课件引例变速直线运动

高阶导高阶导即引例：变速直线运动即

v a

(s)

2定义如果

x)的导

在点x处可,(f(

f(x

x)

f(x)

fx))

x)x d2

d2

(x)记作fxy

dx2

dx2二阶导数的导数称为

d3

d3

(x) (

y,dx3,

dx3三阶导数的导数称为(4)

(4)

d4

d4f(x) (x),

,dx4

dx4

的n阶导数的导数称函 n阶导数,记(n)

(n)

dn

dnf(x) (

,dxn

dxn 二阶和二阶以上的导数统称

fx)

fx)D(a,b)Dn(a,b)Cn(a,b)

(a,b)内全体可导函数的集合(a,b)内全体有n阶导函数的集合;(ab)内全体n阶导函数连续的集合.5直接由高阶导数的定义逐步求高阶导y

f(x)

f解y

y

)

2y

1

x22 x2)2

1x

x2)22(1

x2

(2x)2(1

x2)(2

2(3x

f(0)

2

x2

2(3x

x2)3

x2

x

(0)

x2

x例yx(R求y(n解

x(x1)

1)x2y

1)x2

2)x3y(n)

n1)xn若为自然n,y(n

(xn)(n)

y(

例yax求y(n解

axax

lnln2y

ax

ln3 (ax)(n)

ax

lnn

(ex)(

ex11

ln(a

x)(x

求y(n)解y

a

(a

x)1,

y

(a

x)2

(a

x)2y

y(4) (ax)3

(ax)4

x)](

(n

(n

0!y

1a

(a,求y(n

x)n

a

(a

x)n1

sinkx求y(ny

kcos

ksin(kx2y

k2cos(kx

)k2

sin(kx

)

k2sin(kx

22y

k3cos(kx

2)

k3sin(kx

3 y(

kn

sin(kx

n2

即(sin

kn

sin(kx

n2

kn

n2注注求n阶导数时,关键要寻找规律,的规律性,写出n阶导数.例阶

(x)

3x3x2

x,求

f(n)

存在的最分析

(x)

4x32x3

xx

f(x)

12x26x2

xxf(0)

f(x)

f(0)

2x30 x

x

x

f(x)

f

4x30

f(0)x

x

x 又f(0)

f(x)

f

6x20

xx

xf(x)f(0)x0

xx

12x2x

0

f(0)例2阶2

(x)

3x3x2

x,求

f(n)

存在的最f(x)

12x26x2

xx

f(x)

24x12x

xxf(0)f

(0)

x

f(x)f(0)x

x

12xx

x

f(x)f(0)x0

x

24xx

f(0)

0不存在设

x)y

fx2 y

f(x2)(x2

(x2y[

(x2[2x]

f(x2)

f(x22

(x2)

2xf

(x2)(x22

(x2)

4x2

f(x2 求其中f二阶可导. 设函u和v具有n阶导,

(u

v)(n)

v(n)

(2)

(Cu)(n)

Cu(n)(uv)

(u

uv

(a

a2b0

uva0b2

2uv

n(an

b)n

C0anb0

C1an1b

C2an2b2nnnnCkankbknnnn

Cna0bn

nn

n)(nCkunk)vk 公nk间接

利用已知的高阶导数公式,通过四则代换等方n阶导数.常用高阶导

(ax

ax

lnn

(a

(e

)(

e(2)(3)

kx)(n)(coskx)(n)

kk

sin(kxcos(kx

n)2n)2

(x

)(

1)(

n1)xn

(n1)!(6)

)(

(x

a)nx

(x

a)n1例y

x23

求y(n解y

(x1)(x

(x1)(x

(x1)(x

x x

)(n)

22xa(2

(y

(1)(2)

1

1x

x

(x

(xy(

(x

(x(1)nn!

(

(x1)n1例y

x

x求y(n解若直接求导,将是很复杂的,且不易找出规律,所ysin4xcos4

x

x)2

xcos2 1

11cos4x1

sin2x2

2 34

1cos4(cos(coskx)(n)kncos(kxn2y(

14

cos4x

n2y

x2e2x求y(20)C20解 2x C20

公式y(20)

2C(eC(e

)(20)x2

1(e2x

(x2)C 2C20

)(18)

(x2)220e2

x2

20219e2x

2x

20

218e2x(uv)(n)C0(uv)(n)C0(n)nvC1(nvC2(n2)nvCku(nk)v(k)Cnuv(n)nn

20x

设yx2)(2x3)2(3x4)3求y(6分析此函数是6次多项式6阶导数,

yx(2x)2(3x)3

108x6p(x)5其中p5x为x的5次多项式y(6)

108fx

(x2

3x

x2n nf(n)(2)

n!x2x23x(x2)(x (x

cosπx22x2nC0[(n

2)n](

[(

]x2nC1[(n

2)n](

[(

n![(

cos

x]

(x2)g(x)22n阶导数

1x1解y

2(x

1

1

n

(1

n1xx)

(xa)

2(1)2y1

(1x)3y(n)

2(1)n

x)n1求函数n阶导数

xyx23x2解y

x(x23x2)3(x23x2)7xx23xx3

x3

x

x3

(x1)(

x3

x

x x

1

(x

(x(n)

8(1)n

(1)n (x

2)n1

(x

(ngx连续

f(x)

(x

a)2

g(x)求解gx)

f(a)可f(x)

a)g(x)(

g(x)g(x)