1chapter25導數(shù)與微分習題課

Chapter

2(5)導數(shù)與微分(習題課)一.定義一階導數(shù)的定義高階導數(shù)的定義微分的定義二.基本公式基本初等函數(shù)的導數(shù)公式高階導數(shù)公式基本初等函數(shù)的微分公式三.求導法則函數(shù)的和、差、積、商的求導法則;反函數(shù)的求導法則;復合函數(shù)的求導法則;隱函數(shù)求導法則;對數(shù)求導法;參變量函數(shù)的求導法則;分段函數(shù)的求導法.四.函數(shù)極限存在、函數(shù)連續(xù)、函數(shù)可導、函數(shù)可微的關系函數(shù)極限存在函數(shù)連續(xù)函數(shù)可導函數(shù)可微五.常見題型1.

求導數(shù)用定義求導數(shù)復合函數(shù)求導數(shù)分段函數(shù)的導數(shù)表達式中含有絕對值或最值的函數(shù)的導數(shù)參變量函數(shù)的導數(shù)隱函數(shù)的導數(shù)表達式中含有冪指函數(shù)的導數(shù)求高階導數(shù)求微分函數(shù)連續(xù)性與可導性的討論利用導數(shù)定義證明結(jié)論成立求待定參數(shù)求曲線的切線與法線習題2.1/7,P120f

(0)

=

0,xfi

0-xfi

0-2=

0,f

(0

-

0)

=

lim

f

(

x)

=

lim

1

-

cos

ax

=

limx(ax)2xfi

0-xxln(b

+

x2

)f

(0

+

0)

=

lim

f

(

x)

=

limxfi

0+

xfi

0+由可導必連續(xù),有:f

(0

-0)=f

(0

+0)=f

(0),xln(b

+

x2

)從而,

limxfi

0+=

0,

b

=

1.xx1

(1

-

cos

ax)

-

0xfi

0--又f

￠(0)

=

lim

f

(

x)

-

f

(0)

=

lim

x21

-

cos

ax=

limxfi

0-

x2x2xfi

0-(ax)2=

limxfi

0-2a2=

,xx1

ln(1

+

x2

)

-

0f

(

x)

-

f

(0)

=

lim

xf

￠(0)

=

limxfi

0+xfi

0++2=

1,ln(1

+

x2

)=

limxfi

0+

x+-f

(0),由可導,有:

f

(0)

=2a

=

1,

a

=

–

2.2習題2.1/9,P120x

-

a x

-

axfi

axfi

a(1)

f

￠(a)

=

lim

f

(

x)

-

f

(a)

=

lim

(

x

-

a)j

(

x)=

lim

j

(

x)

=

j

(a).xfi

a\f

(x)在x

=a處可導.x

-

a x

-

axfi

a-xfi

a--(2)g￠(a)

=

lim

g(

x)

-

g(a)

=

lim

|

x

-

a

|

j

(

x)x

-

a x

-

axfi

a+xfi

a++=

lim

-j

(

x)

=

-j

(a),xfi

a-g￠(a)

=

lim

g(

x)

-

g(a)

=

lim

|

x

-

a

|

j

(

x)=

lim

j

(

x)

=

j

(a),xfi

a+\當j

(a)=0時,g(x)在x

=a處可導,且g

(a)=0.當j

(a)?0時,g(x)在x

=a處不可導.Dx習題2.1/10,P120(2)

f

(-

x)

=

-

f

(

x),f

￠(-

x)

=

lim

f

(-

x

+

Dx)

-

f

(-

x)Dxfi

0Dx=

lim

f

[-(

x

-

Dx)]

-

f

(-

x)Dxfi

0Dx=

lim

-

f

(

x

-

Dx)

+

f

(

x)Dxfi

0-

DxDxfi

0=

lim

f

(

x

-

Dx)

-

f

(

x)

=

f

(

x).習題2.3/4,P134

f

(

x0

)

=

j

(

x0

),f

(

x0

-

0)

=

lim

j

(

x)

=

j

(

x0

)xfi

x0f

(

x0

+

0)

=

lim

[a(

x

-

x0

)2

+

b(

x

-

x0

)

+

c]

=

cxfi

x0由f

(x0

+0)=f

(x0

-0)=f

(x0

)得,c

=j

(x0

)0000000￠=

j

(

x

)x

-

xj

(

x)

-j

(

x

)=

limx

-

xf

(

x)

-

f

(

x

)xfi

x-xfi

x

--

0又f

￠(

x

)

=

lim00x

-

xf

(

x)

-

f

(

x0

)f

￠(

x

)

=

limxfi

x++

0=

bx

-

xxfi

x+0=

lim0a(

x

-

x0

)2

+

b(

x

-

x0

)

+

c

-j

(

x0

)由f+(x0

)=f-(x0

)得,b

=j

(x0

)

0000

2a(

x

-

x

)

+

b,

x

>

xx

=

xx

<

x0\

f

￠(

x)

=

j

￠(

x

),

j

(

x),0000000￠=

j

(

x

)x

-

xj

(

x)

-j

(

x

)=

limx

-

xf

(

x)

-

f

(

x

)f

￠(

x

)

=

limxfi

x-xfi

x--

000x

-

xf

(

x)

-

f

(

x0

)f

￠(

x

)

=

limxfi

x++

000x

-

xxfi

x+=

lim

2a(

x

-

x0

)

+

b

-j

(

x0

)

=

2a20+

0

-

0由f

￠(

x

)

=

f

￠(

x

)得,a

=

1j￠(

x

)

習題2.4/9,P147利用一階微分形式不變性，有

dx

=

6tdt

+

2dte

ydy

sin

t

+

e

y

cos

tdt

-

dy

=

0

dx

dt

=6t

+

21從而

dte

y

cos

t1

-

e

y

sin

t

dy

=e

y

cos

t\

dy

=

dx.(1

-

e

y

sin

t

)(6t

+

2)習題2/2(6),P157=

0,xxx2

cos

1

-

0xxfi

0xfi

0

g￠(0)

=

lim

g(

x)

-

g(0)

=

lim=

f

￠[

g(0)]

g￠(0)

=

0.dx\

d

{

f

[

g(

x)]}x=02

,1-1=

ex2ln

cos

x=

lim

exfi

0+=

lim

(cos

x)

x2xfi

0+習題2/7,P157

lim

(cos

x)x-2xfi

0+lim

(cos

x)x-2

=

0,xfi

0+而f

(0)=b,f

(0

+

0)

=

lim

x(cos

x)x-2

=

lim

xxfi

0+

xfi

0+f

(0

-

0)

=

lim

(ax

+

b)

=

b,xfi

0-由f

(x)連續(xù)有,f

(0

+0)=f

(0

-0)=f

(0),\

b

=

0;xxf

(

x)

-

f

(0)x(cos

x)x-2

-

0=

limxfi

0+xfi

0+又f+￠(0)=lim2

,-1x-2=

lim

(cos

x)

=

exfi

0+x

xf

￠(0)

=

lim

f

(

x)

-

f

(0)

=

lim

ax

-

0

=

a,xfi

0-xfi

0--由f

(x)可導有,f+(0)=f-(0),2

.-1\

a

=

e習題2/10,P1583-12y

=

(1

+

x)3

-

(1

+

x)3

](n)2

-1y(n)

=

[(1

+

x)3

](n)

-[(1

+

x)=2-n-

n

+

1)(1

+

x)32323(

-

1)(3

32

2-

2)

(11

1

1-

(-

1)(-

-

1)(-

-

2)

(-

-

n

+

1)(1

+

x)-3-n3

3

3

3=2-n)(1

+

x)333n

-

53

3

32

1

4(-

)(-

)

(-13

3

3-

(-

1)(-

4)

(-

3n

-

2)(1

+

x)-3-n=1+n3n

(1

+

x)3(-1)n-11

4

7

(3n

-

5)(3n

+

2

x)習題2/11,P15821

-

u2令u

=

e-

x

,

則y

=

uarcsin

u

+

1

ln(1

-

u2

),2)23

,23dudy

arcsin

u2arcsin

e-

x(1

-

e-2

x=(1

-

u2

)2由于

=dxdu

=

-2

xe-

x2

,.3)22dy

dy

dudx

du

dx2

22

xe-

x

arcsin

e-

x(1

-

e-2

x=

-=所以習題2/15,P158

f

(

x

+

5)

=

f