chapter 3. 导数 derivative 导数 可导的 derivable 可导的 可导性 derivability 可导性...
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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
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