basic rules of differentiation the product and quotient rules the chain rule

33

Basic Rules of DifferentiationBasic Rules of Differentiation The Product and Quotient RulesThe Product and Quotient Rules The Chain RuleThe Chain Rule Marginal Functions in EconomicsMarginal Functions in Economics Higher Order DerivativesHigher Order Derivatives

Differentiation Differentiation

3.13.1Basic Rules of DifferentiationBasic Rules of Differentiation

1.1. Derivative of a ConstantDerivative of a Constant

2.2. The Power RuleThe Power Rule

3.3. Derivative of a Constant Multiple FunctionDerivative of a Constant Multiple Function

4.4. The Sum RuleThe Sum Rule

Four Basic RulesFour Basic Rules

We’ve learned that to find the rule for the We’ve learned that to find the rule for the derivativederivative f ′of a of a function function ff, we first find the , we first find the difference quotientdifference quotient

But this method is But this method is tedious tedious andand time consuming time consuming, even for , even for relatively simple functions.relatively simple functions.

This chapter we will develop rules that will This chapter we will develop rules that will simplify the simplify the processprocess of finding the derivative of a function. of finding the derivative of a function.

0

( ) ( )limh

f x h f x

h

Rule 1:Rule 1: Derivative of a Constant Derivative of a Constant

We will use the notation We will use the notation

To mean To mean “the derivative of“the derivative of f f with respect towith respect to xx at at xx..””

Rule 1:Rule 1: Derivative of a constant Derivative of a constant

The derivative of a constant function is equal to zero.The derivative of a constant function is equal to zero.

( )d

f xdx

0d

cdx

0d

cdx


We can see We can see geometricallygeometrically why the derivative of a constant why the derivative of a constant must be zero.must be zero.

The graph of a The graph of a constant functionconstant function is a is a straight line parallel straight line parallel to theto the xx axisaxis..

Such a line has a Such a line has a slopeslope that is constant with a value of that is constant with a value of zerozero.. Thus, the derivative of a constant must be zero as well.Thus, the derivative of a constant must be zero as well.

f(x) = c

y

x


We can use the We can use the definition of the derivativedefinition of the derivative to to demonstrate this:demonstrate this:

0

0

0

( ) ( )( )

0

0

lim

lim

lim

h

h

h

f x h f xf x

hc c

h

Rule 2:Rule 2: The Power Rule The Power Rule

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

1n ndx nx

dx 1n nd

x nxdx


Lets verify this rule for the special case of Lets verify this rule for the special case of n n = 2= 2.. If If ff((xx) = ) = xx22, then, 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:Practice Examples: If If ff((xx) = ) = xx, then, then

If If ff((xx) = ) = xx88, then, then

If If ff((xx) = ) = xx5/25/2, then, 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

Example 2, page 159


Practice Examples:Practice Examples: Find the derivative ofFind 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

Example 3, page 159


Practice Examples:Practice Examples: Find the derivative of Find the derivative of

3

1( )f x

x

1/3

3

1( )

d df x x

dx dxx

1/3

3

1( )

d df x x

dx dxx

1/3 11

3x 1/3 11

3x

4/34/3

1 1

3 3x

x 4/3

4/3

1 1

3 3x

x

Example 3, page 159

Rule 3:Rule 3: Derivative of a Constant Multiple Function Derivative of a Constant Multiple Function

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

( ) ( )d d

cf x c f xdx dx

( ) ( )d d

cf x c f xdx dx


Practice Examples:Practice Examples: Find the derivative ofFind the derivative of 3( ) 5f x x 3( ) 5f x x

3

3

( ) 5

5

df x x

dxd

xdx

3

3

( ) 5

5

df x x

dxd

xdx

2

2

5 3

15

x

x

2

2

5 3

15

x

x

Example 4, page 160


Practice Examples:Practice Examples: Find the derivative ofFind the derivative of

3( )f x

x

3( )f x

x

1/ 2( ) 3d

f x xdx

1/ 2( ) 3d

f x xdx

3/ 2

3/ 2

13

2

3

2

x

x

3/ 2

3/ 2

13

2

3

2

x

x

Example 4, page 160

Rule 4:Rule 4: The Sum Rule The Sum Rule


( ) ( ) ( ) ( )d d d

f x g x f x g xdx dx dx

( ) ( ) ( ) ( )d d d

f x g x f x g xdx dx dx


Practice Examples:Practice Examples: Find the derivative ofFind the derivative of 5 4 2( ) 4 3 8 3f x x x x x 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

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

4 3

4 3

4 5 3 4 8 2 1 0

20 12 16 1

x x x

x x x

Example 5, page 161



2

3

5( )

5

tg t

t

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

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

4

5

4 4

12 5 3

5

2 15 2 75

5 5

t t

t t

t t

Example 5, page 161

Applied Example:Applied Example: Conservation of a Species Conservation of a Species

A group of marine biologists at the Neptune Institute of A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of Oceanography recommended that a series of conservation conservation measuresmeasures be carried out over the next decade to save a be carried out over the next decade to save a certain species of whale from extinction.certain species of whale from extinction.

After implementing the conservation measure, the After implementing the conservation measure, the population of this speciespopulation of this species is expected to be is expected to be

where where NN((tt)) denotes the denotes the populationpopulation at the end of year at the end of year tt.. Find the Find the rate of growthrate of growth of the whale population when of the whale population when

tt = 2= 2 and and tt = 6= 6. . How large will the whale How large will the whale populationpopulation be be 88 years after years after

implementing the conservation measures?implementing the conservation measures?

3 2( ) 3 2 10 600 (0 10) N t t t t t 3 2( ) 3 2 10 600 (0 10) N t t t t t

Applied Example 7, page 162


SolutionSolution The The rate of growthrate of growth of the of the whale populationwhale population at any time at any time tt is is

given bygiven by

In particular, for In particular, for tt = 2 = 2, we have, we have

And for And for t t = 6= 6, we have, we have

Thus, the whale population’s Thus, the whale population’s rate of growthrate of growth will be will be 3434 whales per year after whales per year after 22 years and years and 338338 per year after per year after 66 years. years.

2( ) 9 4 10N t t t 2( ) 9 4 10N t t t

2(2) 9 2 4 2 10 34N 2(2) 9 2 4 2 10 34N

2(6) 9 6 4 6 10 338N 2(6) 9 6 4 6 10 338N



SolutionSolution The The whale populationwhale population at the end of the at the end of the eighth yeareighth year will be will be

3 28 3 8 2 8 10 8 600

2184 whales

N

3 28 3 8 2 8 10 8 600

2184 whales

N


3.23.2The Product and Quotient RulesThe 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:Rule 5: The Product Rule The Product Rule

The derivative of the product of two differentiable The derivative of the product of two differentiable functions is given byfunctions is given by

( ) ( ) ( ) ( ) ( ) ( )d


( ) ( ) ( ) ( ) ( ) ( )d



Practice Examples:Practice Examples: Find the derivative ofFind 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

Example 1, page 172


Practice Examples:Practice Examples: Find the derivative ofFind 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

Example 2, page 172

Rule 6:Rule 6: The Quotient Rule The Quotient Rule

The derivative of the quotient of two differentiable The derivative of the quotient of two differentiable functions is given byfunctions is given by

2

( ) ( ) ( ) ( ) ( ) 0

( ) ( )

d f x g x f x f x g xg x

dx g x g x

2

( ) ( ) ( ) ( ) ( ) 0

( ) ( )

d f x g x f x f x g xg x

dx g x g x


Practice Examples:Practice Examples: Find the derivative ofFind 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

Example 3, page 173



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

Example 4, page 173

Applied Example:Applied Example: Rate of Change of DVD Sales Rate of Change of DVD Sales

The The salessales ( in millions of dollars) of DVDs of a hit movie ( in millions of dollars) of DVDs of a hit movie tt years from the date of release is given by years from the date of release is given by

Find the Find the raterate at which the at which the sales are changingsales are changing at time at time tt.. How fast are the sales changing at:How fast are the sales changing at:

✦ The time the DVDs are released The time the DVDs are released ((t t = 0)= 0)? ?

✦ And two years from the date of release And two years from the date of release ((t t = 2)= 2)??

2

5( )

1

tS t

t



SolutionSolution The The rate of changerate of change at which the at which the sales are changingsales are changing at at

time time tt is given by is given by

2

22

1 5 5 2

1

t t t

t

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

22 2

2 22 2

5 15 5 10

1 1

tt t

t t



SolutionSolution The The rate of changerate of change at which the sales are changing at which the sales are changing when when

the DVDs are releasedthe DVDs are released ( (tt = 0 = 0) is) is

That is, sales are That is, sales are increasingincreasing by by $5 million$5 million per year. per year.

2

2 22

5 1 0 5 1(0) 5

10 1S



SolutionSolution The The rate of changerate of change two years after the DVDs are two years after the DVDs are

releasedreleased ( (tt = 2 = 2) is) is

That is, sales are That is, sales are decreasingdecreasing by by $600,000$600,000 per year. per year.

2

2 22

5 1 2 5 1 4 15 3(2) 0.6

25 54 12 1S


3.33.3The Chain RuleThe Chain Rule

dy dy du

dx du dx

( ) ( ) ( ) ( )d

h x g f x g f x f xdx

Deriving Composite FunctionsDeriving Composite Functions

Consider the functionConsider the function

To compute To compute hh′′((xx)), we can first , we can first expandexpand hh((xx))

and then and then derivederive the resulting polynomial the resulting polynomial

But how should we derive a function like But how should we derive a function like HH((xx))??

22( ) 1h x x 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

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 3 2( ) 4 6 6 2h x x x x

1002( ) 1H x x x 1002( ) 1H x x x


Note that Note that is a is a composite functioncomposite function::

HH((xx)) is composed of two simpler functions is composed of two simpler functions

So that So that

We can use thisWe can use this to find the derivative of to find the derivative of HH((xx))..

1002( ) 1H x x x 1002( ) 1H x x x

2 100( ) 1 ( )f x x x g x x and 2 100( ) 1 ( )f x x x g x x and

100100 2( ) ( ) ( ) 1H x g f x f x x x 100100 2( ) ( ) ( ) 1H x g f x f x x x


To To find the derivativefind the derivative of the of the composite functioncomposite function HH((xx))::

We let We let u = fu = f((xx)) = = xx22 + + xx + 1 + 1 and and y = gy = g((uu) = ) = uu100100..

Then we Then we find the derivativesfind the derivatives of each of these functions of each of these functions

The The ratiosratios of these derivatives suggest that of these derivatives suggest that

SubstitutingSubstituting xx22 + + xx + 1 + 1 for for u u we getwe 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:Rule 7: The Chain Rule The Chain Rule

If If h(x) = g[f(x)], then, then

Equivalently, if we write Equivalently, if we write y = h(x) = g(u), , where where u = f(x), then, then

dy dy du

dx du dx

dy dy du

dx du dx

( ) ( ) ( ) ( )d


( ) ( ) ( ) ( )d


The Chain Rule for Power FunctionsThe Chain Rule for Power Functions

Many Many composite functionscomposite functions have the have the special formspecial form

h(x) = g[f(x)]

where where gg is defined by the rule is defined by the rule

gg((xx) = ) = xxnn ((nn,, a real number) a real number) so thatso that

h(x) = [f(x)]nn

In other words, In other words, the functionthe function hh is given by the is given by the power of a power of a functionfunction ff..

Examples:Examples:

1002 233

1( ) 1 ( ) ( ) 2 3

5h x x x H x G x x

x

1002 2

33

1( ) 1 ( ) ( ) 2 3

5h x x x H x G x x

x

The General Power RuleThe General Power Rule

If the function If the function f is differentiable and is differentiable and

h(x) = [f(x)]n ((nn,, a real number)a real number),,

thenthen

1( ) ( ) ( ) ( )

n ndh x f x n f x f x

dx 1

( ) ( ) ( ) ( )n nd

h x f x n f x f xdx



SolutionSolution Rewrite as a Rewrite as a power functionpower function:: Apply the Apply the general power rulegeneral 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

Example 2, page 184



SolutionSolution Apply the Apply the product ruleproduct rule and the and the general power rulegeneral 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

Example 3, page 185



SolutionSolution Rewrite as a Rewrite as a power functionpower function:: Apply the Apply the general power rulegeneral 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

Example 5, page 186



SolutionSolution Apply the Apply the general power rulegeneral power rule and the and the quotient rulequotient 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

Example 6, page 186

Applied Problem:Applied Problem: Arteriosclerosis Arteriosclerosis

Arteriosclerosis begins during childhood when Arteriosclerosis begins during childhood when plaqueplaque forms in the arterial walls, forms in the arterial walls, blockingblocking the flow of blood the flow of blood through the through the arteriesarteries and leading to heart attacks, stroke and leading to heart attacks, stroke and gangrene.and gangrene.



Suppose the idealized Suppose the idealized cross section of the aortacross section of the aorta is circular is circular with radius with radius aa cmcm and by year and by year tt the the thickness of the plaquethickness of the plaque is is

h = g(t) cmcm

then the then the area of the openingarea of the opening is given by is given by

A = (a – h)2 cm2

Further suppose the Further suppose the radiusradius of an individual’s artery is of an individual’s artery is 11 cmcm ((aa = 1 = 1) and the ) and the thickness of the plaquethickness of the plaque in year in year t t is given byis given by

h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm



Then we can use these functions for Then we can use these functions for hh and and AA

h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = (1 – h)2

to find a function that gives us the to find a function that gives us the raterate at which at which AA is is changingchanging with respect to with respect to timetime by applying the by applying the chain rulechain 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



For example, at age For example, at age 5050 ( (tt = 50 = 50),),

So thatSo that

That is, the That is, the area of the arterial openingarea of the arterial opening is is decreasing at the decreasing at the raterate of of 0.030.03 cmcm22 per year for a typical per year for a typical 5050 year old. 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


3.43.4Marginal Functions in EconomicsMarginal Functions in Economics

Percentage change in quantity

demanded

( ) ( )100

( )

100

f p h f pf p

hp

( )E p Percentage

change in price

Marginal AnalysisMarginal Analysis

Marginal analysisMarginal analysis is the study of the is the study of the rate of change of rate of change of economic quantitieseconomic quantities..

These may have to do with the behavior of costs, revenues, These may have to do with the behavior of costs, revenues, profit, output, demand, etc.profit, output, demand, etc.

In this section we will discuss the marginal analysis of In this section we will discuss the marginal analysis of various functions related to:various functions related to:✦ CostCost✦ Average CostAverage Cost✦ RevenueRevenue✦ ProfitProfit✦ Elasticity of DemandElasticity of Demand

Applied Example:Applied Example: Rate of Change of Cost Functions Rate of Change of Cost Functions

Suppose the Suppose the total costtotal cost in dollars incurred each week by in dollars incurred each week by Polaraire for manufacturing Polaraire for manufacturing x x refrigerators is given by the refrigerators is given by the total cost functiontotal cost function

CC((xx) = 8000 + 200) = 8000 + 200xx – 0.2 – 0.2xx22 (0 (0 xx 400) 400)

a.a. What is the actual cost incurred for manufacturing the What is the actual cost incurred for manufacturing the 251251stst refrigerator?refrigerator?

b.b. Find the Find the rate of changerate of change of the total cost function with of the total cost function with respect to respect to xx when when xx = 250 = 250..

c.c. Compare the results obtained in parts Compare the results obtained in parts (a)(a) and and (b)(b)..



SolutionSolution

a.a. The cost incurred in producing the The cost incurred in producing the 251251stst refrigerator is refrigerator is

CC(251) – (251) – CC(250)(250) = [8000 + 200(251) – 0.2(251)= [8000 + 200(251) – 0.2(251)22]]

– – [8000 + 200(250) – 0.2(250)[8000 + 200(250) – 0.2(250)22]]

= 45,599.8 – 45,500= 45,599.8 – 45,500

= 99.80= 99.80

or or $99.80$99.80..



SolutionSolution

b.b. The The rate of changerate of change of the total cost function of the total cost function

CC((xx) = 8000 + 200) = 8000 + 200xx – 0.2 – 0.2xx22

with respect to with respect to xx is given by is given by

CC´́((xx) ) = 200 – 0.4= 200 – 0.4xx

So, when production is So, when production is 250250 refrigerators, the rate of refrigerators, the rate of change of the total cost with respect to change of the total cost with respect to xx is is

CC´́((xx) ) = 200 – 0.4(250)= 200 – 0.4(250)

= 100= 100

or or $100$100..



SolutionSolution

c.c. Comparing the results from Comparing the results from (a)(a) and and (b)(b) we can see they are we can see they are very similarvery similar: : $99.80$99.80 versus versus $100$100..✦ This is because This is because (a)(a) measures the measures the average rate of changeaverage rate of change

over the interval over the interval [250, 251][250, 251], while , while (b)(b) measures the measures the instantaneous rate of changeinstantaneous rate of change at exactly at exactly xx = 250 = 250..

✦ TheThe smallersmaller thethe interval usedinterval used, the , the closercloser the the average rate average rate of changeof change becomesbecomes to the to the instantaneous rate of changeinstantaneous rate of change..



SolutionSolution The actual cost incurred in producing an The actual cost incurred in producing an additional unitadditional unit

of a good is called the of a good is called the marginal costmarginal cost.. As we just saw, the As we just saw, the marginal costmarginal cost is is approximatedapproximated by the by the

rate of changerate of change of the of the total costtotal cost function. function. For this reason, economists define the For this reason, economists define the marginal cost marginal cost

function function as the as the derivativederivative of the of the total cost functiontotal cost function..


Applied Example:Applied Example: Marginal Cost Functions Marginal Cost Functions

A subsidiary of Elektra Electronics manufactures a A subsidiary of Elektra Electronics manufactures a portable music player.portable music player.

Management determined that the Management determined that the daily total costdaily total cost of of producing these players (in dollars) is producing these players (in dollars) is

CC((xx) = 0.0001) = 0.0001xx33 – 0.08 – 0.08xx22 + 40 + 40xx + 5000 + 5000

where where xx stands for the stands for the number of playersnumber of players produced. produced.

a.a. Find the marginal cost function.Find the marginal cost function.

b.b. Find the marginal cost for Find the marginal cost for xx = 200, 300, 400, and 600 = 200, 300, 400, and 600..

c.c. Interpret your results.Interpret your results.



SolutionSolution

a.a. If the If the total cost functiontotal cost function is: is:

CC((xx) = 0.0001) = 0.0001xx33 – 0.08 – 0.08xx22 + 40 + 40xx + 5000 + 5000

then, then, itsits derivativederivative is the is the marginal cost functionmarginal cost function::

CC´́((xx) = 0.0003) = 0.0003xx22 – 0.16 – 0.16xx + 40 + 40



SolutionSolution

b.b. The The marginal costmarginal cost for for xx = 200, 300, 400, and 600 = 200, 300, 400, and 600 is: is:

CC´́(200) = 0.0003(200)(200) = 0.0003(200)22 – 0.16(200) + 40 = 20 – 0.16(200) + 40 = 20

CC´́(300) = 0.0003(300)(300) = 0.0003(300)22 – 0.16(300) + 40 = 19 – 0.16(300) + 40 = 19

CC´́(400) = 0.0003(400)(400) = 0.0003(400)22 – 0.16(400) + 40 = 24 – 0.16(400) + 40 = 24

CC´́(600) = 0.0003(600)(600) = 0.0003(600)22 – 0.16(600) + 40 = 52 – 0.16(600) + 40 = 52

oror $20/unit $20/unit,, $19/unit $19/unit,, $24/unit $24/unit, and, and $52/unit $52/unit, respectively., respectively.



SolutionSolution

c.c. From part From part (b)(b) we learn that we learn that at firstat first the the marginal costmarginal cost is is decreasingdecreasing, but , but as output increasesas output increases, the , the marginal costmarginal cost increasesincreases as well. as well.

This is a This is a common phenomenoncommon phenomenon that occurs because of that occurs because of several factorsseveral factors, such as excessive costs due to , such as excessive costs due to overtimeovertime and and high high maintenance costsmaintenance costs for keeping the plant running at for keeping the plant running at such a fast rate.such a fast rate.


Applied Example:Applied Example: Marginal Revenue Functions Marginal Revenue Functions

Suppose the relationship between the unit Suppose the relationship between the unit priceprice pp in in dollars and the dollars and the quantity demandedquantity demanded xx of the of the Acrosonic Acrosonic model F loudspeaker systemmodel F loudspeaker system is given by the equation is given by the equation

pp = – 0.02 = – 0.02xx + 400 (0 + 400 (0 xx 20,000) 20,000)

a.a. Find the Find the revenue functionrevenue function RR..b.b. Find the Find the marginal revenue functionmarginal revenue function RR′′..c.c. Compute Compute RR′′(2000)(2000) and interpret your result. and interpret your result.



SolutionSolution

a.a. The The revenue functionrevenue function is given by is given by

RR((xx)) == px px

= (– 0.02= (– 0.02xx + 400) + 400)xx

= – 0.02= – 0.02xx22 + 400 + 400xx (0 (0 xx 20,000) 20,000)



SolutionSolution

b.b. Given the Given the revenue functionrevenue function

RR((xx)) = – 0.02= – 0.02xx22 + 400 + 400xx

We find its derivative to obtain the We find its derivative to obtain the marginal revenue marginal revenue functionfunction::

RR′′((xx)) = – 0.04= – 0.04xx + 400 + 400



SolutionSolution

c.c. When When quantity demandedquantity demanded is is 2000 2000, the , the marginalmarginal revenue revenue will be:will be:

RR′′(2000)(2000) = – 0.04(2000) + 400= – 0.04(2000) + 400

= 320= 320

Thus, the Thus, the actual revenueactual revenue realized from the sale of the realized from the sale of the 2001st 2001st loudspeaker systemloudspeaker system is approximately is approximately $320 $320..


Applied Example:Applied Example: Marginal Profit Function Marginal Profit Function

Continuing with the last example, suppose the Continuing with the last example, suppose the totaltotal costcost (in dollars) of producing (in dollars) of producing xx units of the units of the Acrosonic model F Acrosonic model F loudspeaker systemloudspeaker system is is

CC((xx)) = 100= 100xx + 200,000 + 200,000

a.a. Find the Find the profit functionprofit function PP..b.b. Find the Find the marginal profit functionmarginal profit function PP′′..c.c. Compute Compute PP′′ (2000)(2000) and interpret the result. and interpret the result.



SolutionSolution

a.a. From last example we know that the From last example we know that the revenue functionrevenue function is is

RR((xx)) = – 0.02= – 0.02xx22 + 400 + 400xx✦ ProfitProfit is the is the differencedifference between between total revenuetotal revenue and and total total

costcost, so the profit function is, so the profit function is

PP((xx)) == R R((xx) –) – C C((xx))

= (– 0.02= (– 0.02xx22 + 400 + 400xx) – (100) – (100xx + +

200,000)200,000)

= – 0.02= – 0.02xx22 + 300 + 300xx – 200,000 – 200,000



SolutionSolution

b.b. Given the Given the profit functionprofit function

PP((xx)) = – 0.02= – 0.02xx22 + 300 + 300xx – 200,000 – 200,000

we find its we find its derivativederivative to obtain the to obtain the marginal profit marginal profit functionfunction::

PP′′((xx)) == – 0.04– 0.04xx + 300 + 300



SolutionSolution

c.c. When producing When producing xx = 2000 = 2000, the , the marginal profitmarginal profit is is

PP′′(2000)(2000) == – 0.04(2000) + 300– 0.04(2000) + 300

= 220= 220

Thus, the Thus, the profitprofit to be made from to be made from producingproducing thethe 20012001stst loudspeakerloudspeaker is is $220$220..


Elasticity of DemandElasticity of Demand

Economists are frequently concerned with Economists are frequently concerned with how stronglyhow strongly do do changes in changes in pricesprices causecause quantity demandedquantity demanded to changeto change..

The measure of the strength of this reaction is called the The measure of the strength of this reaction is called the elasticity of demandelasticity of demand, which is given by, which is given by

percentage change in quantity demanded( )

percentage change in priceE p

percentage change in quantity demanded( )

percentage change in priceE p

Note:Note: Since the ratio is Since the ratio is negativenegative, economists use , economists use the negative the negative of the ratioof the ratio, to make the elasticity be a , to make the elasticity be a positivepositive number. number.


Suppose the Suppose the priceprice of a good of a good increasesincreases by by hh dollars from dollars from p p to to ((p + hp + h)) dollars. dollars.

The The percentage changepercentage change of theof the price price is is

The The percentage changepercentage change inin quantity demanded quantity demanded is is

Change in price

Price= 100 100

h

p Change in price

Price= 100 100

h

pPercentage

change in pricePercentage

change in price

100 100Percentage

change in quantity demanded


demanded Quantity demanded at price p

Quantity demanded at price p

Change in quantity demanded

Change in quantity demanded

( ) ( )100

( )

f p h f p

f p

( ) ( )100

( )

f p h f p

f p


One good way to measure the effect that a percentage One good way to measure the effect that a percentage change in price has on the percentage change in the change in price has on the percentage change in the quantity demanded is to look at the ratio of the latter to the quantity demanded is to look at the ratio of the latter to the former. We findformer. We find


demanded

( ) ( )100

( )

100

f p h f pf p

hp

( )E p Percentage

change in price

( ) ( )( )

f p h f pf p

hp

( ) ( )

( )

f p h f ph

f pp

Elasticity of DemandElasticity of Demand We haveWe have

If If ff is is differentiabledifferentiable at at pp, then, when , then, when hh is small, is small,

Therefore, if Therefore, if hh is small, the ratio is approximately equal to is small, the ratio is approximately equal to

Economists call the Economists call the negativenegative of this quantity the of this quantity the elasticity of demandelasticity of demand..

( ) ( )( )

f p h f pf p

h

( ) ( )

( )( )

f p h f phE p

f pp

( ) ( )( )

( ) ( )

f p pf pE p

f p f pp


Elasticity of DemandElasticity of Demand IfIf f is a is a differentiable demand functiondifferentiable demand function defined by defined by

x = f(p) , then the , then the elasticity of demandelasticity of demand at price at price p is is given bygiven by

( )( )

( )

pf pE p

f p

Note:Note: Since the ratio is Since the ratio is negativenegative, economists use , economists use the negative of the negative of the ratiothe ratio, to make the elasticity be a , to make the elasticity be a positivepositive number. number.

Applied Example:Applied Example: Elasticity of Demand Elasticity of Demand

Consider the Consider the demand equationdemand equation for the for the Acrosonic model F Acrosonic model F loudspeaker systemloudspeaker system::

pp = – 0.02= – 0.02xx + 400 + 400 (0 (0 xx

20,000)20,000)

a.a. Find the Find the elasticity of demandelasticity of demand EE((pp))..

b.b. Compute Compute EE(100)(100) and and interpret your resultinterpret your result..

c.c. Compute Compute EE(300)(300) and and interpret your resultinterpret your result..



SolutionSolution

a.a. Solving the Solving the demand equationdemand equation for for x in terms of in terms of p, we get, we get

x = f(p) = – 50p + 20,000

From which we see thatFrom which we see that

f ′′(p) = – 50

Therefore,Therefore,( ) ( 50)

( )( ) 50 20,000

400

pf p pE p

f p p

p

p



SolutionSolution

b.b. When When pp = 100 = 100 the the elasticity of demandelasticity of demand is is

✦ This means that for every This means that for every 1%1% increaseincrease in in priceprice we can we can expect to see a expect to see a 1/3%1/3% decreasedecrease in in quantity demandedquantity demanded..

✦ Because the Because the responseresponse (change in (change in quantity demandedquantity demanded) is ) is lessless than the than the actionaction (change in (change in priceprice), we say demand is ), we say demand is inelasticinelastic..

✦ Demand is said to be Demand is said to be inelasticinelastic whenever whenever EE((pp) < 1) < 1..

100 1(100)

400 100 3E



SolutionSolution

c.c. When When pp = 300 = 300 the the elasticity of demandelasticity of demand is is

✦ This means that for every This means that for every 1%1% increaseincrease in in priceprice we can we can expect to see a expect to see a 3%3% decreasedecrease in in quantity demandedquantity demanded..

✦ Because the Because the responseresponse (change in (change in quantity demandedquantity demanded) is ) is greatergreater than the than the actionaction (change in (change in priceprice), we say demand ), we say demand is is elasticelastic..

✦ Demand is said to be Demand is said to be elasticelastic whenever whenever EE((pp) > 1) > 1..

✦ Finally, demand is said to be Finally, demand is said to be unitaryunitary whenever whenever EE((pp) = 1) = 1..

300(300) 3

400 300E


3.53.5Higher Order DerivativesHigher Order Derivatives

2

28 8

dv d ds d s da t

dt dt dt dt dt

7/3 7/3

2 3

2 4 8 8( )

9 3 27 27f x x x

x x

Higher-Order DerivativesHigher-Order Derivatives

The derivative The derivative f f ′′ of a function of a function ff is also a functionis also a function.. As such, As such, f f ′′ may also be differentiatedmay also be differentiated.. Thus, the function Thus, the function f f ′′ has a derivative has a derivative f f ″″ at a point at a point xx in the in the

domain ofdomain of f f if the limit of the quotient if the limit of the quotient

exists as exists as h h approaches zero.approaches zero. The function The function f f ″″ obtained in this manner is called the obtained in this manner is called the

second derivativesecond derivative of the function of the function ff, just as the derivative , just as the derivative f f ′′ of of ff is often called the is often called the first derivativefirst derivative of of ff..

By the same token, you may consider the By the same token, you may consider the thirdthird, , fourthfourth, , fifthfifth, etc. , etc. derivatives of a function derivatives of a function ff..

( ) ( )f x h f x

h


Practice Examples:Practice Examples: Find the Find the third derivativethird derivative of the function of the function ff((xx)) = = xx2/32/3 and and

determine its determine its domaindomain..

SolutionSolution We haveWe have andand

So the required derivative isSo the required derivative is

The The domaindomain of the of the third derivativethird derivative is the set of is the set of all real all real numbers exceptnumbers except xx = 0 = 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

Example 1, page 208


Practice Examples:Practice Examples: Find the Find the second derivativesecond derivative of the function of the function ff((xx)) = (2= (2xx22 +3) +3)3/23/2

SolutionSolution Using the Using the general power rulegeneral power rule we get the we get the first derivativefirst derivative::

1/2 1/22 23( ) 2 3 4 6 2 3

2f x x x x x

Example 2, page 209


Practice Examples:Practice Examples: Find the Find the second derivativesecond derivative of the function of the function ff((xx)) = (2= (2xx22 +3) +3)3/23/2

SolutionSolution Using the Using the product ruleproduct rule we get the we get the second derivativesecond 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

Example 2, page 209

Applied Example:Applied Example: Acceleration of a Maglev Acceleration of a Maglev

The The distancedistance s s (in (in feetfeet) covered by a maglev moving along ) covered by a maglev moving along a straight track a straight track tt secondsseconds after starting from rest is given after starting from rest is given by the function by the function

ss = 4 = 4tt22 (0 (0 t t 10) 10) What is the maglev’s What is the maglev’s accelerationacceleration after after 3030 seconds? seconds?

SolutionSolution The The velocityvelocity of the maglev of the maglev tt seconds from rest is given by seconds from rest is given by

The The accelerationacceleration of the maglev of the maglev tt seconds from rest is given seconds from rest is given by the by the rate of changerate of change of the of the velocityvelocity of of tt, given by, given by

or or 8 8 feet per second per second feet per second per second ((ft/secft/sec22))..

24 8ds d

v t tdt dt

2

28 8

d d ds d s da v t

dt dt dt dt dt


3.63.6Implicit Differentiation and Related RatesImplicit Differentiation and Related Rates

yy

4000 ft4000 ft

SpectatorSpectator

Launch PadLaunch Pad

RocketRocket

xx

Differentiating ImplicitlyDifferentiating Implicitly

Up to now we have dealt with functions in the form Up to now we have dealt with functions in the form

yy = = ff((xx)) That is, the That is, the dependent variabledependent variable yy has been expressed has been expressed

explicitlyexplicitly in terms of the in terms of the independent variableindependent variable xx. . However, However, not all functions not all functions areare expressedexpressed explicitly explicitly.. For example, considerFor example, consider

xx22yy + + yy – – x x22 + 1 = 0 + 1 = 0 This equation expresses This equation expresses yy implicitlyimplicitly as a function of as a function of xx.. Solving for Solving for yy in terms of in terms of xx we get we get

which expresses which expresses yy explicitlyexplicitly..

2 2

2

2

( 1) 1

1( )

1

x y x

xy f x

x


Now, consider the equationNow, consider the equation

yy44 – – yy33 – – y + y + 22xx33 – – xx = 8 = 8 With certain restrictions placed on With certain restrictions placed on yy and and xx, this equation , this equation

definesdefines yy as a function ofas a function of xx.. But in this case it is But in this case it is difficultdifficult to to solve forsolve for yy in order to in order to

expressexpress the function the function explicitlyexplicitly.. How do we compute How do we compute dy/dx dy/dx in this case?in this case? The The chain rulechain rule gives us a way to do this. gives us a way to do this.


Consider the equation Consider the equation yy22 = = x x.. To find To find dy/dxdy/dx, we differentiate both sides of the equation:, we differentiate both sides of the equation:

Since Since yy is a function of is a function of xx, we can rewrite , we can rewrite yy = = ff((xx)) and find: and find:

2d dy x

dx dx 2d d

y xdx dx

Using chain ruleUsing chain rule

22 ( )

2 ( ) ( )

2

d dy f x

dx dxf x f x

dyy

dx

22 ( )

2 ( ) ( )

2

d dy f x

dx dxf x f x

dyy

dx

Example 1, page 216


Therefore the above equation is equivalent to:Therefore the above equation is equivalent to:

Solving for Solving for dy/dxdy/dx yields: yields:

2 1dy

ydx

2 1dy

ydx

1

2

dy

dx y

1

2

dy

dx y

Example 1, page 216

Steps for Differentiating ImplicitlySteps for Differentiating Implicitly

To find To find dy/dx dy/dx by implicit differentiation:by implicit differentiation:

1.1. Differentiate both sidesDifferentiate both sides of the equation with of the equation with respect torespect to xx..

(Make sure that the derivative of any term (Make sure that the derivative of any term involvinginvolving y y includes the factor includes the factor dy/dxdy/dx))

2.2. SolveSolve the resulting equation the resulting equation forfor dy/dxdy/dx in in terms of terms of xx and and yy..

Differentiating ImplicitlyDifferentiating ImplicitlyExamplesExamples Find Find dy/dxdy/dx for the equation for the equationSolutionSolution Differentiating both sidesDifferentiating both sides and and solvingsolving for for dy/dx dy/dx we getwe get

3 32 8y y x x 3 32 8y y x x

3 32 8d d

y y x xdx dx

3 32 8d d

y y x xdx dx

2 2

2

2

3 1 1 6

1 6

3 1

dyy x

dx

dy x

dx y

2 2

2

2

3 1 1 6

1 6

3 1

dyy x

dx

dy x

dx y

3 32 8d d d d d

y y x xdx dx dx dx dx

3 32 8d d d d d

y y x xdx dx dx dx dx

2 23 6 1 0dy dy

y xdx dx

2 23 6 1 0dy dy

y xdx dx

Example 2, page 216

Differentiating ImplicitlyDifferentiating ImplicitlyExamplesExamples Find Find dy/dxdy/dx for the equation for the equation Then, find the value of Then, find the value of dy/dx dy/dx when when y y = 2= 2 and and xx = 1 = 1..SolutionSolution

2 3 26 12x y x y 2 3 26 12x y x y

2 3 26 12d d d d

x y x ydx dx dx dx

2 3 26 12d d d d

x y x ydx dx dx dx

2 2 33 2 12dy dy

x y xy xdx dx

2 2 33 2 12dy dy

x y xy xdx dx

2 2 3

3

2 2

3 1 2 12

2 12

1 3

dyx y xy x

dx

dy xy x

dx x y

2 2 3

3

2 2

3 1 2 12

2 12

1 3

dyx y xy x

dx

dy xy x

dx x y

2 3 3 2 12d d dy

x y y x xdx dx dx

2 3 3 2 12d d dy

x y y x xdx dx dx

Example 4, page 217

Differentiating ImplicitlyDifferentiating ImplicitlyExamplesExamples Find Find dy/dxdy/dx for the equation for the equation Then, find the value of Then, find the value of dy/dx dy/dx when when y y = 2= 2 and and xx = 1 = 1..SolutionSolution SubstitutingSubstituting y y = 2= 2 and and xx = 1 = 1 we find: we find:

2 3 26 12x y x y 2 3 26 12x y x y

3

2 2

2 12

1 3

dy xy x

dx x y

3

2 2

2 12

1 3

dy xy x

dx x y

3

2 2

2(1)(2) 12(1)

1 3(1) (2)

3

2 2

2(1)(2) 12(1)

1 3(1) (2)

16 12

1 1228

11

16 12

1 1228

11

Example 4, page 217

Differentiating ImplicitlyDifferentiating ImplicitlyExamplesExamples Find Find dy/dxdy/dx for the equation for the equationSolutionSolution

2 2 2 5x y x 2 2 2 5x y x

1/22 2 2 2 4dy

x y x y xdx

1/22 2 2 2 4dy

x y x y xdx

1/22 22dy

y x x y xdx

1/22 22dy

y x x y xdx

1/22 2 2 5d d d

x y xdx dx dx

1/22 2 2 5d d d

x y xdx dx dx

1/22 212 2 2 0

2

dyx y x y x

dx

1/22 212 2 2 0

2

dyx y x y x

dx

1/22 22dy

x y x x ydx

1/22 22dy

x y x x ydx

1/22 22x x y xdy

dx y

1/22 22x x y xdy

dx y

Example 5, page 219

Related RatesRelated Rates

Implicit differentiation is a useful technique for solving a Implicit differentiation is a useful technique for solving a class of problems known as class of problems known as related-rate problemsrelated-rate problems..

Here are some Here are some guidelinesguidelines to solve related-rate problems: to solve related-rate problems:

1.1. Assign a Assign a variablevariable to each quantity. to each quantity.2.2. Write the given Write the given valuesvalues of the variables and their of the variables and their rate rate

of changeof change with respect to with respect to tt..3.3. Find an Find an equationequation giving the giving the relationshiprelationship between the between the

variables.variables.4.4. DifferentiateDifferentiate both sides of the equation both sides of the equation implicitlyimplicitly with with

respect to respect to tt..5.5. ReplaceReplace the the variablesvariables and their and their derivativesderivatives by the by the

numerical data found in numerical data found in step 2step 2 and solve the equation and solve the equation for the required rate of change.for the required rate of change.

Applied Example:Applied Example: Rate of Change of Housing Starts Rate of Change of Housing Starts

A study prepared for the National Association of Realtors A study prepared for the National Association of Realtors estimates that the number of estimates that the number of housing startshousing starts in the in the southwest, southwest, NN((tt)) (in millions), over the next (in millions), over the next 55 years is years is relatedrelated to the to the mortgage ratemortgage rate rr((tt)) (percent per year) by the (percent per year) by the equationequation

99nn22 + + rr = 36 = 36

What is the rate of What is the rate of changechange of the number of of the number of housing startshousing starts with respect to time when the with respect to time when the mortgage ratemortgage rate is is 11%11% per per year and is year and is increasingincreasing at the rate of at the rate of 1.5%1.5% per year? per year?


1.1. SubstituteSubstitute rr = 11 = 11 into the given equation: into the given equation:

Applied Example:Applied Example: Rate of Change of Housing Starts Rate of Change of Housing StartsSolutionSolution We are given that We are given that rr = 11% = 11% and and dr/dt dr/dt = 1.5 = 1.5 at a certain at a certain

instant in time, and we are required to find instant in time, and we are required to find dN/dtdN/dt..

2

2

9 11 36

25

95

3

N

N

N

(rejecting the (rejecting the negative root)negative root)


2.2. DifferentiateDifferentiate the given equation the given equation implicitlyimplicitly on both on both sides sides with respect towith respect to tt::

Applied Example:Applied Example: Rate of Change of Housing Starts Rate of Change of Housing StartsSolutionSolution We are given that We are given that rr = 11% = 11% and and dr/dt dr/dt = 1.5 = 1.5 at a certain at a certain


29 36

18 0

d d dN r

dt dt dtdN dr

Ndt dt


SolutionSolution We are given that We are given that rr = 11% = 11% and and dr/dt dr/dt = 1.5 = 1.5 at a certain at a certain


Thus, at the time under consideration, the number of Thus, at the time under consideration, the number of housing starts is housing starts is decreasingdecreasing at rate of at rate of 50,000 50,000 units per year.units per year.

3.3. SubstituteSubstitute N N = 5/3= 5/3 andand dr/dtdr/dt = 1.5= 1.5 into this equation and into this equation and solvesolve for for dN/dtdN/dt::

Applied Example:Applied Example: Rate of Change of Housing Starts Rate of Change of Housing Starts

518 1.5 0

3

30 1.5

1.5

30

0.05

dN

dt

dN

dtdN

dtdN

dt


Applied Example:Applied Example: Watching a Rocket Launch Watching a Rocket Launch

At a distance of At a distance of 40004000 feet from the launch site, a spectator feet from the launch site, a spectator is observing a rocket being launched.is observing a rocket being launched.

If the rocket lifts off If the rocket lifts off verticallyvertically and is rising at a and is rising at a speedspeed of of 600600 feet per second when it is at an feet per second when it is at an altitudealtitude of of 30003000 feet, feet, how how fastfast is the is the distancedistance between the rocket and the between the rocket and the spectator changing at that instant?spectator changing at that instant?

yy

4000 ft4000 ft

SpectatorSpectator


RocketRocket

xx



SolutionSolution

1.1. LetLet

yy = = altitude of the rocketaltitude of the rocketxx = = distance between the rocket and the spectatordistance between the rocket and the spectator

at any time at any time tt..

2.2. We are told that at a certain instant in timeWe are told that at a certain instant in time

and are asked to find and are asked to find dx/dtdx/dt at that instant. at that instant.

3000 600and dy

ydt



SolutionSolution

3.3. Apply the Pythagorean theorem to the right triangle we Apply the Pythagorean theorem to the right triangle we find thatfind that

Therefore, when Therefore, when yy = 3000 = 3000,,

2 2 24000x y 2 23000 4000 5000x

yy

4000 ft4000 ft

SpectatorSpectator


RocketRocket

xx



SolutionSolution

4.4. Differentiate Differentiate with respect to with respect to tt, , obtainingobtaining

5.5. Substitute Substitute x x = 5000= 5000, , yy = 3000 = 3000, and , and dy/dtdy/dt = 600 = 600, to find, to find

Therefore, the Therefore, the distancedistance between the rocket and the between the rocket and the spectator is spectator is changingchanging at a rateat a rate of of 360 360 feet per second.feet per second.

2 2 24000x y

2 2dx dy

x ydt dt

2 5000 2 3000 600

360

dx

dtdx

dt


3.73.7DifferentialsDifferentials

x

y

x x + x

f(x + x)

f(x)

y

x

dyP

T

IncrementsIncrements

Let Let xx denote a variable quantity and suppose denote a variable quantity and suppose x x changeschanges from from xx11 to to xx22..

This change in This change in xx is called the is called the incrementincrement in in xx and is denoted and is denoted by the symbol by the symbol xx (read “ (read “deltadelta xx”).”).

Thus, Thus, x = x = xx22 – – xx11

Examples:Examples: Find the Find the incrementincrement in in xx as as x x changeschanges from from 33 to to 3.23.2..

SolutionSolution Here, Here, xx11 = 3 = 3 and and xx22 = 3.2 = 3.2, so, so

x = x = xx22 – – xx11 = 3.2 – 3 = 0.2 = 3.2 – 3 = 0.2

Example 1, page 227


Let Let xx denote a variable quantity and suppose denote a variable quantity and suppose x x changeschanges from from xx11 to to xx22..

This change in This change in xx is called the is called the incrementincrement in in xx and is denoted and is denoted by the symbol by the symbol xx (read “ (read “deltadelta xx”).”).

Thus, Thus, x = x = xx22 – – xx11

Examples:Examples: Find the Find the incrementincrement in in xx as as x x changeschanges from from 33 to to 2.72.7..

SolutionSolution Here, Here, xx11 = 3 = 3 and and xx22 = 2.7 = 2.7, so, so

x = x = xx22 – – xx11 = 2.7 – 3 = – = 2.7 – 3 = – 0.3 0.3

Example 1, page 227


Now, suppose Now, suppose two quantitiestwo quantities,, x x and and yy, , are relatedare related by an by an equation equation y = fy = f((xx)), where , where ff is a function. is a function.

If If xx changeschanges from from xx to to x + x + xx, then the corresponding , then the corresponding change inchange in yy is called the is called the increment inincrement in yy..

It is denoted It is denoted y y and is defined byand is defined by

y = fy = f((x + x + xx) – ) – ff((xx))

xx

yy

xx x + x + xx

ff((x + x + xx))

ff((xx))

yy

xxExample 1, page 227

ExampleExample Let Let yy = = xx33. . Find Find xx and and yy when when xx changes changes

a.a. from from 22 to to 2.012.01, and , and b.b. from from 22 to to 1.981.98..

SolutionSolution

a.a. Here,Here,xx = 2.01 – 2 = 0.01 = 2.01 – 2 = 0.01

Next,Next,

b.b. Here,Here,xx = 1.98 – 2 = – 0.02 = 1.98 – 2 = – 0.02

Next,Next,

3 3

( ) ( ) (2.01) (2)

(2.01) 2 8.120601 8 0.120601

y f x x f x f f

3 3

( ) ( ) (2.01) (2)

(2.01) 2 8.120601 8 0.120601

y f x x f x f f

3 3

( ) ( ) (1.98) (2)

(1.98) 2 7.762392 8 0.237608

y f x x f x f f

3 3

( ) ( ) (1.98) (2)

(1.98) 2 7.762392 8 0.237608

y f x x f x f f

Example 2, page 228

DifferentialsDifferentials

We can obtain a relatively We can obtain a relatively quick and simple wayquick and simple way of of approximatingapproximating yy, the change in , the change in yy due to small change due to small change xx..

Observe below that Observe below that nearnear the the point of tangencypoint of tangency PP, the , the tangent linetangent line TT is is closeclose to the graph of to the graph of ff..

Thus, if Thus, if xx is is smallsmall, then , then dydy is a is a good approximationgood approximation of of yy..

xx

yy

xx x + x + xx

ff((x + x + xx))

ff((xx))

yy

xx

dydyPP

TT

DifferentialsDifferentials

Notice that the slope of Notice that the slope of TT is given by is given by dy/dy/x x (rise over run)(rise over run).. But the slope of But the slope of TT is given by is given by f f ′′((xx)), so we have , so we have

dy/dy/x x == f f ′′((xx)) or or dy dy == f f ′′((xx) ) xx Thus, we have the Thus, we have the approximationapproximation

y ≈y ≈ dy dy == f f ′′((xx))xx The quantity The quantity dydy is called the is called the differential ofdifferential of yy..

xx

yy

xx x + x + xx

ff((x + x + xx))

ff((xx))

yy

xx

dydyPP

TT

The DifferentialThe Differential

Let Let y y = = ff((xx)) define a differentiable function define a differentiable function xx. Then. Then

1.1. The The differentialdifferential dxdx of the of the independentindependent variable variable xx is is

dx = dx = xx

2.2. The The differentialdifferential dydy of the of the dependentdependent variable variable yy is is

dy dy == ff ′′((xx))x x = = f f ′′((xx))dx dx

ExampleExample

Approximate the value of Approximate the value of using differentials..

SolutionSolution Let’s consider the function Let’s consider the function y y = = ff((xx)) == . . Since Since 2525 is the number nearest is the number nearest 26.5 26.5 whose square root is whose square root is

readily recognized, let’s take readily recognized, let’s take xx = 25 = 25.. We want to know the change in We want to know the change in yy, , yy, as , as xx changes from changes from

xx = 25= 25 to to xx = 26.5= 26.5, an increase of , an increase of xx = 1.5= 1.5.. So we findSo we find

Therefore,Therefore,

26.5

x

25

1 1( ) 1.5 1.5 0.15

102 x

y dy f x xx

25

1 1( ) 1.5 1.5 0.15

102 x

y dy f x xx

26.5 25 0.15

26.5 25 0.15 5.15

y

26.5 25 0.15

26.5 25 0.15 5.15

y

Example 4, page 229

Applied Example:Applied Example: Effect of Speed on Vehicular Operating Effect of Speed on Vehicular Operating

The total cost incurred in operating a certain type of truck The total cost incurred in operating a certain type of truck on a on a 500-mile500-mile trip, traveling at an trip, traveling at an average speedaverage speed of of vv mph, mph, is estimated to be is estimated to be

dollars. dollars. Find the approximate Find the approximate change in the total operating costchange in the total operating cost

when the when the average speed isaverage speed is increasedincreased from from 5555 to to 58 58 mph.mph.

4500( ) 125C v v

v

4500( ) 125C v v

v


Applied Example:Applied Example: Effect of Speed on Vehicular Operating Effect of Speed on Vehicular Operating

SolutionSolution Total operating cost is given byTotal operating cost is given by

With With vv = 55 = 55 an an vv = = dvdv = 3 = 3, we find, we find

so the so the total operating costtotal operating cost is found to is found to decreasedecrease by by $1.46$1.46.. This might explain why so many independent truckers This might explain why so many independent truckers

often exceed the often exceed the 55 55 mph speed limit.mph speed limit.

255

4500( ) 1 3

45001 3 1.46

3025

v

C dC C v dvv

255

4500( ) 1 3

45001 3 1.46

3025

v

C dC C v dvv

4500( ) 125C v v

v

4500( ) 125C v v

v


End of End of Chapter Chapter

basic rules of differentiation the product and quotient rules the chain rule

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