一、 The definition of higher derivatives

Q:the acceleration of the moving objects?

.])([)()( tftvta

Def.

).(sec))((,

)()(lim))((..,

)()(

0

xfofderivativeondthecalledisxfthenexistx

xfxxfxfeixatderivative

thehasxfofderivativethexfif

x

The instantaneous rate of change of velocity with respect to time is the acceleration a(t) of the object. Therefore,

)()(),( tftvthentfsif

denoted .)(

,),(2

2

2

2

dxxfd

ordxyd

yxf

.)(

,),( )()(

n

n

n

nnn

dxxfd

ordxyd

yxf

The derivative of the third derivative is called the fourth derivative,

The second and up derivatives is called higher derivatives.

.)(;)(,Re

derivativefirstthecalledisxfderivativzeroththecalledisxflatively

.,),(3

3

dx

ydyxf

The derivative of the second derivative is called the third derivative,

.,),(4

4)4()4(

dx

ydyxf

In general, the derivative of the (n-1)th derivative is called the nth derivative,

二、 The examples of finding higher derivative

Eg.1 ).0(),0(,arctan fffindxyif

Solution 21

1

xy

)

1

1(

2

xy

22 )1(

2

x

x

))1(

2(

22

x

xy

32

2

)1(

)13(2

x

x

022 )1(

2)0(

xx

xf

032

2

)1(

)13(2)0(

xx

xf;0 .2

1.Dir.: Get the higher derivative according the def..

Eg.2 .),( )(nyfindRxyif

Solution1 xy

)( 1 xy 2)1( x

3)2)(1( x))1(( 2 xy

)1()1()1()( nxny nn

thennegernatrualtheisif ,int)()( )( nnn xy ,!n )!()1( ny n .0

Eg.3 .),1ln( )(nyfindxyif Solution

xy

1

12)1(

1

xy

3)1(

!2

xy

4)4(

)1(

!3

xy

)1!0,1(

)1(

)!1()1( 1)(

nx

ny

n

nn

Tip: when finding the nth derivative,first finding the first to the third or fourth derivative, analyze the regular pattern, write out the nth derivative (prove it by using mathematical induction)

Eg.4 .,sin )(nyfindxyif Solution

xy cos )2

sin(

x

)2

cos(

xy )22

sin(

x )2

2sin(

x

)2

2cos(

xy )2

3sin(

x

)2

sin()( nxy n

)2

cos()(cos )( nxx n

Using the same way, we get

Eg.5 .),tan,(sin )(nax yfindtconsisbabxeyif

Solution bxbebxaey axax cossin

)cossin( bxbbxae ax

)arctan()sin(22

a

bbxbae ax

)]cos()sin([22 bxbebxaebay axax

)2sin(2222 bxbaeba ax

)sin()( 222)( nbxebay ax

nn )arctan(

a

b

2. The operation rule of the higher derivative:

thenderivativenththehavevandufunctionif ,)()()()()1( nnn vuvu

)()()()2( nn CuCu

)()(

0

)()()(

)2()1()()(

!

)1()1(!2

)1()()3(

kknn

k

k

n

nkkn

nnnn

vuC

uvvuk

knnn

vunn

vnuvuvu

leibniz formula

莱布尼兹公式

Eg.6 ., )20(22 yfindexyif xSolution

,,, 22 formulaleibnizfromthenxveuif x

0)()(!2

)120(20

)()(20)(

2)18(2

2)19(22)20(2)20(

xe

xexey

x

xx

22!2

1920

22202

218

2192220

x

xx

e

xexe

)9520(2 2220 xxe x

3.indirec.:

Constantly used formulas for find higher derivatives:

nn xnx )1()1()()4( )(

nnn

x

nx

)!1()1()(ln)5( 1)(

)2

sin()(sin)2( )( nkxkkx nn

)2

cos()(cos)3( )( nkxkkx nn

)0(ln)()1( )( aaaa nxnx xnx ee )()(

Using the known formula and the

1)( !

)1()1

( n

nn

x

n

x

methods of operations, instead the variable etc., finding the nth derivative.

Eg.7 .,1

1 )5(

2yfind

xyif

Solution

)1

1

1

1(

2

1

1

12

xxx

y

])1(

!5

)1(

!5[

2

166

)5(

xx

y

])1(

1

)1(

1[60

66

xx

Eg.8 .,cossin )(66 nyfindxxyif

Solution 3232 )(cos)(sin xxy

)coscossin)(sincos(sin 422422 xxxxxx

xxxx 22222 cossin3)cos(sin

x2sin4

31 2

2

4cos1

4

31

x

x4cos8

3

8

5

).2

4cos(483)(

nxy nn

三、 ConclusionThe def. of the higher derivative and the physics meaning;

The higher derivative operational rule

(the leibniz formula);The methods of finding the nth derivative;

1.direct.; 2.indirect..

思考题

设 连续，且 ，)(xg )()()( 2 xgaxxf

求 .)(af

思考题解答)(xg 可导

)()()()(2)( 2 xgaxxgaxxf

)(xg 不一定存在 故用定义求 )(af

)(af axafxf

ax

)()(lim 0)( af

axxf

ax

)(lim )]()()(2[lim xgaxxg

ax

)(2 ag

一 、 填 空 题 ：

1 、 设 te

ty

sin 则 y =_ _ _ _ _ _ _ _ _ .

2 、 设 xy tan , 则 y =_ _ _ _ _ _ _ _ _ .3 、 设 xxy arctan)1( 2 ， 则 y =_ _ _ _ _ _ _ _ .

4 、 设2xxey , 则 y =_ _ _ _ _ _ _ _ _ .

5 、 设 )( 2xfy , )( xf 存 在 ， 则 y =_ _ _ _ _ _ _ _ _ .6 、 设 6)10()( xxf , 则 )2(f =_ _ _ _ _ _ _ _ _ .7 、 设 nn

nnn axaxaxax

12

21

1 ( naaa ,,, 21 都 是 常 数 ) ， 则 )( ny =_ _ _ _ _ _ _ _ _ _ _ .8 、 设 )()2)(1()( nxxxxxf , 则 )()1( xf n =_ _ _ _ _ _ _ _ _ _ _ _ .

练 习 题

二 、 求 下 列 函 数 的 二 阶 导 数 ：

1、 x

xxy

42 3 ；

2、 xxy lncos 2 ；

3、 )1ln( 2xxy .

三 、 试 从ydy

dx

1， 导 出 ：

1、 32

2

)( y

y

dy

xd

；

2、 5

2

3

3

)()(3y

yyydy

xd

.

四、验证函数 xx ececy 21 (, 1c , 2c是常数） 满足关系式 02 yy .

五 、 下 列 函 数 的 n 阶 导 数 ： 1 、 xey x cos ；

2 、 x

xy

1

1；

3 、 232

3

xx

xy ;

4 、 xxxy 3sin2sinsin .

一、1、 te t cos2 ； 2、 xx tansec2 2 ；

3、 21

2arctan2

x

xx

； 4、 )23(2 22

xxe x ；

5、 )(4)(2 222 xfxxf ； 6、207360； 7、 !n ； 8、 )!1( n .

二、1、 32

5

84

34

xx ；

2、 2

2cos2sin2ln2cos2

x

x

x

xxx ；

3、2

32 )1( x

x

.

练习题答案

五 、 1 、 )4

cos()2(

nxe xn ；

2 、 1)1(

!2)1(

nn

x

n；

3 、 )2(],)1(

1

)2(

8![)1(

11

n

xxn

nnn ；

4 、 )2

2sin(2[41

n

xn

+ )]2

6sin(6)2

4sin(4

n

xn

x nn .