Post on 28-Dec-2015
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 .