Chapter 2

Differentiation:

Basic Concepts

1

The Concepts

The Derivative

Product and Quotient Rules

Higher-Order Derivatives

The Chain Rule

Marginal Analysis

Implicit Differentiation

2

2.1 The Derivative

3

Why We Learn Differentiation?

Calculus is the mathematics of change, and the primary tool for studying change is a procedure called differentiation.

In this section, we will introduce this procedure and examine some of its uses, especially in computing rates of change.

Rate of changes, for example velocity, acceleration, the rate of growth of a population, and many others, are described mathematically by derivatives.

4

Illustration

If air resistance is neglected, an object dropped from a

great height will fall feet in t seconds.

What is the object’s instantaneous velocity after t=2 seconds?

Average rate of change of s(t) over the time period [2,2+h] is

Compute the instantaneous velocity by the limit

That is, after 2 seconds, the object is traveling at the rate of 64 feet per second.

5

216)( tts

hh

h

h

shsVave

1664)2(16)2(16

2)2(

)2()2(

timeelapsed

traveleddistance

22

64)1664(limlim00

hVVh

aveh

ins

6

Rates of Change

How to determine instantaneous rate of change or rate

of change of f(x) at x=c ?

Find the average rate of change of f(x) as x varies from

x=c to x=c+h

h

cfhcf

chc

cfhcf

x

xfrateave

)()(

)(

)()()(

Compute the instantaneous rate of change of f(x) at x=c by

finding the limiting value of the average rate as h tends to 0

h

cfhcfraterate

have

hins

)()(limlim

00

7

Rates of Change (Linear function)

A linear function y(x)=mx+b changes at the constant rate m

with respect to the independent variable x. That is the rate of

change of y(x) is given by the slope or steepness of its graph.

12

12

x in change

y in change

change of rateSlope

xx

yy

x

y

8

Rates of Change (Non-Linear function)

For the function that is

nonlinear, the rate of change

is not constant but varies

with x.

In particular, the rate of

change at x=c is given by the

steepness of the graph of f(x)

at the point (c,f(c)), which

can be measured by the slope

of the tangent line to the

graph at p.

9

The Slope of Secant LineThe secant line is a line that intersects the curve at the

point x and point x+h

The average rate of change can be interpreted geometrically as

the slope of the secant line from the point (x,f(x)) to the point

(x+h,f(x+h)).

10

The DerivativeA difference quotient for the function f(x) is the

expression

The derivative of a function f(x) with respect to x is the

function f’(x) given by

The process of computing the derivative is called

differentiation. f(x) is differentiable at x=c if f’(c) exists

h

xfhxfxf

h

)()(lim)(

0

h

xfhxf )()(

Example

11

Find the derivative of the function 35162)( 2 xxxf

The difference quotient for f(x) is

16241624

3516235)(16)(2)()(

2

22

hxh

hhxh

h

xxhxhx

h

xfhxf

Thus, the derivative of f(x) is the function

164)1624(lim)()(

lim)(00

xhxh

xfhxfxf

hh

12

Slope as a Derivative: The slope of the tangent

line to the curve y=f(x) at the point (c,f(c)) is

Instantaneous Rate of Change as a Derivative:

The rate of change of f(x) with respect to x

when x=c is given by f’(c)

)(tan cfm

Remarks: Since the slope of the tangent line at (a,f(a))

is f’(a), the equation of the tangent line is

))(()( axafafy The point-slope form

Example

13

What is the equation of the tangent line to the curve

at the point where x=4 ?xy According to the definition of derivative, we have

xxhxxhxh

xhxxhx

h

xhx

h

xfhxfxf

hh

hh

2

11lim

)(

))((lim

lim)()(

lim)(

00

00

Thus, the slope of the tangent line to the curve at the

point where x=4 is f’(4)=1/4 . So the point-slope

form is 14

1)4(

4

12 xyxy

Example

14

A manufacturer determines that when x thousand units of a

particular commodity are produced, the profit generated will be

120006800400)( 2 xxxp

dollars. At what rate is profit changing with respect to the level of

production x when 9 thousand units are produced?

We find that

68008006800800400

lim

)120006800400(12000)(6800)(400lim

)()(lim)(

2

0

22

0

0

xh

hhxh

h

xxhxhx

h

xphxpxp

h

h

h

15

Thus, when the level of production is x=9, the profit is changing

at the rate of dollars per

thousand units.

4006800)9(800)9( p

Which means that the tangent line to the profit curve y=p(x) is

sloped downward at point Q where x=9. Therefore, the profit

curve must be falling at Q and profit must be decreasing when

9 thousand units are being produced.

Significance of the Sign of the Derivative

16

If the function f is differentiable at x=c, then

f is increasing at x=c if f’(c)>0 and

f is decreasing at x=c if f’(c)<0

17

Alternative Derivative Notation

Given a function y=f(x) all of the following are equivalent

and represent the derivative of f(x) with respect to x.

If we want to evaluate the derivative at x=a all of the

following are equivalent

18

Differentiability and Continuity

If the function f(x) is differentiable at x=c, then it is also

continuous at x=c.

Notice that a continuous function may not be differentiable

Graphs of four functions that are not differentiable at x=0.

2.2 Techniques of Differentiation

19

20

For any constant c, we have

That is, the derivative of a constant is zero

0cdx

d

Since f(x+h)=c for all x

0lim)()(

lim)(00

h

cc

h

xfhxfxf

hh

The Constant Rule

The Power Rule

Examples.

21

For any real number n1][ nn nxx

dx

d

2

1

2

1

23

2

1)()(

3)(

xxdx

dx

dx

d

xxdx

d

The Constant Multiple Rule

22

If c is a constant and f(x) is differentiable then so is

cf(x) and

That is, the derivative of a multiple is the multiple of

the derivative

)()( xfdx

dcxcf

dx

d

3344 12)4(3)(3)3( xxxdx

dx

dx

d

2/32/32/1

2

7)

2

1(7)7()

7(

xxxdx

d

xdx

d

The Sum Rule

23

If f(x) and g(x) are differentiable then so is the sum of

s(x)=f(x)+g(x) and

That is, the derivative of a sum is the sum of the separate

derivative

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

dxf

dx

dxgxf

dx

d

3322 20)(2)7()()7( xxdx

dx

dx

dx

dx

d

84

847575

2110

)7(3)5(2)(3)(2)32(

xx

xxxdx

dx

dx

dxx

dx

d

Example

24

Find the equation of the tangent line to the function

at x=16.xxxf 84)(

We know that the equation of a tangent line is given by

So, we will need the derivative of the function

Now we need to evaluate the function and the derivative.

Thus, the point-slope form for a tangent line is:

163)16(332 xyxy

Example

25

It is estimated that x months form now, the

population of a certain community will be

800020)(2 xxxp

a. At what rate will the population be changing with

respect to time 15 months from now?

b. By how much will the population actually change

during the 16th month?

26

a. The rate of change of the population with respect to time

is the derivative of the population function. That is

202)(change of Rate xxp

The rate of change of the population 15 months from now

will be month per people 5020)15(2)15( p

b. The actual change in the population during the 16th month

is people 5185258576)15()16( pp

Note: Since the rate varies during the month, the actual

change in population during 16th month differs from the

monthly rate of the change at the beginning of the month.

27

The Derivative as a Rate of Change (Review)

28

The relative rate of change of a quantity Q(x) with

respect to x is given by the ratio

The corresponding percentage rate of change of Q(x)

with respect to x is

Q

dxdQ

xQ

xQ

Q(x)

/

)(

)(

of change

of rate Relative

)(

)(100

)( of change of

rate Percentage

xQ

xQ

xQ

Relative and Percentage Rates of Change

Example

29

The gross domestic product (GDP) of a certain country was

billion dollars t years after 1995.

a. At what rate was the GDP changing with respect to time in 2005?

b. At what percentage rate was the GDP changing with respect to

time in 2005?

1065)( 2 tttN

a. The rate of change of the GDP is the derivative N’(t)=2t+5.

The rate of change in 2005 was N’(10)=2(10)+5=25 billion

dollars per year.

b. The percentage rate of change of the GDP in 2005 was

yearper %77.9256

25100

)10(

)10(100

N

N

Example

30

Let , find all x where p’(x)>0, p’(x)=0 and

p’(x)<0.

51232)( 23 xxxxp

)2)(1(6

1266

012)2(3)3(2)(

2

2

xx

xx

xxxp

Based on the techniques of

differentiation, we have

So the solution is

p’(x)=0 at x=-1 and 2

p’(x)>0 at x<-1 and x>2

p’(x)<0 at -1<x<2

Rectilinear Motion

31

Motion of an object along a straight-line in some way. If the

position at time t of an object moving along a straight line is

given by s(t), then the object has

and

The object is advancing when v(t)>0, retreating when v(t)<0,

and stationary when v(t)=0. It is accelerating when a(t)>0

and decelerating when a(t)<0

dt

dststv )()( velocity

dt

dvtvta )()(on accelerati

Example

32

The position at time t of an object moving along a line is given by

a. Find the velocity of the object and discuss its motion between

times t=0 and t=4.

b. Find the total distance traveled by the object between times

t=0 and t=4.

c. Find the acceleration of the object and determine when the

object is accelerating and decelerating between times t=0 and

t=4.

596)( 23 tttts

33

The motion of an object:

596)(23 tttts

Interval Sign

of v(t)

Description

of Motion

0<t<1 +

Advancing

from s(0)=5 to s(1)=9

1<t<3 -

Retreating

from s(1)=9 to s(3)=5

3<t<4 +

Advancing

from s(3)=5 to s(4)=9

a. The velocity is . The object will be

stationary when

9123)( 2 ttdt

dstv

0)3)(1(39123)( 2 tttttv

that is, at times t=1 and t=3. Otherwise, the object is either

advancing or retreating, as described in the following table.

34

b. The object travels from s(0)=5 to s(1)=9, then back to

s(3)=5, and finally to s(4)=9. Thus the total distance

traveled by the object is

12|59||95||59|

433110

ttt

D

c. The acceleration of the object is

)2(6126)( ttdt

dvta

The object will be accelerating [a(t)>0] when 2<t<4 and

decelerating [a(t)<0] when 0<t<2.

2.3 Product and Quotient Rules; Higher-Order Derivative

35

36

The derivative of a product of functions is not the product of

separate derivative!! Similarly, the derivative of a quotient of

functions is not the quotient of separate derivative.

Product and Quotient Rules

Suppose we have two function f(x)=x3 and g(x)=x6

The Product Rule

37

If the two functions f(x) and g(x) are differentiable at x, then we

have the derivative of the product P(x)=f(x)g(x) is

or equivalently,

)]([)()]([)()()( xfdx

dxgxg

dx

dxfxgxf

dx

d

fggffg )(

Use the product rule to find the derivative of the function

3

5

3

2

3

5

3

2

3

5

3

2

3 223

1

3

8

3

1022

3

2

3

4

)22()2(3

2)(

xxxxxx

xxxxxxy

)2( 23 2 xxxy

Example

38

A manufacturer determines that t months after a new

product is introduced to the market, hundred

units can be produced and then sold at a price of

dollars per unit .

a. Express the revenue R(t) for this product as a function of

time .

b. At what rate is revenue changing with respect to time after 4

months? Is revenue increasing or decreasing at this time?

tttx 3)( 2

302)( 2

3

ttp

39

a. The revenue is given by

)302)(3()()()( 2/32 ttttptxtR

hundred dollars.

b. The rate of change of revenue R(t) with respect to time is given

by the derivative R’(t), which we find using the product rule:

]32)[302()2

3(2)3(

]3[)302(302)3()(

2/32/12

22/32/32

ttttt

ttdt

dtt

dt

dtttR

At time t=4, the revenue is changing at the rate R’(4)=-14

Thus, after 4 months, the revenue is changing at the rate of 14

hundred dollars per month. It is decreasing at that time since

R’(4) is negative.

The Quotient Rule

40

If the two functions f(x) and g(x) are differentiable at x, then the

derivative of the quotient Q(x)=f(x)/g(x) is given by

or equivalently,

0)( if )(

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

])(

)([

2

xgxg

xgdx

dxfxf

dx

dxg

xg

xf

dx

d

2

''')(

g

fggf

g

f

Example.

Use the quotient rule to find the derivative of the function

22 )2(

15

)2(

)1)(93()2(3)(

zz

zzzW

A Word of Advice

41

The quotient rule is somewhat cumbersome, so don’t use it

unnecessarily.

Differentiate the function . x

xx

xy

1

5

4

33

22

Don’t use the quotient rule! Instead, rewrite the function as

12 15

4

3

1

3

2 xxxy

and then apply the power rule term by term to get

23

23

23

1

3

1

3

4

3

1

3

4

)1(003

1)2(

3

2

xxxx

xxdx

dy

42

The Second Derivative

In applications, it may be necessary to compute the rate of change of a function that is itself a rate of change. For example, the acceleration of a car is the rate of change with respect to time of its

velocity.

The second derivative of a function is the derivative of its derivative.

If y=f(x), the second derivative is denoted by

The second derivative gives the rate of change of the rate of change

of the original function.

)(or 2

2

xfdx

yd

Note: The derivative f ’(x) is called the first derivative.

Alternate Notation

43

There is some alternate notation for higher order derivatives as

well.2

2

)( )(dx

fdxf

dx

dfxf

Note: Before computing the second derivative of a function, always

take time to simplify the first derivative as much as possible.

Example.

An efficiency study of the morning shift at a certain factory

indicates that an average worker who arrives on the job at 8:00

A.M. will have produced units t hours later.

a. Compute the worker’s rate of production at 11:00 A.M.

b. At what rate is the worker’s rate of production changing with

respect to time at 11:00 A.M.?

ttttQ 246)( 23

44

a. The worker’s rate of production is the first derivative

24123)()( 2 tttQtR

of the output Q(t). The rate of production at 11:00 A.M. (t=3) is

hour per units 33)3()3( QR

b. The rate of change of the rate of production is the second

derivative of the output function. At 11:00 A.M.,

this rate is

The minus sign indicates that the worker’s rate of production is

decreasing; that is, the worker is slowing down. The rate of this

decrease in efficiency at 11:00 A.M. is 6 units per hour per hour.

126)()( ttQtR

hour per hour per units 6)3()3( QR

Acceleration as Second Order Derivative

45

The acceleration a(t) of an object moving along a straight line is

the derivative of the velocity v(t). Thus, the acceleration may be

thought of as the second derivative of position; that is 2

2

)(dt

sdta

If the position of an object moving along a straight line is given

by at time t, find its velocity and acceleration. 43)( 23 ttts

Example.

The velocity of the object is

and its acceleration is

463)( 2 ttdt

dstv

66)(2

2

tdt

sd

dt

dvta

The nth Derivative

46

For any positive integer n, the nth derivative of a function is

obtained from the function by differentiating successively n times.

If the original function is y=f(x) the nth derivative is denoted by

)(or )( xfdx

yd n

n

n

Example.

Find the fifth derivative of the function y=1/x

6

65

5

5

5

54

4

4

4

43

3

3

3

32

2

2

2

21

120120)24(

2424)6(

66)2(

22)(

1)(

xxx

dx

d

dx

yd

xxx

dx

d

dx

yd

xxx

dx

d

dx

yd

xxx

dx

d

dx

yd

xxx

dx

d

dx

dy

2.4 The Chain Rule

47

48

Illustration

)unitper dollars( output respect towith

cost of change of rate

dq

dC

hour)per (units timerespect to with

output of change of rate

dt

dq

Suppose the total manufacturing cost at a certain factory is a function

of the number of units produced, which in turn is a function of the

number of hours the factory has been operating. If C, q, t, denote the

cost, units produced and time respectively, then

The product of these two rates is the rate of change of cost with

respect to time that is

hour)per (dollars dt

dq

dq

dC

dt

dC

The Chain Rule

49

If y=f(u) is a differentiable function of u and u=g(x) is

in turn a differentiable function of x, then the composite

function y=f(g(x)) is a differentiable function of x

whose derivative is given by the product

or, equivalently, by dx

du

du

dy

dx

dy

)())(( xgxgfdx

dy

Example

50

Find if dx

dy.1)2(3)2( 2232 xxy

We rewrite the function as ,

where . Thus,

13 23 uuy

22 xu

xdx

duuu

du

dy2 and 63 2

and according to the chain rule,

)2(6)2]()[2(3

2 )2)](2(6)2(3[

)2)(63(

2322

2222

2

xxxxx

xwithureplacexxx

xuudx

du

du

dy

dx

dy

Example

51

The cost of producing x units of a particular commodity

is

dollars, and the production level t hours into a particular

production run is

5343

1)( 2 xxxC

units 03.02.0)( 2 tttx

At what rate is cost changing with respect to time after 4

hours?

52

We find that

So according to the chain rule, we have

03.04.0 and 43

2 t

dt

dxx

dx

dC

)03.04.0(43

2

tx

dt

dx

dx

dC

dt

dC

When t=4, the level of production is x(4)=3.32 units,

and by substituting t=4 and x=3.32 into the formula for

, we get dt

dC1277.10]03.0)4(4.0[4)32.3(

3

2

4

tdt

dC

Thus, after 4 hours, cost is increasing at the rate of

approximately $10.13 per hour.

53

Let’s look at the functions

then we can write the function as a composition.

differentiate a composition function using the Chain Rule.

The derivative is then

54

In general, we differentiate the outside function

leaving the inside function alone and multiple all

of this by the derivative of the inside function

function insideof derivative

alonefunction inside leave

function outside of derivative

)( ))(( )( xgxgfxF

Example.

a. Based on the chain rule, find the derivative of

with respect to x

b. Find the derivative of

32 )()( xxxf

11

1)(

2

xxf

The General Power Rule

55

For any real number n and differentiable function h, we

have)]([)]([)]([ 1 xh

dx

dxhnxh

dx

d nn

Think of [h(x)]n as the composite function

nn uxhgxh g where)]([)]([

By the chain rule, we have

)]([)]([)()]([)]([)]([ 1 xhdx

dxhnxhxhgxhg

dx

dxh

dx

d nn

2.5 Marginal Analysis and Approximations Using Increments

56

57

For instance, suppose C(x) is the total cost of producing x units of

a particular commodity. If units are currently being produced,

then the derivative

Marginal Analysis

In economics, the use of the derivative to approximate

the change in a quantity that results from a 1-unit

increase in production is called marginal analysis

0x

h

xChxCxC

h

)()(lim)( 00

00

is called the marginal cost of producing units. The limiting

value that defines this derivative is approximately equal to the

difference quotient of C(x) when h=1; that is

0x

)()1(1

)()1()( 00

000 xCxC

xCxCxC

Marginal Cost

58

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

+1, assuming >>1.

0x )( 0xC

)()1( 00 xCxC

0x0x

0x

59

The marginal revenue is , it approximates

, the additional revenue generated by

producing one more unit.

)( 0xR

)()1( 00 xRxR

)( 0xP

)()1( 00 xPxP

The marginal profit is , it approximates

, the additional profit obtained by

producing one more unit, assuming >>1.

Marginal Revenue and Marginal Profit

Suppose R(x) is the revenue function generated when x

units of a particular commodity are produced, and P(x) is

the corresponding profit function, when x=x0 units are

being produced, then

0x

Example

60

A manufacturer estimates that when x units of a particular commodity

are produced, the total cost will be dollars, and

furthermore, that all x units will be sold when the price is

dollars per unit.

9838

1)( 2 xxxC

)75(3

1)( xxp

a. Find the marginal cost and the marginal revenue.

b. Use marginal cost to estimate the cost of producing the ninth unit.

c. What is the actual cost of producing the ninth unit?

d. Use marginal revenue to estimate the revenue derived from the

sale of the ninth unit.

e. What is the actual revenue derived from the sale of the ninth unit?

61

a. The marginal cost is C’(x)=(1/4)x+3. The total revenue is

, the marginal revenue is

R’(x)=25-2x/3.

b. The cost of producing the ninth unit is the change in cost as x

increases from 8 to 9 and can be estimated by the marginal cost

C’(8)=8/4+3=$5.

c. The actual cost of producing the ninth unit is C(9)-C(8)=$5.13

which is reasonably well approximated by the marginal cost C’(8)

d. The revenue obtained from the sale of the ninth unit is

approximated by the marginal revenue R’(8)=25-2(8)/3=$19.67

e. The actual revenue obtained from the sale of the ninth unit is

R(9)-R(8)=$19.33.

3/25)3/)75(()()( 2xxxxxxpxR

62

Marginal analysis is an important example

of a general incremental approximation

procedure

Approximation by Increment

63

Based on the fact that since h

xfhxfxf

h

)()(lim)( 00

00

then for small h, the derivative is approximately equal

to the difference quotient . We indicate this

approximation by writing

)( 0xf

h

xfhxf )()( 00

hxfxfhxfh

xfhxfxf )()()( ly,equivalent or,

)()()( 000

000

Or, equivalently, if , then

If f(x) is differentiable at x=x0 and △x is a small

change in x, then xxfxfxxf )()()( 000

)()( 00 xfxxff

xxff )( 0

Percentage of Change

64

The percentage of change of a quantity expresses the

change in that quantity as a percentage of its size prior to

the change. In particular,

quantity of size

quantityin change100change of percentage

Example

65

The GDP of a certain country was billion dollars t

years after 1997. Use calculus to estimate the percentage change in

the GDP during the first quarter of 2005.

2005)( 2 tttN

Use the formula

)(

)(100in change Percentage

tN

ttNN

with t=8, △t=0.25 and N’(t)=2t+5, to get

%73.1200)8(5)8(

)25.0](5)8(2[100

)8(

25.0)8(100in change Percentage

2

N

NN

Differentials

66

The differential of x is dx=△x, and if y=f(x) is a

differentiable function of x, then dy=f’(x)dx is the

differential of y.

Example

67

In each case, find the differential of y=f(x).

27)( 23 xxxfa.

b. )23)(5()( 22 xxxxf

a.

b. By the product rule,

dxxxdxxxdxxfdy )143()]2(73[)( 22

dxxxx

dxxxxxxdxxfdy

)51438(

)]23)(2()41)(5[()(

23

22

2.6 Implicit Differentiation and

Related Rates

68

69

Explicit and Implicit Forms of Functions

23

2 1 and 32

1 13 xy

x

xyxxy

So far the functions have all been given by equations of the form

y=f(x). A function in this form is said to be in explicit form.

are all functions in explicit form.

Sometimes practical problems will lead to equations in which the

function y is not written explicitly in terms of the independent

variable x.

yxyyxxyyyx 232 and 56 32332

are said to be in implicit form.

70

Example.

Find dy/dx if 322 xyyx

We are going to differentiate both sides of the given equation with

respect to x. Firstly, we temporarily replace y by f(x) and rewrite the

equation as . Secondly, we differentiate both

sides of this equation term by term with respect to x:

322 ))(()( xxfxfx

)(

2

]))([()]([

22

322

3

22

3)(2)()(

][]))(()([

xdx

d

xfdx

dxfx

dx

d

xdx

dfxfx

dx

dxf

dx

dfx

xdx

dxfxfx

dx

d

To be continued

71

Thus, we have

dx

df

xfx

xxfx

dx

df

xxfxdx

dfxfx

xxfxdx

dfxf

dx

dfx

dx

dfx

dx

dfxfxxf

dx

dfx

for solve )(2

)(23

termscombine )(23)](2[

equation theof side oneon )(23)(2

terms allgather 3)(2)2)((

2

2

22

22

22

Finally, replace f(x) by y to get

yx

xyx

dx

dy

2

232

2

Implicit Differentiation

72

Suppose an equation defines y implicitly as a

differentiable function of x. To find df/dx

1. Differentiate both sides of equation with respect to x.

remember that y is really a function of x and use the

chain rule when differentiating terms containing y.

2. Solve the differentiated equation algebraically for

dy/dx.

Example.

Find dy/dx by implicit differentiation where xyyx 33

Example

2522 yx

y

x

dx

dy

dx

dyyx 022

4

3

4

3

)4,3(

dx

dy

73

Thus, the slope at (3,4) is

The slope at (3,-4) is

Differentiating both sides of the equation with respect to x, we have

Find the slope of the tangent line to the circle at the

point (3,4). What is the slope at the point (3,-4)?

4

3

)4,3(

dx

dy

An Application

3232 yyxxQ

74

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.

75

If output is to be maintained at the current level, which is the value of Q when

x=30 and y=20, the relationship between skilled labor x and unskilled labor y is

given by the equation 3232)20,30(000,80 yyxxQ

The goal is to estimate the change in y that corresponds to a 1-unit increase in x

when x and y are related by above equation. As we know, the change in y caused

by a 1-unit increase in x can be approximated by the derivative dy/dx. Using

implicit differentiation, we have

22

2

222

222

3

26

26)3(

3260

yx

xyx

dx

dy

xyxdx

dyyx

dx

dyyxy

dx

dyxx

Now evaluate this derivative when x=30 and y=20 to conclude that

hours 14.3)20(3)30(

)20)(30(2)30(6in Change

22

2

20

30

y

xdx

dyy

Related Rates

76

In certain practical problems, x and y are related by an

equation and can be regarded as a function of a third variable t,

which often represents time. Then implicit differentiation can

be used to relate dx/dt to dy/dt. This kind of problem is said to

involve related rates.

A procedure for solving related rates problems

1. Find a formula relating the variables.

2. Use implicit differentiation to find how the rates are related.

3. Substitute any given numerical information into the

equation in step 2 to find the desired rate of change.

Example

77

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?

42753 32 qC

78

We want to find dC/dt when q=15 and dq/dt=0.2. Differentiating

the equation implicitly with respect to time, we

get 42753 32 qC

0332 2

dt

dqq

dt

dCC

so that

dt

dq

C

q

dt

dC

2

9 2

When q=15, the cost C satisfies

12014400)15(342754275)15(3 3232 CCC

and by substituting q=15, C=120 and dq/dt=0.2 into the formula

for dC/dt, we obtain

per week. dollars nd thousa6875.1)2.0()120(2

)15(9 2

dt

dC

Example

79

A storm at sea has damaged an oil rig. Oil spills from

the rupture at the constant rate of 60 , forming a

slick that is roughly circular in shape and 3 inches thick.

a. How fast is the radius of the slick increasing when the

radius is 70 feet?

b. Suppose the rupture is repaired in such a way that flow

is shut off instantaneously. If the radius of the slick is

increasing at the rate of 0.2 ft/min when the flow stops.

What is the total volume of oil that spilled onto the sea?

min/3ft

80

We can think of the slick as a cylinder of oil of radius r feet and

thickness h=3/12=0.25 feet. Such a cylinder will have volume 322 ft 25.0 rhrV

Differentiating implicitly in this equation with respect to time t,

and since dV/dt=60 at all times, we get

dt

drr

dt

drr

dt

drr

dt

dV 5.0605.0225.0

a. We want to find dr/dt when r=70, so that

ft/min 55.0)70()5.0(

60

dt

dr

b. Since dr/dt=0.2 at that instant, we have feet 191)2.0(5.0

60

r

Therefore, the total amount of oil spilled is 32 ft 28652)191(25.0 V

81

Summary

Definition of the Derivative

h

xfhxfxf

h

)()(lim)(

0

Interpretation of the Derivative

Slope as a Derivative:

The slope of the tangent line to the curve y=f(x) at the

point (c,f(c)) is

Instantaneous Rate of Change as a Derivative:

The rate of change of f(x) with respect to x when x=c is

given by f’(c)

)(tan cfm

82

Sign of The Derivative

If the function f is differentiable at x=c, then

f is increasing at x=c if >0)(cf

f is decreasing at x=c if <0)(cf

Techniques of Differentiation

0cdx

d 1][ nn nxxdx

d )()( xfdx

dcxcf

dx

d

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

dxf

dx

dxgxf

dx

d

)]([)()]([)()()( xfdx

dxgxg

dx

dxfxgxf

dx

d The Product Rule

0)( if )(

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

])(

)([

2

xgxg

xgdx

dxfxf

dx

dxg

xg

xf

dx

dThe Quotient Rule

83

The Chain Rule

dx

du

du

dy

dx

dy )())(( xgxgf

dx

dy

)]([)]([)]([ 1 xhdx

dxhnxh

dx

d nn The General Power Rule

The Higher -order Derivative

Applications of Derivative

Tangent line, Rectilinear Motion

)(or 2

2

xfdx

yd

)(or )( xfdx

yd n

n

n

The Second Derivative

The nth Derivative

84

The marginal cost is , it approximates ,

the additional cost generated by producing one more unit.

)( 0xC )()1( 00 xCxC

Marginal Analysis and Approximation by Increments

h

xChxCxCxCxC

xCxC

h

)()(lim)()()1(

1

)()1( 00

0000

00

xxfxfxxf )()()( 000Approximation by Increment

Marginal Revenue Marginal Profit

Implicit Differentiation and Related Rate