copyright © by houghton mifflin company, all rights reserved. calculus concepts 2/e latorre,...

Copyright © by Houghton Mifflin Company, All rights reserved.Copyright © by Houghton Mifflin Company, All rights reserved.

Calculus Concepts 2/eCalculus Concepts 2/eLaTorre, Kenelly, Fetta, Harris, and CarpenterLaTorre, Kenelly, Fetta, Harris, and Carpenter

Chapter 4Chapter 4Determining Change: DerivativesDetermining Change: Derivatives


Chapter 4 Key ConceptsChapter 4 Key Concepts• Numerically Estimating Rates of Change Numerically Estimating Rates of Change

• The Four-Step MethodThe Four-Step Method

• Simple Derivative FormulasSimple Derivative Formulas

• More Simple Derivative FormulasMore Simple Derivative Formulas

• Chain RuleChain Rule

• Product RuleProduct Rule


Numerically Estimating Rates of ChangeNumerically Estimating Rates of Change• The slope of the tangent line is the limiting The slope of the tangent line is the limiting

value of the slopes of nearby secant linesvalue of the slopes of nearby secant lines

• Slopes of piecewise continuous graphs at Slopes of piecewise continuous graphs at discontinuities may be estimated with discontinuities may be estimated with symmetric difference quotientsymmetric difference quotient


Estimating Rates: ExampleEstimating Rates: Example

8x4,534.38x629.15x086.1

4x1,1.2x8.8x45.3x45.0)x(E

2

23

8x4,534.38x629.15x086.1

4x1,1.2x8.8x45.3x45.0)x(E

2

23

The number of Comcast employees (in thousands) x The number of Comcast employees (in thousands) x years after 1990 may be modeled byyears after 1990 may be modeled by

The graph is discontinuous at 4 so The graph is discontinuous at 4 so the derivative at x = 4 does not exist. the derivative at x = 4 does not exist. However, we can estimate the rate of However, we can estimate the rate of change.change.

6.32

4.55.12

35

)3(E)5(E

6.32

4.55.12

35

)3(E)5(E


Estimating Rates: Exercise 4.1 #3Estimating Rates: Exercise 4.1 #3Numerically estimate the limit of slopes of secant Numerically estimate the limit of slopes of secant lines on the graph of h(x) = xlines on the graph of h(x) = x22 + 16x between x = 2 + 16x between x = 2 and close points to the right of x = 2.and close points to the right of x = 2.

x h(x)

2.01

2.001

2.0001

2

36.2

36.02

36.002

36

01.2001.0

362.36

201.2

)2(h)01.2(h

01.2001.0

362.36

201.2

)2(h)01.2(h

001.20001.0

3602.36

2001.2

)2(h)001.2(h

001.20001.0

3602.36

2001.2

)2(h)001.2(h

0001.200001.0

36002.36

20001.2

)2(h)0001.2(h

0001.200001.0

36002.36

20001.2

)2(h)0001.2(h

h'(2) = 20h'(2) = 20


The Four-Step MethodThe Four-Step Method• To find f '(x),To find f '(x),

– Begin with a point (x, f(x))Begin with a point (x, f(x))– Choose a close point (x + h, f(x + h))Choose a close point (x + h, f(x + h))– Write the formula for the slope of the secant line Write the formula for the slope of the secant line

between the two pointsbetween the two points– Evaluate the limit of the slope as h nears 0 Evaluate the limit of the slope as h nears 0

h

)x(f)hx(flim

x)hx(

)x(f)hx(flim)x('f

0h

0h

h

)x(f)hx(flim

x)hx(

)x(f)hx(flim)x('f

0h

0h


The Four-Step Method: ExampleThe Four-Step Method: ExampleCalculate f '(3) for f(x) = xCalculate f '(3) for f(x) = x22 - x. - x.

5hh

66h5h

h

)3(f)h3(f.3

2

5hh

66h5h

h

)3(f)h3(f.3

2

1. (3, f(3)) = (3, 6)1. (3, f(3)) = (3, 6)

2. (3 + h, f(3 + h)) 2. (3 + h, f(3 + h)) = (3 + h, (3 + h)= (3 + h, (3 + h)22 - (3 + h)) - (3 + h))= (3 + h, h= (3 + h, h22 + 5h + 6) + 5h + 6)

5)5h(lim.40h

5)5h(lim.40h


Four-Step Method: Exercise 3.2 #11Four-Step Method: Exercise 3.2 #11Calculate f '(x) for f(x) = 3x - 2Calculate f '(x) for f(x) = 3x - 2

3h

)2x3(2h3x3h

)x(f)hx(f.3

3h

)2x3(2h3x3h

)x(f)hx(f.3

1. (x, f(x)) = (x, 3x - 2)1. (x, f(x)) = (x, 3x - 2)

2. (x + h, f(x + h)) 2. (x + h, f(x + h)) = (x + h, 3(x + h) - 2)= (x + h, 3(x + h) - 2)= (x + h, 3x + 3h - 2)= (x + h, 3x + 3h - 2)

33lim.40h

33lim.40h


Four-Step Method: Exercise 3.2 #13Four-Step Method: Exercise 3.2 #13Calculate f '(x) for f(x) = 3xCalculate f '(x) for f(x) = 3x22

h3x6h

x3h3xh6x3

h

)x(f)hx(f.3

222

h3x6h

x3h3xh6x3

h

)x(f)hx(f.3

222

1. (x, f(x)) = (x, 3x1. (x, f(x)) = (x, 3x22))

2. (x + h, f(x + h)) 2. (x + h, f(x + h)) = (x + h, 3(x + h)= (x + h, 3(x + h)22))= (x + h, 3x= (x + h, 3x22 + 6xh + 3h + 6xh + 3h22))

x6)h3x6(lim.40h

x6)h3x6(lim.40h


Simple Derivative FormulasSimple Derivative Formulas• Constant RuleConstant Rule

– If f(x) = b, f '(x) = 0.If f(x) = b, f '(x) = 0.

• Linear Function RuleLinear Function Rule– If f(x) = ax + b, f '(x) = a.If f(x) = ax + b, f '(x) = a.

• Simple Power RuleSimple Power Rule– If f(x) = xIf f(x) = xnn, f '(x) = nx, f '(x) = nxn-1n-1..

• Constant Multiplier RuleConstant Multiplier Rule– If f(x) = k g(x), f '(x) = k g'(x).If f(x) = k g(x), f '(x) = k g'(x).

• Sum RuleSum Rule– If f(x) = g(x) + h(x), f '(x) = g'(x) + h'(x) If f(x) = g(x) + h(x), f '(x) = g'(x) + h'(x)


Simple Derivatives: ExamplesSimple Derivatives: Examples• Constant RuleConstant Rule

– If f(x) = 5, f '(x) = 0.If f(x) = 5, f '(x) = 0.

• Linear Function RuleLinear Function Rule– If f(x) = -3x + 4, f '(x) = -3.If f(x) = -3x + 4, f '(x) = -3.

• Simple Power RuleSimple Power Rule– If f(x) = xIf f(x) = x44, f '(x) = 4x, f '(x) = 4x33..


Simple Derivatives: ExamplesSimple Derivatives: Examples• Constant Multiplier RuleConstant Multiplier Rule

– If f(x) = 4xIf f(x) = 4x33, f '(x) = 4(3x, f '(x) = 4(3x22) = 12x) = 12x22

– If f(x) = -3(2x - 1), f '(x) = -3(2) = -6If f(x) = -3(2x - 1), f '(x) = -3(2) = -6

• Sum RuleSum Rule– If f(x) = 4xIf f(x) = 4x3 3 - 3(2x - 1), f '(x) = 12x- 3(2x - 1), f '(x) = 12x22 - 6 - 6


Simple Derivatives: Exercises 4.3 #7, 12Simple Derivatives: Exercises 4.3 #7, 127. Calculate y' for y = 7x7. Calculate y' for y = 7x22 - 12x + 13 - 12x + 13

Using the Sum Rule, Power Rule, Constant Rule, Using the Sum Rule, Power Rule, Constant Rule, and Constant Multiple Rule we getand Constant Multiple Rule we gety' = 14x - 12y' = 14x - 12

12. Calculate y' for y = 3x12. Calculate y' for y = 3x-2-2

Using the Power Rule and Constant Multiple Rule Using the Power Rule and Constant Multiple Rule we getwe gety' = -6xy' = -6x-3-3


More Simple Derivative FormulasMore Simple Derivative Formulas• Exponential RuleExponential Rule

– If f(x) = bIf f(x) = bx x with b > 0, f '(x) = ln(b) bwith b > 0, f '(x) = ln(b) bxx

• eexx Rule Rule– If f(x) = eIf f(x) = ex x with b > 0, f '(x) = ewith b > 0, f '(x) = exx

• Natural Log RuleNatural Log Rule– If f(x) = ln(x), f '(x) = xIf f(x) = ln(x), f '(x) = x-1-1 for x > 0 for x > 0


More Simple Derivatives: ExampleMore Simple Derivatives: ExampleCalculate y' for y = 7xCalculate y' for y = 7x22 - 3 - 3xx

Using the Sum Rule, Power Rule, Exponential Rule, Using the Sum Rule, Power Rule, Exponential Rule, and Constant Multiple Rule we getand Constant Multiple Rule we gety' = 14x - (ln3)3y' = 14x - (ln3)3xx

Calculate y' for y = eCalculate y' for y = exx + 4ln(x) + 4ln(x)

Using the eUsing the exx Rule and Natural Log Rule we get Rule and Natural Log Rule we gety' = ey' = exx + 4x + 4x-1-1


More Derivatives: Exercise 4.4 #11, 14More Derivatives: Exercise 4.4 #11, 1411. Calculate y' for y = 100,000(1 + 0.05/12)11. Calculate y' for y = 100,000(1 + 0.05/12)12x12x

14. Calculate y' for y = 9 - 4.2 lnx + 3.3(2.9)14. Calculate y' for y = 9 - 4.2 lnx + 3.3(2.9)xx

x

x

x

x12

)05116.1(612.4989

)05116.1)(05116.1ln(000,100'y

)05116.1(000,100

)00417.1(000,100y

x

x

x

x12

)05116.1(612.4989

)05116.1)(05116.1ln(000,100'y

)05116.1(000,100

)00417.1(000,100y

x

x

)9.2(5136.3x

2.4

)9.2)(9.2ln(3.3x

12.4'y

x

x

)9.2(5136.3x

2.4

)9.2)(9.2ln(3.3x

12.4'y


More Derivatives: Exercise 4.4 #19More Derivatives: Exercise 4.4 #19The weight of a laboratory mouse between 3 and 11 The weight of a laboratory mouse between 3 and 11 weeks of age can be modeled by the equationweeks of age can be modeled by the equation

w(t) = 11.3 + 7.37 ln(t) gramsw(t) = 11.3 + 7.37 ln(t) gramswhere the age of the mouse is (t + 2) weeks after where the age of the mouse is (t + 2) weeks after birth (thus for a 3-week old mouse, t = 1.) How birth (thus for a 3-week old mouse, t = 1.) How rapidly is the weight of a 9-week old mouse rapidly is the weight of a 9-week old mouse changing?changing?

weekpergrams05286.17

37.7)7('w

t

37.7)t('w

weekpergrams05286.17

37.7)7('w

t

37.7)t('w

Note: 9 weeks implies t = 7Note: 9 weeks implies t = 7


Chain RuleChain Rule• Form 1: If C is a function of p and p is a Form 1: If C is a function of p and p is a

function of t, thenfunction of t, then

dt

dp

dp

dC

dt

dC

dt

dp

dp

dC

dt

dC

)x('g))x(g('h)x('fdx

df )x('g))x(g('h)x('f

dx

df

• Form 2: If f(x) = (h g)(x) = h(g(x)) thenForm 2: If f(x) = (h g)(x) = h(g(x)) then


Chain Rule: Example (Form 1)Chain Rule: Example (Form 1)

))t(p(Cy

thent)t(pande)p(CLet

ey2p

t2

))t(p(Cy

thent)t(pande)p(CLet

ey2p

t2

t2e

t2e

dt

dp

dp

dC

dt

dC

2t

p

t2e

t2e

dt

dp

dp

dC

dt

dC

2t

p


Chain Rule: Example (Form 2)Chain Rule: Example (Form 2)

))x(h(gy

thenx)x(hande)x(gLet

ey2x

x2

))x(h(gy

thenx)x(hande)x(gLet

ey2x

x2

2

2

x

x

)x(h

xe2

x2e

x2e

)x('h))x(h('g'y

2

2

x

x

)x(h

xe2

x2e

x2e

)x('h))x(h('g'y


Chain Rule: Exercise 4.5 #3Chain Rule: Exercise 4.5 #3An investor buys gold at a constant rate of 0.2 ounce An investor buys gold at a constant rate of 0.2 ounce per day. The investor currently has 400 troy ounces per day. The investor currently has 400 troy ounces of gold. If gold is currently worth $395.70 per troy of gold. If gold is currently worth $395.70 per troy ounce, how quickly is the value of the investor’s gold ounce, how quickly is the value of the investor’s gold increasing? (Use Form 1)increasing? (Use Form 1)

ownedareouncesgwhendollarsg70.395)g(V

nowfromdaystownedounces400t2.0)t(g



dayperdollars14.79

2.070.395

dt

dg

dg

dV

dt

dV

dayperdollars14.79

2.070.395

dt

dg

dg

dV

dt

dV


Chain Rule: Exercise 4.5 #3Chain Rule: Exercise 4.5 #3An investor buys gold at a constant rate of 0.2 ounce An investor buys gold at a constant rate of 0.2 ounce per day. The investor currently has 400 troy ounces per day. The investor currently has 400 troy ounces of gold. If gold is currently worth $395.70 per troy of gold. If gold is currently worth $395.70 per troy ounce, how quickly is the value of the investor’s gold ounce, how quickly is the value of the investor’s gold increasing? (Use Form 2)increasing? (Use Form 2)

)400t2.0(70.395))t(g(V



)400t2.0(70.395))t(g(V



dayperdollars14.792.070.395

)t('g))t(g('V)t(g(Vdt

d

dayperdollars14.792.070.395

)t('g))t(g('V)t(g(Vdt

d


Product RuleProduct Rule• Product RuleProduct Rule

If f(x) = g(x) • h(x) If f(x) = g(x) • h(x)

then f '(x) = g'(x) • h(x) + g(x) • h'(x)then f '(x) = g'(x) • h(x) + g(x) • h'(x)

• Example:Example:

f(x) = (xf(x) = (x33 + 1)(2 + 1)(2xx))

f '(x) = 3xf '(x) = 3x22 • 2 • 2xx + (x + (x33 + 1) • (ln2)2 + 1) • (ln2)2xx


Product Rule: ExampleProduct Rule: ExampleA music store has determined from a customer A music store has determined from a customer survey that when the price of each CD is $x, the survey that when the price of each CD is $x, the number of CDs sold monthly can be modeled by number of CDs sold monthly can be modeled by

N(x) = 6250 (0.92985)N(x) = 6250 (0.92985)x x CDsCDsFind and interpret the rate of change of revenue Find and interpret the rate of change of revenue when the CDs are priced at $10.when the CDs are priced at $10.

R(x) = N(x) • x = 6250 (0.92985)R(x) = N(x) • x = 6250 (0.92985)xx • x • xR'(x) R'(x) = 6250 ln(0.92985)(0.92985)= 6250 ln(0.92985)(0.92985)xx • x + • x +

6250 (0.92985) 6250 (0.92985)xx • 1 • 1= - 454.575 x(0.92985)= - 454.575 x(0.92985)xx + 6250 (0.92985) + 6250 (0.92985)xx

R'(10) = 823 means revenue is increasing by $823 R'(10) = 823 means revenue is increasing by $823 per $1 of CD price when the price is $10 per CDper $1 of CD price when the price is $10 per CD


Product Rule: Exercise 4.6 #28Product Rule: Exercise 4.6 #28A music store has determined that the number of A music store has determined that the number of CDs sold monthly can be modeled by CDs sold monthly can be modeled by

N(x) = 6250 (0.9286)N(x) = 6250 (0.9286)x x CDsCDswhere x is the price in dollars. Find the rate at which where x is the price in dollars. Find the rate at which the revenue is changing when x = $20.the revenue is changing when x = $20.

R(x) = N(x) • x = 6250 (0.9286)R(x) = N(x) • x = 6250 (0.9286)xx • x • xR'(x) R'(x) = 6250 ln(0.9286)(0.9286)= 6250 ln(0.9286)(0.9286)xx • x + • x +

6250 (0.9286) 6250 (0.9286)x x • 1• 1= - 462.983 x(0.9286)= - 462.983 x(0.9286)xx + 6250 (0.9286) + 6250 (0.9286)xx

R'(20) = -684.10 means revenue is decreasing by R'(20) = -684.10 means revenue is decreasing by $684.10 per $1 of CD price when the price is $20 $684.10 per $1 of CD price when the price is $20 per CDper CD

copyright © by houghton mifflin company, all rights reserved. calculus concepts 2/e latorre,...

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x ln33 x

right of x

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