Chapter 3Chapter 3

alsDifferenti and sDerivative

Derivative 导数导数

Derivable 可导的可导的

Derivability 可导性可导性

One-sided derivative 单侧导数单侧导数

Left-hand derivative 左导数左导数

Right-hand derivative 右导数右导数

Secant line 割线 割线 Tangent line 切线切线

Instantaneous rate of change 瞬时变化率 瞬时变化率

Vocabulary

微积分的创立

微积分 (Calculus) 是微分学 (Differential calculus)和积分学 (Integral calculus) 的总称， 它是由牛顿与莱布尼兹在研究物理和几何的过程中总结前人的经验，于十七世纪后期建立起来的。

牛顿 [ 英国 ] (Isaac Newton) 1642—1727

莱布尼兹 [ 德国 ] (G.W.Leibniz) 1646—1716

关于牛顿

牛顿的三大成就： 流数术（微积分） 万有引力定律 光学分析的基本思想

我不知道在别人看来，我是什么样的人；但在我自己看来，我不过就象是一个在海滨玩耍的小孩，为不时发现比寻常更为光滑的一块卵石或比寻常更为美丽的一片贝壳而沾沾自喜，而对于展现在我面前的浩瀚的的真理的海洋，却全然没有发现。 —— 牛顿

1. IntroductionOne of central ideas of calculus is the notion of derivative. The derivative originated from a problem in geometry, that is, the problem of finding the tangent line at a point of a curve. It was soon found that it also provides a way to calculate velocity and, more generally, the rate of change of a function. These problems contain the essential features of the derivative concept and may help to motivate the general definition of derivative which is given in this section.

(1) The problem of finding the tangent line at a point of a curve

See figureP

)(xfy

x

y

o

0P

0x xx 0

M

? curve a of point

at line tangent the

ofequation find To

0

xfy

P

is and endpoints with interval over the

respect to with of change of rate average The

00 xxx

xy

Solution

xxfxxf

xy

)()( 00

. of ),( and ),(

points thejoining linesecant theof slope theis This

00000 fyyxxPyxP

is

),(at )( of line tangent theof slope The 000 yxPxfy

x

xfxxfk

x

)()(limtan 00

0

00

0

is curve a

of point at line tangent theofequation theThus,

xxkyy

xfy

P

is ],[ interval the

over respect to with of speed average theSince,

00 ttt

ttf

t

tfttf

t

f

)()(

interval timeoflength

interval timeduring distancein change

00

(2) The problem involving velocity

To find speed of an object moving on a line whose distant at time t is given by f(t)?

Solution

is at speed ousinstantane theThus, 0tt

t

tfttf

t

ftv

tt

)()(limlim)( 00

000

2. The Definition of Derivatives

0at function a of derivative The (1) xx

Definition 1

if , containsthat

intervalopen an in definedfunction a be )(Let

0x

xfy

x

xfxxf

x

yxx

)()(limlim 00

00

Notes

0

derivablenot is then exist, doeslimit thisIf a)

x

x f

.at derivable

be tosaid is function The . denoted

is and at of derivative thecalled isit exists,

0

0

0

x

xfxf

xxf

)( then infinity, islimit thisIf b) 0xf

.or ,|or ,

denotedcommonly also is )( derivative The c)

0

0

0

0

xxxx

xx dx

dfy

dx

dy

xf

x

xfxxfxf

x

)()(lim)( d) 00

00

0

0 )()(lim

0

0

xx

xfxfxx

xxx

h

xfhxfh

xh )()(lim 00

0

(2) The derivative of a function defined on an open interval I

Definition 2

is at

derivative theand . offunction derivable a be tosaid

is then ,point every at derivable is

if ,open an in definedfunction a be )(Let

xxf

I

xfIxxf

Ixfy

x

xfxxf

dx

dyxf

x

)()(lim)(

0

Note:

0)()(

then, and ,on derivable is If a)

0

0

xxxfxf

IxIxf

(3) One-sided derivative

Left-hand derivative

x

xfxxf

xy

xfxx

)()(limlim)( 00

000

Right-hand derivative

x

xfxxf

xy

xfxx

)()(limlim)( 00

000

Remarks:

)()(exists )( a) 000 xfxfxf

.at derivablenot is then exist,not does

and of oneor , If b)

0

0000

xxf

xfxfxfxf

Example 1

?limits

following in the isat explain wh toderivative

of definition theuse exists, that Suppose 0

A

xf

A

x

xfxxfx

00

0lim 1

exists 0 and ,00 where,lim 20

ffAx

xfx

A

h

hxfhxfh

00

0lim 3

constants. are

and where,lim 4 00

0

Ah

hxfhxfh

Solution: According to the definition of derivative

000

0

00

0

lim

lim 1

xfx

xfxxfx

xfxxfA

x

x

00

0limlim 2

00f

x

fxf

x

xfA

xx

000

0000

0

00

0

2

lim

lim 3

xfxfxf

h

xfhxf

h

xfhxfh

hxfhxfA

h

h

0

00

0

00

0

00

0

limlim

lim 4

xf

h

xfhxf

h

xfhxfh

hxfhxfA

hh

h

The above example illustrate the concept of the derivative.

3. Examples for Finding Derivatives

)( find then ,)( If 1. Example xfCxf

Solution

Using definition of derivative, we have at any point x

x

xfxxfxf

x

)()(lim)(

00lim

0

xCC

x

0)( Therefore C

)( find then ,)( If 2. Example xfxxf n Solution

By definition of the derivative,

x

xxx

x

xfxxfxf

nn

xx

)(lim

)()(lim)(

00

x

xxxxCxxCx nnnn

nn

n

x

)()(lim

22211

0

x

xxxCxxC nnn

nn

x

)()(lim

22211

0

1)( Therefore nn nxx

])([lim 12211

0

nn

nn

nx

xxxCxC

111 nnn nxxC

Note

1|))(1( nax

n nax

)( ))(2( 1 Rax

x

xxx

x

xfxxfxf

xx

cos)cos(lim

)()(lim)(

00

x

xxx

x

2

sin)2

sin(2lim

0 x

xx

xx

2

sin)

2sin(2lim

0

xsin

xx sin)(cos Therefore

xx cos)(sin Similarly

)( find then ,cos)( If 3. Example xfxxf

Solution By definition of the derivative,

)( find then ,)( If 4. Example xfaxf x

Solution

x

xfxxfxf

x

)()(lim)(

0 xaa xxx

x

0lim

xaa xx

x

)1(lim

0 xe

aax

x

x

1lim

ln

0

xax

ax

x

lnlim

0aa x ln

,ln)( Therefore aaa xx xx ee )(

By definition of the derivative,

1,0log)( if )( Find 5. Example aaxxfxf a

Solution

x

h

hh

xhxx a

h

aa

ha 1log

1lim

logloglimlog

00

By definition of the derivative,

x

x

h

ha

h

ha x

h

x

h

11

0

1

01limlog1limlog

axe

xe a

xa ln

1log

1log

1

.1

ln andx

x

.)( if )0( Find 6. Example xxff

Solution

x

x

x

x

x

fxf ||

0

0||

0

)0()( Since

1lim)0()(

lim)0( Thus,00

x

x

x

fxff

xx

1lim)0()(

lim)0( and00

xx

xfxf

fxx

exist.not does )0( Hence, f

4. Geometric Interpretation of the Derivative

is that ),,(

at )( of line tangent theof slope theis

at respect to with of derivative theSince

00

0

yxP

xfyx

xxfy

x

xfxxfxfk

x

)()(lim)( 00

00

Then equation of tangent line is

))(( 000 xxxfyy

and equation of normal line is

)()(

10

00 xx

xfyy

used. islimit sided-one eappropriat

then the, ofdomain theofendpoint an is If 1 0 fx

Remark:

Example 7

. of 9,3 and 1,1 points thejoining line

secant the toparallel is at lineangent in which t

,at of line tangent theofequation theFind

2

0

02

xyQP

x

xxy

Solution

By definition of the derivative, we have

00 2)( xxy

The slope of the secant line PQ is 41319

k

,2 thus,

,42 have we,assumptiongiven Under

0

0

x

x

4 and 0 y

44

is,that

),2(44 is

2at of line tangent ofequation theThus, 0

xy

xy

xxf

5. The Derivative and Continuity

Theorem 1

.at continuous isit then ,at derivable is If 00 xxf

Proof

have we,derivative theof definition

theusing ,at derivable is that Suppose 0xxf

)]()([lim 00

xfxfxx

)()()(

lim 00

0

0

xxxx

xfxfxx

)(lim)()(

lim 00

0

00

xxxx

xfxfxxxx

00)( 0 xf

.at continuous

is is, that ),()(lim Hence,

0

00

x

xfxfxfxx

The converse of theorem 1 is not true, that is, a continuous function needs not always be derivable.

Note:

0.at derivablenot isit show,

6 example asbut 0,at continuous is function The x

Solution

.0at 0for

0for 1

cos)(

ofty derivabili and continuity Discuss

2

x

xx

xx

xxf

Example 8

and ,0)0( f

01

coslim)(lim)00( 2

00

xxxff

xx

0lim)(lim)00(00

xxff

xx

0.at

continuous is is, that ),0(0)(lim Hence0

xffxfx

But we have

,10

lim0

)0()(lim)0(

00

x

xx

fxff

xx

01

coslim

01

coslim

0

)0()(lim)0(

0

2

00

xx

xx

x

x

fxff

x

xx

)0()0( Thus, ff

0.at derivable

not is Hence exist.not does )0( is,that xff

Example 9

.1at derivable and continuous is

1for

1for )( if , and Find

2

x

xbax

xxxfba

Solution

1lim)(lim)01( 2

11

xxff

xx

babaxxffxx

)(lim)(lim)01(11

Also .1,1at continous is Since baxxf

ax

aaxx

bax

x

fxff

x

ba

xx

1lim

1

1lim

1

)1()(lim)1(

1

1

11

and ,1)1( f

.1,2 Thus,

).1()1(,1at derivative is Since

.2)1(lim1

1lim

1

)1()(lim)1(

1

2

11

ba

ffxxf

xx

x

x

fxff

x

xx

Example 10

?2

at

derivable and continuous is |cos|)(Wether

x

xxf

Solution and ,0|2

cos|)2

(

f

02

cos)cos(lim|cos|lim)(lim

222

xxxfxxx

02

coscoslim|cos|lim)(lim

222

xxxfxxx

But .2

at

continuous is thus,),2

(0)(lim 2

x

xffxfx

2

coslim

2

)2

()(lim)

2(

22

x

x

x

fxff

xx

1sin

lim)

2cos(

lim00

2

tt

t

t

tt

tx

2

coslim

2

)2

()(lim)

2(

22

x

x

x

fxff

xx

1sin

lim)

2cos(

lim00

2

tt

t

t

tt

tx

exist.not does )2

( thus,),2

()2

(

fff

.2

at derivablenot is Thus,

xxf

Example 11

?0 find

,at derivable and ,on defined

is where),()()(Let

f

ax

xbxabxaxf

Solution By definition of the derivative,

xfxf

fx

)0()(lim)0(

0

x

bxabxax

)()(lim

0

)()(lim

)()(lim

00 xabxa

xabxa

xx

bxabxa

bbx

abxab

xx

)()(lim

)()(lim

00

)(2 ab

Example 12

. find ,11 and

,0,0any for

and ,,0on defined is that Suppose

xff

yxyfxfxyf

xf

Solution

and ,01 is, that ,121obtain we

, from 1 Take

fff

yfxfxyfyx

x

xfxxfxf

x

)()(lim

0

.11

11

)1(1lim

)(1

lim

0

0

xxf

xxx

fxx

f

x

xfxx

xf

x

x