高数上2二章3节

2.3

高阶导数引例、y

=f

(x)=e2x+3，求：(1)

y

'(x)

=

j(x),

(2)j

'

(x)

=

w

(x).解：(1)

y

=e2

x+3

\y'(x)=j(x)=2e2

x+3(2)

j(x)

=

2e2

x+3\

j'

(x)

=w

(x)

=

2 2e2x+3

=

22

e2x+3(n)(n)dn

y

dn

f

(x)dxndxdy(n-1)y

=

=

f

(x)

=

=dxn注(1)f

(0)(x)=f

(x)(2)n

‡2时f

(n)(x)统称高阶导数'x[

f

(x)](n-1)一、y

=f

(x)的n阶导数：(n

‡1,n˛

Z)y

=f

(x)的n

-1阶导数的导数称为y

=f

(x)的n阶导数.记作：32d

2ydx(2)若f

二阶可导,

y

=

f

(x

),

求：例1(1)y

=f

(x)=e2x+3

,求：y(3)(x)、y(3)(0)、y(n)(x)例1(1)y

=f

(x)=e2x+3

,求：y(3)(x)、y(3)(0)、y(n)(x)解：

y

=e2

x+3

y'(x)=2e2

x+3

y''

(x)

=

2 2e2x+3

=

22

e2x+3\

y(3)

(x)

=

23

e2x+3

y(3)

(0)

=

23

e3

=

8e3

(数归法)

\

y(n)

(x)

=

2n

e2x+3解：

y

=f

(x3

)

y'=3x2

f

'(x3

)''2d

2

y

dy'\

y

=

=dx

dx=

6x f

'

(x3

)

+3x2

3x2

f

''

(x3

)=

6x f

'

(x3

)

+9x4

f

''

(x3

)32d

2ydx(2)若f

二阶可导,

y

=

f

(x

),求：xeax+b(1)(eax+b

)(n)

=

anxf

(n)

(ax

+b)(2)[

f

(ax

+b)](n)

=

an重要结论：二、设u(x)、v(x)均n阶可导,则：1、(u

–

v)(n)

=

u(n)

–

v(n)ni

(i)n(n)C

u

v(n-i)

i=02、(uv)

=(Cu)(n)

=

Cu(n)(Leibniz公式)例2(1)f

(x)=2e3x-8

+5e7

x+2

,求：f

(n+1)(x)解：f

(n+1)(x)=(2e3x-8

+5e7x+2

)(n+1)=(2e3x-8)(n+1)

+(5e7x+2)(n+1)=3n+1

2e3x-8

+7n+1

5e7x+2(2)y

=x2e3x，求y(20)解：记u(x)=x2

,v(x)=e3

x

u(x)

=

x2

,

v(20)

(x)

=

320

e3xu'

(x)

=

2x,

v(19)

(x)

=

319

e3xu''

(x)

=

2,

v(18)

(x)

=

318

e3xu(k

)

(x)

=

0,

(k

‡

3)220(20)C

u

vi

(i)

(20-i)

\

y

=i=0=

320

x2e3x

+20·2·319

xe3x

+

190·2·318

e3x2(2)(sinx)(n)

=sin(1

np

+

x)(奇变偶不变,符号看象限)xlnn

a

(ex

)(n)

=ex(1)(ax

)(n)

=ax2

(cosx)(n)

=cos(1

np

+

x)三、常用导函数(四)：ax+b=

a(n)

n

eax+bx

(e

)a-x(a-x)n+1

(

1

)(n)

=

n!

；1(3)(=

n+1(a+x)(-1)n

n!)(n)a+xn-1[ln(a

+

x)](n)

=

(-1)

(n-1)!(a+x)n注：结合例1结论(2)可得更多有用结论.推导a+x

y

=

1

=(a

+

x)-1\

y'

=(-1)

(a

+

x)-2y''

=(-1)

(-2)

(a

+x)-3y(3)

=(-1)

(-2)

(-3)

(a

+x)-4

(数归法)

y(n)

=(-1)n

n!(a

+x)-(n+1)例3(1)f

(x)=sin2

x,求f

(10)(x)2解：

f

(x)=1

(1-cos2x)\

f

(10)

(x)

=(1

-

1

cos2x)(10)2

2=-

1

cos(10p

+2x)

2102

2=

29

cos2xex：f

(x)=cos2

x,求f

(100)(x)2解：

f

(x)=1

(1+cos2x)\

f

(100)

(x)

=(1

+

1

cos

2x)(100)2

2=

1

cos(100p

+2x)

21002

2=

299

cos2x\

y(n)

=

1

[(

1

)(n)

+(

1

)(n)

]2

1-x

1+x解：

y

=

1

(

1

+

1 )2

1-x

1+x(-1)n

n!=2(1-x)n+1

+

2(1+

x)n+1n!1-x2(2)y

=

1

,求：y(n)小结：1、例1注(1,2)及其应用：

2、Leibniz公式及其应用：3、常用导数(四)及其应用：作业2-3：103页1(3,4,6,10),3(1,2),10(1或2),11(2)1、y

=cos2

x,求y(n)2解：

y

=1

(1+cos2x)\

y(n)

=(1

+

1

cos2x)(n)2

2=

1

cos(np

+2x)

2n2

22=

2n-1

cos(np

+2x)课后练习：1-x22、y=

x

,求y(n)(-1)nn!=

n+1

-

n+12(1-

x)

2(1+

x)n!解：

y

=

1

(

1

-

1 )2

1-x

1+x\

y(n)

=

1

[(

1

)(n)

-(

1

)(n)

]2

1-x

1+x1+x3、y

=1-x

,求y(n)1+x解：

y

=

2

-11+x\

y(n)

=

2

(

1

)(n)=

n+1(1+x)2

(-1)n

n!x2