9 Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: Higher-Order Derivatives The Chain.

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


• 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


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


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



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


• If c is any constant real number, then

( ) ( )d d

cf x c f xdx dx



3( ) 5f x x

3

3

( ) 5

5

df x x

dxd

xdx

2

2

5 3

15

x

x



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


( ) ( ) ( ) ( )d d d

f x g x f x g xdx dx dx



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



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


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


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



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



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



( )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



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


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


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


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


3xy e

3

3

3

( 3 )

( 3)

3

x

x

x

dy de x

dx dx

e

e


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


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


( )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


( ) lnf x x x

( ) (ln ) ln ( )

1ln (1)

1 ln

d df x x x x x

dx dx

x xx

x


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


2( ) ln( 1)f x x

2

2

2

1( )

12

1

dx

dxf xxx

x


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).


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.


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.


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.


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.


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


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.


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.


Solutiona. The revenue function is given by

R(x) = px

= (– 0.02x + 400)x

= – 0.02x2 + 400x

(0 x 20,000)


Solutionb. Given the revenue function

R(x) = – 0.02x2 + 400xWe find its derivative to obtain the marginal revenue function:

R′(x) = – 0.04x + 400


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

9 Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: Higher-Order Derivatives The Chain.

Documents

9 Limits One-sided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules: Higher-Order Derivatives The Chain.