高階導(dǎo)數(shù)與高階微分

設(shè)y

f

(x)在(a,b)內(nèi)可導(dǎo),則它的導(dǎo)函數(shù)y

f

(x)和微分函數(shù)dy

df

(x)

f

(x)dx仍然是(a,

b)上的函數(shù),因此可以繼續(xù) 它們的可導(dǎo)性與可微性,

這就產(chǎn)生了高階與高階微分.一、高階導(dǎo)數(shù)00dx2

dx2d

2

y

d

2

f,

或

表示f

(x)在x

的二階導(dǎo)數(shù).

用f

(x

)或y

,

或x

x0

x

x0x

x00點x0

處二階可導(dǎo),且稱y

f

(x)在點x0

處的導(dǎo)數(shù)為函數(shù)f

(x)在點x0

的二階導(dǎo)數(shù).x定義3.4:如果y

f

(x)在x

點處可導(dǎo),即:limf

(x

0

0

x)

f

(x

)存在,則稱y

f

(x)在x0xlimx00f

(x

x)

f

(x0

)0即:

f

(x

)

3.5、高階導(dǎo)數(shù)與高階微分2018/11/5Edited

by

Lin

Guojian1y(n)

f

(n)

(x)

lim

f

(n1)

(x

x)

f

(n1)

(x)xx0(n)x0

f

(x0

)

limf

(n1)

(x0

x)

f

(n1)

(x

)0x(n)由定義3.4知:y0x

x通常稱二階及二階以上的導(dǎo)數(shù)為高階導(dǎo)數(shù).(n1)

(x)]

df

(n1)

(x)

,

n

1,2,3,,2018/11/5Edited

by

Lin

Guojian2(

n

)即

:

f

(x)

[

fdx類似地可以定義三階導(dǎo)數(shù)f

(x0

),四階導(dǎo)數(shù)f

(4)(x

)及n階導(dǎo)數(shù)f

(n)(x

).0

0x

x0

lim

0

.f

(n1)

(x)

f

(n1)

(x

)x

x0例:

設(shè)y

f

(x)

xex

,

求y.解:

y

(xex

)

ex

xex

2ex

xex

.

xexy

(ex

xex

)

ex

ex例:設(shè)y

arctan

x,求y

,y.1

x2解:y

1

(1

x2)221

x

y

1

2x

..2018/11/5Edited

by

Lin

Guojian3(1

x

)2

32(3x2

1)2

2(1

x

)y

2x2018/11/5Edited

by

Lin

Guojian4例:求y

xk

(k為正整數(shù))的n階導(dǎo)數(shù)y(n).解:y

(xk

)

kxk

1y

(kxk

1

)

k

(k

1)xk

2y

[k

(k

1)xk

2

]

k

(k

1)(k

2)xk

3y(k

1)

k(k

1)(k

2)3

2

xy(k

)

k(k

1)(k

2)3

2

1

k!y(k

1)

0,

y(k

2)

0,,

y(k

i

)

0(i

1).0k(n)故y(n)

k

(k

1)(k

n

1)xk

n

,

n

kn

k

x

2018/11/5Edited

by

Lin

Guojian5例:設(shè)y

2x4

4x3

7x2

8x

1,求y(4).解:y(4)

(2x4

4x3

7x2

8x

1)(4)

2

4!

48.例:求n次多項式x

an

,(a0

0)的n階導(dǎo)數(shù).n1y

a xn

a

xn1

a0

1解:y(n)

a

n!0例:設(shè)y

ex

,求y(n)與y(n)(0).解:

y

(ex

)

ex

,

y

(ex

)

ex,,

y(n)

ex

.2018/11/5Edited

by

Lin

Guojian6y(n)

(0)

y(n)

ex

e0

1.x0

x02018/11/5Edited

by

Lin

Guojian7例:設(shè)y

sin

x,求y(n).解:y

(sin

x)

cos

x

sin(x

),22y

(cos

x)

sin

x

sin(x

)

sin(x

2

).設(shè)y(k

)

(sin

x)(k

)

sin(x

k

).2則y(k

1)

(sin

x)(k

1)

[sin(x

k

)]

cos(x

k

)2

2

sin(x

k

)

sin[x

(k

1)

].2

2

2由數(shù)學(xué)歸納法知:(sin

x)(n)

sin[x

n

],(n

1,2,3,).22018/11/5Edited

by

Lin

Guojian82

y

(sin

x)

cos

x

cos(x

)

cos(x

2

)2例:設(shè)y

cos

x,求y(n).解:y

(cos

x)

sin

x

cos(x

)設(shè)y(k

)

(cos

x)(k

)

cos(x

k

)2則y(k

1)

(cos

x)(k

1)

[cos(x

k

)]

sin(x

k

)2

2

cos(x

k

)

cos[x

(k

1)

].2

2

2由數(shù)學(xué)歸納法知:(cos

x)(n)

cos[x

n

],(n

1,2,3,).22018/11/5Edited

by

Lin

Guojian9解:y

[x1

]

x2

.y

[(x2

]

(1)(2)

x3.y

[(1)(2)

x3

]

(1)(2)(3)

x4

.設(shè)y(k

)

(1)(2)(k

)

xk

1

(1)k

k!x(k

1).則y(k

1)[(1)k

k!x(k

1)]

(1)k

k![(k1)]

x(k

2)

(1)k

1

(k

1)!x(k

2).由數(shù)學(xué)歸納法知:y(n)

(x1

)(n)

(1)n

n!x(n1)

,

(n

1,2,).例:設(shè)y

x1

,求y(n).例:設(shè)y

ln(1

x),求y(n).y

[ln(1

x)]

[(1

x)1

]

(1)(1

x)2y

[ln(1

x)]

[(1)(1

x)2

]

(1)(2)(1

x)3設(shè)y(k

)

[ln(1

x)](k

)

(1)(2)(k

1)(1

x)k

(1)k

1

(k

1)!(1

x)k則y(k

1)

[ln(1

x)](k

1)

[(1)k

1(k

1)!(1

x)k

]

(1)k

1(k

1)!(k)(1

x)k

1

(1)k

(k)!(1

x)(k

1).由數(shù)學(xué)歸納法知:y(n)(1)n1

(n

1)!(1

x)n

,(n

1,2,3,).

(1

x)12018/11/5Edited

by

Lin

Guojian1011

x解:y

[ln(1

x)]

例:設(shè)y

ln(x2

3x

4),求y(n

).1

12018/11/5Edited

by

Lin

Guojian11

.x

1

x

4解:y

ln(x

1)

ln(x

4)

y

y(n)

(1)(2)[(n

1)](

x

1)n

(1)(2)[(n

1)](

x

4)n

(1)n1

(n

1)!(x

1)n

(1)n1

(n

1)!(x

4)n

.2

2

1

1

1

1

y

x

1

x

4

x

1

x

4

(1)(x

1)

(1)(x

4)

.y

(1)(2)(

x

1)3

(1)(2)(

x

4)3.求函數(shù)的高階導(dǎo)數(shù)常用以下兩個公式:0n注

:

(a)

:

規(guī)定

:

0!

1,C

1.(1)

:

[

f(x)

g(x)](

n

)

f

(n)(x)

g(n)(x);(2)

:

(公式)

:

[

f(x)

g(x)](

n)

(n

k

)(x),n

k

(k

)nk

0C

f

(x)gf(x).,

f

(0)(x)

n!k!(n

k)!n其中,Ck

(b)

:

如果f(x)

xk

,則當(dāng)n

k時,

f

(n)(x)

0,此時應(yīng)用

公式計算[

f(x)

g(x)](n)時可能有很多項為零.2018/11/5Edited

by

Lin

Guojian122018/11/5Edited

by

Lin

Guojian13例

:

設(shè)y

x3e

x

,求y(30).解:[x3e

x

](30)3030k

f

(k

)(x)g(30

k

)(x)Ck

0

C

0

f

(0)(x)g(30)(x)

C1

f

(x)g(29)(x)

C

2

f

(x)g(28)(x)

C3

f

(x)g(27)(x)30

30

30

3030

30

C

4f

(4)(x)g(26)(x)

C30

f

(30)(x)g(0)(x).由于

:

(x3

)(k

)

0,

k

3,且(e

x

)(

n

)

(1)n

e

x

.因此:[x3e

x

](30)

C

0

f

(0)(x)g(30)(x)

C1

f

(x)g(29)(x)30

30

C

2

f

(x)g(28)(x)

C

3

f

(x)g(27)(x)30

30

C1

3x

2(1)e

x

C

2

6xe

x

C3

6(1)e

x30

30

30

C

0

x3e

x30

90x2e

x

2610xe

x

24360e

x

.

x3e

x例

:

求y

arctan

x在x

0點的n階導(dǎo)數(shù)..(1

x2

)3

2

6x2,

y

1

2x,

y

解

:

y

1

x2

(1

x2

)2在(1

x2

)y

1方程兩邊對x求n

1階導(dǎo)數(shù),即(1

x2

)y(n1)2018/11/5Edited

by

Lin

Guojian14

1(n1).

(1

x2

)y

1.1

x21由y

n

1k

0(n

1

k

)2

(k

)2x

)

(

y)那么:(1(n1)

x

)y

kn

1C

(1

(1

x2

)y(n)

(n

1)2xy(n

1)

(n

1)(n

2)y(n2)

0.2

(1)10 2

(0)(n

3)n

1(n

2)2 2

(2)n

1

C

(1

x

) (

y

)

C(1

x

)

(

y

)(n

1)n

1

C

(1

x

)

(

y

)2018/11/5Edited

by

Lin

Guojian15令x

0可得

:y(n)(0)

(n

1)(n

2)y(n2)(0)

0

y(n)(0)

(n

1)(n

2)y(n2)(0).由于y(0)

1,y(0)

0,

因此,當(dāng)n為偶數(shù),即n

2k，k

1,2,.則y(4)(0)

(4

1（)

4

2)y(2)(0)

0,y(6)(0)

(6

1（)

6

2)y(4)(0)

0,

,

y(2k

)(0)

(2k

1（)

2k

2)y(2k

2)(0)

0.當(dāng)n為奇數(shù),即n

2k

1，k

1,2,.y(2k

1)(0)

(2k

2（)

2k

3)y(2k

3)(0)

(2k

2（)

2k

3)(2k

4（)

2k

5)y(2k

5)(0)

(2k

2（)

2k

3)(2k

4（)

2k

5)(2k

6)(2k

7)y(2k

7)(0)

(1)k

1(2k

2（)

2k

3)(2k

4（)

2k

5)(2k

6)(2k

7)

2

1y(0)

(1)k

1(2k

2)!.2018/11/5Edited

by

Lin

Guojian16二、高階微分定義3.5

:

若y

f

(x)的微分函數(shù)dy

df

(x)

f

(x)dx關(guān)于x可微,則稱y

f

(x)關(guān)于x二階可微,且稱dy

f

(x)dx關(guān)于x的微分為y

f

(x)的二階微分.用d

2

y(或d

2

f

(x))表示y

f

(x)的二階微分.類似地可以定義n階微分d

n

y(或d

n

f

(x)).如果記(dx)n

dxn

.由定義3.5知:d

2

y

d

(dy)

d[

f(x)dx]

dx

d[

f

(x)]

dx

f

(x)dx

f

(x)[dx]2

f

(x)dx2

.設(shè)

:

d

k

y

f

(

k

)

(x)dxk

,則:

d

k

1

y

d

(d

(

k

)

y)

d[

f

(

k

)

(

x)

dxk

]

dxk

f

(k

1)

(x)dxk

1

d[

f

(

k

)

(x)]由數(shù)學(xué)歸納法知:d

n

y

f

(

n

)

(x)dxn

.從而高階導(dǎo)數(shù)可用高階微分定義:通常稱二階及二階以上的微分為高階微分.f

(x)

2018/11/5Edited

by

Lin

Guojian17(n),

f

(x)

.dx2d

2

ydxndn

y2018/11/5Edited

by

Lin

Guojian18例:

設(shè)y

f

(x)

x

sin

x,

求d

2y.解:dy

(x

sin

x)dx

(sin

x

x

cos

x)dx.d

2

y

d[(sin

x

x

cos

x)dx]

dx

d

(sin

x

x

cos

x)

dx

(cos

x

cos

x

x

sin

x)dx

(cos

x

cos

x

x

sin

x)(dx)2

(cos

x

cos

x

x

sin

x)dx2

.例:設(shè)y

x2

,求d

2

y,d

3

y.解:

dy

2xdx

d

2

y

2dx2

.d

3

y

0.2018/11/5Edited

by

Lin

Guojian19注:高階微分沒有形式不變性.事實上:由dy

df

(x)

f

(x)dx.當(dāng)x是自變量時,上式兩邊關(guān)于x求微分,這時dx相對于x是常數(shù).故d

2

y

d

(dy)

d[f

(x)dx]

dx

d[f

(x)]

dx

f

(x)dx

f

(x)dx2

.當(dāng)x不是自變量時,這時x依賴于另一個變量t,即x

x(t).則d

2

y

d

(dy)

d[f

(x)dx]

d[f

(x)]dx

f

(x)d

(dx)

f

(x)

dx

dx

f

(x)d

(dx)

f

(x)dx2

f

(x)d

(dx).這時d

(dx)未必為零.

因為x與dx都是自變量t的函數(shù),且d

(dx)

d

2

x是x關(guān)于自變量t的二階微分.2018/11/5Edited

by

Lin

Guojian20例:設(shè)y

x2

,x

t

2

,求d

2

y.解：由于y

t4

,故dy

4t3dt.故d2

y

d(dy)

d(4t3dt)12t2dt

dt

12t2dt2.另解：由于d

2

y

f

(x)dx2

f

(x)d

2

x

2dx2

2xd2

x而d2x

d(dx)

d(2tdt)

2dtdt

2(dt)2

2dt2,故2xd2

x

2t2

2dt2

4t2dt2.從而d2

y

2dx2

2xd2

x

8t2dt2

4t2dt2

12t2dt2.注：因為x不是自變量,故不能用d

2y

f

(x)dx2求解d

2

y.實際上：d2

y

f

(x)dx2

2dx2

2(dx)2

2[2tdt]2

8t2

(dt)2

8t2dt2

12t2dt2.例:

設(shè)y

y(x)是由方程ey

xy所確定的隱函數(shù),

求y.解:對方程ey

xy兩邊關(guān)于x求導(dǎo),.ey

xy有ey

y

y

xy

y

1)(ey

x)

y(ey

yyyyy(ey

x)2y(ey

x)2

x)

y(e

y

1)

e

x

e

x

有y

y

(e兩邊關(guān)于x求導(dǎo),再對y

yey

x.2018/11/5Edited

by

Lin

Guojian21

2y(ey

x)

y2ey

2xy

2

2xy

xy3(ey

x)3

(xy

x)3.例

:

求由方程x

y

sin

y

0所確定的隱函數(shù)的二階導(dǎo)數(shù)dx2d

2

y.11

cos

y1

y

y

cos

y

0,

y

解

:

在x

y

sin

y

0的兩邊對x求導(dǎo)數(shù),則(1

cos

y)2y

sin

y的兩邊對x求導(dǎo)數(shù),則1

cos

y1在y

y

.2018/11/5E