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9
• 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
The Derivative
9.4Basic Rules of Differentiation
5 4 2
5 4 2
( ) 4 3 8 3
4 3 8 3
df x x x x x
dxd d d d d
x x x xdx dx dx dx dx
4 3
4 3
4 5 3 4 8 2 1 0
20 12 16 1
x x x
x x x
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.
0
( ) ( )limh
f x h f x
h
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.
( )d
f xdx
0d
cdx
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.
f(x) = c
y
x
Rule 1: Derivative of a Constant• We can use the definition of the derivative
to demonstrate this:
0
0
0
( ) ( )( )
0
0
lim
lim
lim
h
h
h
f x h f xf x
hc c
h
Rule 2: The Power Rule
Rule 2: The Power Rule• If n is any real number, then
1n ndx nx
dx
Rule 2: The Power Rule
• Lets verify this rule for the special case of n = 2.• If f(x) = x2, then
2
2 2 2 2 2
0
0 0
( ) ( )( )
( ) 2
lim
lim lim
h
h h
d f x h f xf x x
dx h
x h x x xh h x
h h
2
0 0
0
2 (2 )
(2 ) 2
lim lim
lim
h h
h
xh h h x h
h h
x h x
Rule 2: The Power Rule
Practice Examples:• If f(x) = x, then
• If f(x) = x8, then
• If f(x) = x5/2, then
1 1 0( ) 1 1d
f x x x xdx
8 8 1 7( ) 8 8d
f x x x xdx
5/2 5/2 1 3/25 5( )
2 2
df x x x x
dx
Rule 2: The Power Rule
Practice Examples:• Find the derivative of
( )f x x
1/2( )d d
f x x xdx dx
1/2 11
2x 1/21
2x
1
2 x
Rule 2: The Power Rule
Practice Examples:• Find the derivative of
3
1( )f x
x
1/3
3
1( )
d df x x
dx dxx
1/3 11
3x
4/34/3
1 1
3 3x
x
Rule 3: Derivative of a Constant Multiple Function
Rule 3: Derivative of a Constant Multiple Function
• If c is any constant real number, then
( ) ( )d d
cf x c f xdx dx
Rule 3: Derivative of a Constant Multiple Function
Practice Examples:• Find the derivative of
3( ) 5f x x
3
3
( ) 5
5
df x x
dxd
xdx
2
2
5 3
15
x
x
Rule 3: Derivative of a Constant Multiple Function
Practice Examples:• Find the derivative of
3( )f x
x
1/ 2( ) 3d
f x xdx
3/ 2
3/ 2
13
2
3
2
x
x
Rule 4: The Sum Rule
Rule 4: The Sum Rule
( ) ( ) ( ) ( )d d d
f x g x f x g xdx dx dx
Rule 4: The Sum Rule
Practice Examples:• Find the derivative of
5 4 2( ) 4 3 8 3f x x x x x
5 4 2
5 4 2
( ) 4 3 8 3
4 3 8 3
df x x x x x
dxd d d d d
x x x xdx dx dx dx dx
4 3
4 3
4 5 3 4 8 2 1 0
20 12 16 1
x x x
x x x
Rule 4: The Sum Rule
Practice Examples:• Find the derivative of
2
3
5( )
5
tg t
t
22 3
3
2 3
5 1( ) 5
5 5
15
5
d t dg t t t
dt t dt
d dt t
dt dt
4
5
4 4
12 5 3
5
2 15 2 75
5 5
t t
t t
t t
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?
3 2( ) 3 2 10 600 (0 10) N t t t t t
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.
2( ) 9 4 10N t t t
2(2) 9 2 4 2 10 34N
2(6) 9 6 4 6 10 338N
Applied Example: Conservation of a Species
Solution• The whale population at the end of the eighth
year will be
3 28 3 8 2 8 10 8 600
2184 whales
N
9.5The Product and Quotient Rules
( ) ( ) ( ) ( ) ( ) ( )d
f x g x f x g x g x f xdx
2
( ) ( ) ( ) ( ) ( )
( ) ( )
d f x g x f x f x g x
dx g x g x
Rule 5: The Product Rule
• The derivative of the product of two differentiable functions is given by
( ) ( ) ( ) ( ) ( ) ( )d
f x g x f x g x g x f xdx
Rule 5: The Product Rule
Practice Examples:• Find the derivative of
2 3( ) 2 1 3f x x x
2 3 3 2( ) 2 1 3 3 2 1d d
f x x x x xdx dx
2 2 32 1 3 3 4x x x x
4 2 4
3
6 3 4 12
10 3 12
x x x x
x x x
Rule 5: The Product Rule
Practice Examples:• Find the derivative of
3( ) 1f x x x
3 1/2 1/2 3( ) 1 1d d
f x x x x xdx dx
3 1/2 1/2 211 3
2x x x x
5/2 5/2 2
5/2 2
13 3
27
32
x x x
x x
Rule 6: The Quotient Rule
• The derivative of the quotient of two differentiable functions is given by
2
( ) ( ) ( ) ( ) ( ) 0
( ) ( )
d f x g x f x f x g xg x
dx g x g x
Rule 6: The Quotient Rule
Practice Examples:• Find the derivative of
( )2 4
xf x
x
2
2 4 ( ) 2 4( )
2 4
d dx x x x
dx dxf xx
2
2 4 1 2
2 4
x x
x
2 2
2 4 2 4
2 4 2 4
x x
x x
Rule 6: The Quotient Rule
Practice Examples:• Find the derivative of
2
2
1( )
1
xf x
x
2 2 2 2
22
1 1 1 1( )
1
d dx x x x
dx dxf xx
2 2
22
1 2 1 2
1
x x x x
x
3 3
2 22 2
2 2 2 2 4
1 1
x x x x x
x x
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)?
2
5( )
1
tS t
t
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
2
22
1 5 5 2
1
t t t
t
2
5( )
1
d tS t
dt t
22 2
2 22 2
5 15 5 10
1 1
tt t
t t
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.
2
2 22
5 1 0 5 1(0) 5
10 1S
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.
2
2 22
5 1 2 5 1 4 15 3(2) 0.6
25 54 12 1S
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.
( ) ( )f x h f x
h
Higher-Order DerivativesPractice 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.
1/32( )
3f x x 4/3 4/32 1 2
( )3 3 9
f x x x
7/3 7/37/3
2 4 8 8( )
9 3 27 27f x x x
x
Higher-Order DerivativesPractice 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:
1/2 1/22 23( ) 2 3 4 6 2 3
2f x x x x x
Higher-Order DerivativesPractice Examples:• Find the second derivative of the function f(x) = (2x2 +3)3/2
Solution
• Using the product rule we get the second derivative:
1/2 1/22 2
1/2 1/22 2
( ) 6 2 3 2 3 6
16 2 3 4 2 3 6
2
d df x x x x x
dx dx
x x x x
1/2 1/22 2 2
1/22 2 2
2
2
12 2 3 6 2 3
6 2 3 2 2 3
6 4 3
2 3
x x x
x x x
x
x
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).
24 8ds d
v t tdt dt
2
28 8
d d ds d s da v t
dt dt dt dt dt
9.6The Chain Rule
dy dy du
dx du dx
( ) ( ) ( ) ( )d
h x g f x g f x f xdx
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)?
22( ) 1h x x x
22 2 2
4 3 2
( ) 1 1 1
2 3 2 1
h x x x x x x x
x x x x
3 2( ) 4 6 6 2h x x x x
1002( ) 1H x x x
Deriving Composite FunctionsNote 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).
1002( ) 1H x x x
2 100( ) 1 ( )f x x x g x x and
100100 2( ) ( ) ( ) 1H x g f x f x x x
Deriving Composite FunctionsTo find the derivative of the composite function H(x):
• We let u = f(x) = x2 + x + 1 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
99( ) 2 1 ( ) 100du dy
f x x g u udx du
and
99100 2 1dy dy du
u xdx du dx
992( ) 100 1 2 1dy
H x x x xdx
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
dy dy du
dx du dx
( ) ( ) ( ) ( )d
h x g f x g f x f xdx
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 thath(x) = [f(x)]n
• In other words, the function h is given by the power of a function f.
• Examples:
1002 233
1( ) 1 ( ) ( ) 2 3
5h x x x H x G x x
x
The General Power Rule
• If the function f is differentiable and
h(x) = [f(x)]n (n, a real number),then
1( ) ( ) ( ) ( )
n ndh x f x n f x f x
dx
The General Power RulePractice Examples:• Find the derivative of Solution• Rewrite as a power function:• Apply the general power rule:
2( ) 1G x x
1/22( ) 1G x x
1/22 2
1/22
2
1( ) 1 1
21
1 22
1
dG x x x
dx
x x
x
x
The General Power RulePractice Examples:• Find the derivative of Solution• Apply the product rule and the general power rule:
52( ) 2 3f x x x
5 52 2( ) 2 3 2 3d d
f x x x x xdx dx
4 52
4
4
10 2 3 2 2 3
2 2 3 5 2 3
2 2 3 7 3
x x x x
x x x x
x x x
4 52 5 2 3 2 2 3 2x x x x
The General Power RulePractice Examples:• Find the derivative of
Solution• Rewrite as a power function:• Apply the general power rule:
22
1( )
4 7f x
x
32( ) 2 4 7 8f x x x
22( ) 4 7f x x
32
16
4 7
x
x
The General Power RulePractice Examples:• Find the derivative of Solution• Apply the general power rule and the quotient rule:
32 1
( )3 2
xf x
x
22 1 2 1
( ) 33 2 3 2
x d xf x
x dx x
22
2 4
3 2 12 1 6 4 6 33
3 2 3 2 3 2
xx x x
x x x
2
2
3 2 2 2 1 32 13
3 2 3 2
x xx
x x
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.
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
Applied Problem: Arteriosclerosis • Then we can use these functions for h and A
h = g(t) = 1 – 0.01(10,000 – t2)1/2 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:
( ) ( )dA dA dh
f h g tdt dh dt
1/22
1/22
2
12 (1 )( 1) 0.01 10,000 ( 2 )
2
0.012 (1 )
10,000
0.02 (1 )
10,000
h t t
th
t
h t
t
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.
0.02 (1 0.134)500.03
10,000 2500
dA
dt
1/2(50) 1 0.01(10,000 2500) 0.134h g
9.7Differentiation of the Exponential and Logarithmic Functions
1/212
,e 1/212
,e
1
x
y
– 1 1
1/212
,e 1/212
,e
2
( ) xf x e2
( ) xf x e
Rule 8Derivative of the Exponential Function
• The derivative of the exponential function with base e is equal to the function itself:
x xde e
dx
Examples• Find the derivative of the functionSolution• Using the product rule gives
2( ) xf x x e
2 2 2
2
( )
(2 )
( 2)
x x x
x x
x
d d df x x e x e e x
dx dx dx
x e e x
xe x
Examples• Find the derivative of the functionSolution• Using the general power rule gives
3/2( ) 2tg t e
1/2
1/2
1/2
3( ) 2 2
23
223
22
t t
t t
t t
dg t e e
dt
e e
e e
Rule 9Chain Rule for Exponential Functions
• If f(x) is a differentiable function, then
( ) ( ) ( )f x f xde e f x
dx
Examples• Find the derivative of the functionSolution
2( ) xf x e
2
2
2
( ) 2
(2)
2
x
x
x
df x e x
dx
e
e
Examples• Find the derivative of the functionSolution
3xy e
3
3
3
( 3 )
( 3)
3
x
x
x
dy de x
dx dx
e
e
Examples• Find the derivative of the functionSolution
22( ) t tg t e
2
2
2 2
2
( ) 2
(4 1)
t t
t t
dg t e t t
dt
t e
Examples• Find the derivative of the functionSolution
2 xy xe
2 2
2 2
2 2
2 2
2
2 (1)
( 2)
2
(1 2 )
x x
x x
x x
x x
x
dy d dx e e x
dx dx dxd
x e x edx
xe e
xe e
e x
Examples• Find the derivative of the functionSolution
( )t
t t
eg t
e e
2
2
2 2
2
2
( )
1 1
2
t t t t t t
t t
t t t t t t
t t
t t
t t
t t
d de e e e e e
dt dtg te e
e e e e e e
e e
e e
e e
e e
Rule 10Derivative of the Natural Logarithm
• The derivative of ln x is1
ln ( 0)d
x xdx x
Examples• Find the derivative of the functionSolution
( ) lnf x x x
( ) (ln ) ln ( )
1ln (1)
1 ln
d df x x x x x
dx dx
x xx
x
Examples• Find the derivative of the functionSolution
ln( )
xg x
x
2
2
2
(ln ) ln ( )( )
1ln (1)
1 ln
d dx x x x
dx dxg xx
x xx
xx
x
Rule 11Chain Rule for Logarithmic Functions
• If f(x) is a differentiable function, then
( )ln ( ) [ ( ) 0]
( )
d f xf x f x
dx f x
Examples• Find the derivative of the functionSolution
2( ) ln( 1)f x x
2
2
2
1( )
12
1
dx
dxf xxx
x
Examples• Find the derivative of the functionSolution
2 3 6ln[( 1)( 2) ]y x x
2 3 6
2 3 6
2 3
ln[( 1)( 2) ]
ln( 1) ln( 2)
ln( 1) 6ln( 2)
y x x
x x
x x
2 3
2 3
2
2 3
2
2 3
( 1) ( 2)6
1 2
2 36
1 2
2 18
1 2
d dx xdy dx dx
dx x x
x x
x x
x x
x x
9.8Marginal Functions in Economics
MC(251) = C(251) – C(250)
= [8000 + 200(251) – 0.2(251)2]
– [8000 + 200(250) – 0.2(250)2]
= 45,599.8 – 45,500
= 99.80
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
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) = 8000 + 200x – 0.2x2 (0 x 400)a. What is the actual cost incurred for manufacturing
the 251st refrigerator?b. Find the rate of change of the total cost function
with respect to x when x = 250.c. Compare the results obtained in parts (a) and (b).
Applied Example: Rate of Change of Cost Functions
Solutiona. The cost incurred in producing the 251st refrigerator is
C(251) – C(250) = [8000 + 200(251) –
0.2(251)2] – [8000 + 200(250) – 0.2(250)2]
= 45,599.8 – 45,500= 99.80
or $99.80.
Applied Example: Rate of Change of Cost Functions
Solutionb. The rate of change of the total cost function
C(x) = 8000 + 200x – 0.2x2
with respect to x is given by
C´(x) = 200 – 0.4xSo, when production is 250 refrigerators, the rate of change of the total cost with respect to x is
C´(x) = 200 – 0.4(250)
= 100or $100.
Applied Example: Rate of Change of Cost Functions
Solutionc. 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.
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.
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) = 0.0001x3 – 0.08x2 + 40x + 5000where x stands for the number of players produced.a. Find the marginal cost function.b. Find the marginal cost for x = 200, 300, 400, and
600.c. Interpret your results.
Applied Example: Marginal Cost Functions
Solutiona. If the total cost function is:
C(x) = 0.0001x3 – 0.08x2 + 40x + 5000then, its derivative is the marginal cost function:
C´(x) = 0.0003x2 – 0.16x + 40
Applied Example: Marginal Cost Functions
Solutionb. The marginal cost for x = 200, 300, 400, and 600 is:
C´(200) = 0.0003(200)2 – 0.16(200) + 40 = 20C´(300) = 0.0003(300)2 – 0.16(300) + 40 = 19C´(400) = 0.0003(400)2 – 0.16(400) + 40 = 24C´(600) = 0.0003(600)2 – 0.16(600) + 40 = 52
or $20/unit, $19/unit, $24/unit, and $52/unit,
respectively.
Applied Example: Marginal Cost Functions
Solutionc. 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.
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 + 400 (0 x 20,000)
a. Find the revenue function R.b. Find the marginal revenue function R′.c. Compute R′(2000) and interpret your result.
Applied Example: Marginal Revenue Functions
Solutiona. The revenue function is given by
R(x) = px
= (– 0.02x + 400)x
= – 0.02x2 + 400x
(0 x 20,000)
Applied Example: Marginal Revenue Functions
Solutionb. Given the revenue function
R(x) = – 0.02x2 + 400xWe find its derivative to obtain the marginal revenue function:
R′(x) = – 0.04x + 400
Applied Example: Marginal Revenue Functions
Solutionc. 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.
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 + 200,000a. Find the profit function P.b. Find the marginal profit function P′.c. Compute P′ (2000) and interpret the result.
Applied Example: Marginal Profit FunctionSolutiona. From last example we know that the revenue
function is
R(x) = – 0.02x2 + 400x– Profit is the difference between total revenue and
total cost, so the profit function is
P(x) = R(x) – C(x)
= (– 0.02x2 + 400x) – (100x + 200,000)
= – 0.02x2 + 300x – 200,000
Applied Example: Marginal Profit FunctionSolutionb. Given the profit function
P(x) = – 0.02x2 + 300x – 200,000we find its derivative to obtain the marginal profit function:
P′(x) = – 0.04x + 300
Applied Example: Marginal Profit FunctionSolutionc. When producing x = 2000, the marginal profit
is
P′(2000) = – 0.04(2000) + 300
= 220Thus, the profit to be made from producing the 2001st loudspeaker is $220.
End of Chapter