1 chapter 2 differentiation: basic concepts in this chapter, we will encounter some important...

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1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts. The Derivative Product and Quotient Rules, Higher-Order Derivatives The Chain Rule Marginal Analysis, Implicit Differentiation.

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Page 1: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

1

Chapter 2Differentiation: Basic Concepts

In this Chapter, we will encounter some important concepts.

The Derivative

Product and Quotient Rules, Higher-Order Derivatives

The Chain Rule

Marginal Analysis, Implicit Differentiation.

Page 2: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

1.1. Constant Rule: Constant Rule: derivative of a constant is zeroderivative of a constant is zero

2.2. Power Rule: Power Rule: for any real n, for any real n,

3.3. Constant Multiple Rule: Constant Multiple Rule: for constant c and a for constant c and a differentiable function f(x), differentiable function f(x),

4.4. Sum Rule: Sum Rule: when f(x) and g(x) are both differentiable,when f(x) and g(x) are both differentiable,

0cdx

d

Review: Techniques of Differentiation

1][ nn nxxdx

d

)()( xfdx

dcxcf

dx

d

)]([)]([)]()([ xgdx

dxf

dx

dxgxf

dx

d

Page 3: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise

3

2

23

5141

13

x

xy

x

xy

Page 4: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Solution

Page 5: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise

Find the equation of the line that is tangent to the graph of the function at the point (-1,-8).

135 23 xxxy

Page 6: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Suggested Solution:

Suppose the equation of the tangent line is

m= at the point (-1,-8).

So at the point (-1,-8), Hence, , and therefore -8=(-10)(-1)+b, gives b=2.The equation of the tangent line to this function at point (-1,-8)

is

3103

135

2

23

xxdx

dym

xxxy

103)1(*10)1(*3 2 m

bmxy tt

dx

dy

bxy tt 10

210 tt xy

Page 7: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

1.1. Relative rate of change:Relative rate of change:

2.2. Percentage rate of change:Percentage rate of change:

Review:

Q

dxdQ

xQ

xQ

Q(x)

/

)(

)(

of change

of rate Relative

%)(

)(100

)( of change of

rate Percentage

xQ

xQ

xQ

Page 8: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise1.Find the relative rate of change of with respect to x for the value x=1.

452)( 23 xxxf

2.The gross annual earnings of a certain company were thousand dollars t years after its formation in 2004.

(1)At what rate were the gross annual earnings of the company growing with respect to time in 2008?(2)At what percentage rate were the gross annual earnings of the company growing with respect to time in 2008?

20101.0)( 2 tttA

Page 9: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

1.Suggested Solution: The function is Its rate of change is

Its relative rate of change is

When x=1,

41

4

42*51*2

1*101*6

)1(

)1('

452

106

)(

)('

106)('

452)(

23

2

2

23

f

f

xx

xx

xf

xf

xxxf

xxxf

Page 10: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

2.Suggested Solution:

(1) , which is the rate that company’s gross

annual earning changes (G.A.E) with respect to t years after 2004.

In 2008, t=4, the G.A.E will change at A’(4)=0.2*4+10=10.8 thousand dollars per year.

(2)The percentage rate of that of the company’s G.A.E changes is expressed as . So in year 2008, this value is

This means the company’s G.A.E increases 17.53% per year in the year 2008.

102.0)('

20101.0)( 2

ttA

tttA

%100*)(

)('

tA

tA

%53.17%100*204*104*1.0

8.10%100*

)4(

)4('2

A

A

Page 11: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Product Rules: For function y=f(x) and z=g(x). If they are both

differentiable at x, then the derivative of their product is

Quotient Rules: For function y=f(x) and z=g(x). If they are both differentiable at x, then the derivative of their quotient is

iff z=g(x) ≠ 0

dx

dzyz

dx

dyzy

dx

d

xgxfxgxfxgxf

**)*(

)(')()()(')]'()([

2

2

**)(

)]([

)(')()()(']'

)(

)([

z

ydxdz

zdxdy

z

y

dx

d

xg

xgxfxgxf

xg

xf

Page 12: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Second Order Derivative: is the derivative of the first derivative.

Notation as follow:

Example: Function S(t) PositionFirst Order Derivative S’(t)=V(t) VelocitySecond Order Derivative S”(t)=V’(t)=A(t) Acceleration

2

2

)('' )('dx

fdxf

dx

dfxf

Page 13: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise:An efficiency study of the morning shift at a certain

factory indicates that an average worker arriving on the job at 8:00am will have produced units t hours later.

(a)Compute the worker’s rate of production.

(b)At what rate is the worker’s rate of production changing with respect to time at 9:00am?

ttttQ 158)( 23

Page 14: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Suggested Solution(a)Worker’s rate of production is

(b) The rate that worker’s rate of production changes is

So at 9:00am, t=1

hourper units 15163)(')( 2 tttQtP

hourper hour per units 166)(")(' ttQtP

hourper hour per units 10166)1(")1(' QP

Page 15: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review: Chain Rule: For functions y=f(u), which is differentiable at

u, and z=g(x), which is differentiable at x.

The composition function y=f(z)=f(g(x)) is differentiable at x, which is

!!! Pay Attention to the phrase “WITH RESPECT TO”

)('*))(('))]'(([ xgxgfxgfdx

dz

dz

dy

dx

dy

Page 16: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise Differentiate

Suggested Solution

12

13)(

x

xxf

2/32/1

22

)12()13(2

5)('

rule,chain by then

)12(

5

)12(

)13(2)12(3 and

1213

2

1

2

1then

12

13 ,

12

13)(

xxdx

du

du

dy

dx

dyxf

xx

xx

dx

dux-

xudu

dyx

xuwhereu

x

xyxf

Page 17: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Marginal Cost: If C(x) is the total cost of producing x units of a commodity.

Then the marginal cost of producing units is the derivative ,

which approximates the additional cost incurred when the level of production is increased by one unit, from to

Marginal Revenue :

Marginal Profit :

Review:Marginal Cost: If C(x) is the total cost of producing x units of a commodity.

Then the marginal cost of producing units is the derivative ,

which approximates the additional cost incurred when the level of production is increased by one unit, from to

Marginal Revenue :

Marginal Profit :

10 x

)( 0xC

)()1( 00 xCxC

0x

0x

)(' 0xR

)(')(')(' 000 xCxRxP

Page 18: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Approximation by Increment: If y=f(x) is differentiable

at , and △x is a small change in x, then

Approximation Percentage of change: if △x is a small change in x, then

xxfxfxxff

xxfxfxxf

)(')()(

)(')()(

000

000

oxx

)(

)('100

)(100in change Percentage

xf

xxf

xf

ff

Page 19: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise

Use Incremental Approximation to approximate the value of 1.00380. Remember f(x+x) – f(x) = f f’(x)x

2

2

)0('')0(')0()( x

fxffxf

Page 20: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Suggested solution

• Let f(x) = x80, x=1, x = 0.003• f’(x) = 80x79.• When x = 1, f’(x) = 80• f 80x = 0.24• f(x+x) – f(x) = 0.24.• 1.00380

• =1+0.24=1.24

Page 21: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

21

The ratio of Errors (output to input)

• Suppose x is the correct (or precise) input• And x+x is the incorrect (or approximate)

input.• Then x is called the error of the input.• and y = f(x +x) - f(x) is called the error of

the output• If x is small, then y’ y/ x.• Or y = y’ x.

Page 22: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Example 23

During a medical procedure, the size of a roughly spherical During a medical procedure, the size of a roughly spherical tumor is estimated by measuring its diameter and using the tumor is estimated by measuring its diameter and using the formula to compute its volume. If the diameter is formula to compute its volume. If the diameter is measured as measured as 2.52.5 cm with a maximum error of cm with a maximum error of 2%,2%, how how accurate is the volume measurement?accurate is the volume measurement?

3

3

4RV

Solution:Solution:

A sphere of radius A sphere of radius RR and diameter and diameter x=2Rx=2R has volume has volume

33333 cm 181.8)5.2(6

1

6

1)

2(

3

4

3

4 x

xRV

The error made in computing this volume using the diameter 2.5 , while the actual

diameter is 2.5+△x , is

(2.5 ) (2.5) (2.5)V V x V V x 22

Page 23: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise

At a certain factory, the daily output is units, where K denotes the capital investment measured in units of $1,000. The current capital investment is $900,000. Estimate the effect that an additional capital investment of $800 will have on the daily output.

2/1600)( KKQ

Page 24: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Suggested SolutionThe current capital investment K₀=900 thousand dollars.The increase in capital investment K=0.8 thousand dollars.△To estimate the effect of this K on the daily output:△

This means an additional capital investment of 800 dollars would increase the daily output by 8 units.

units 88.0*)900(3008.0)900('

80900for

300K )('

2/1

-1/2

QQ

., Δ K

KKKQQ

Page 25: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Differentials: • Differentials of x is dx= x△ (small increment)• If y=f(x) is a differentiable function of x, then the differential of y

is dxxfdy )('

Page 26: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Review:Implicit Differentiation: • Explicit Form: a function that can be written as y=f(x) “y is solved, and given by an equation of x”• Implicit Form: cannot express y as a equation of x.• How to differentiate an implicit equation? (1) Differentiate both side with respect to x you now have a function containing x and y and y’. (2) Express y’ in terms of x and y.

Exercise: Differentiate implicitly with respect to x.132 22 xxyy

Page 27: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Exercise: Differentiate implicitly with respect to x.

Suggested Solution:

132 22 xxyy

xyy

yy

yxyyy

xyyyyy

42

23'

23)42('

:equation above thegrearranginBy

3'42'2

:have weside,both on ateDifferenti

2

2

2

Page 28: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

28

Example

Suppose the output at a certain factory is units, where x is the number of hours of skilled labor used and y is the number of hours of unskilled labor. The current labor force consists of 30 hours of skill labor and 20 hours of unskilled labor.

Question: Use calculus to estimate the change in unskilled labor y that should be made to offset a 1-hour increase in skilled labor x so that output will be maintained at its current level.

3232 yyxxQ

Page 29: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

Solution:

Page 30: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

30

The manager of a company determines that when q hundred units of a particular commodity are produced, the cost of production is C thousand dollars, where . When 1500 units are being produced, the level of production is increasing at the rate of 20 units per week.

What is the total cost at this time and at what rate is it changing?

Example

42753 32 qC

Page 31: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

31

312 22 ppxx

Exercise

When the price of a certain commodity is p dollars per unit, the manufacturer is willing to supply x thousand units, where

How fast is the supply changing when the price is $9 per unit and is increasing at the rate of 20 cents per week?

Page 32: 1 Chapter 2 Differentiation: Basic Concepts In this Chapter, we will encounter some important concepts.  The Derivative  Product and Quotient Rules,

32

Exercise

A lake is polluted by waste from a plant located on its shore. Ecologists determine that when the level of pollutant is x parts per million (ppm), there will be F fish of a certain species in the lake, where

When there are 4000 fish left in the lake, the pollution is increasing at the rate of 1.4ppm/year. At what rate is the fish population changing at this time?

xF

3

32000