the profit to be made from producing the 2001st loudspeaker is $220.

87. End of Chapter