1
第四节 高阶导数
习题 2-4
1,求下列函数的二阶导数,
(1)
22
3eln
x
yx x=++; (2) cosyx x= ;
(3) e sin
t
yt
= ; (4)
2
(1 ) arctanyx x=+,
解 (1)
22
2
11
62e,64e
xx
yx y
x x
′′′=+ + =+?,
(2) cos sin,sin sin cos 2sin cosyxxxy xxxx xxx=? = =,
(3) e sin e cos
tt
ytt

′=? +,
esin ecos ecos esin 2ecos
tt t t t
ytttt t

′′ =? =?,
(4)
2
2
2 arctan 1,2arctan
1
x
yx x y x
x
′′′=+=+
+
,
2,设
6
() ( 10)fx x=+,求 (2)f ′′′,
解 由于
54
( ) 6( 10),( ) 30( 10)fx x fx x′′′=+ = +,
3
( ) 120( 10)fx x′′′ =+,所以
3
(2) 12 120 207360f ′′′ =?=,
3,设 ()f x′′ 存在,求下列函数 y 的二阶导数
2
2
d
d
y
x
,
(1) (e )
x
yf
= ; (2) ln ( )yfx=,
解 (1)
2
(e )e,e (e ) e (e )
x xxxxx
yf y f f

′ ′ ′′ ′′ ′=? = +,
(2)
2
2
() () () ()
,
() ()
f xfxfxfx
yy
fx fx
′′′′?
′′′==,
4,试从
d1
d
x
yy
=
′
导出,
2
(1)
2
23
d
d()
x y
yy
′′
=?
′; (2)
32
35
d3()
d()
x yyy
yy
′′ ′ ′′′?
=
′
,
证 (1)
2
223
dd1d1d 1
() ()
dddd()()
x xy y
yy xy y y
′′′′
== =?=?
′′ ′′ ′
,
(2)
32 223 2
32 3 6 5
d dd d d 3()( ) ()1 3( )
() ( )
ddd d () () ()
x xyxyyy yy
yx y yyy y y y
′′ ′ ′′ ′′′ ′ ′′ ′ ′′′
==?= =
′′′
,
5,验证函数 esin
x
yx= 满足关系式
220yyy′′′? +=,
解 因为
e sin e cos e (sin cos )
xx x
yxxxx′=+=+,
e (sin cos ) e (cos sin ) 2e cos
xx x
yxxxx x′′ =++?=,
故而
2 2 2e cos 2e (sin cos ) 2e sin 0
xx x
yyy x x x x′′ ′?+=? + + =,
6,求下列函数的 n阶导数的表达式,
(1)
2
sinyx= ; (2) lnyxx= ;
(3) e
x
yx= ; (4)
22
1
y
x a
=;
(5)
1
ln
1
x
x
+
,
解 (1)
2
1
sin (1 cos 2 )
2
yx x==?,
() 1 1
2 ππ π
cos(2 ) 2 cos(2 ) 2 sin[2 ( 1) ]
22 2 2
n
nnn
yxn xnxn

=? + =? + = +?,
(2)
1
ln 1,yxy
x
′′′=+ =,
() ( 2) 2
11
111
() (1) ( 2)! (1)( 2)! ( 2)
nn n n
nn
yn n
x xx


= = = ≥,
(3)
()
e e ( 1)e,e ( 1)e ( 2)e,( )e
x xxxx xn x
yxx y x x yxn′′′=+ =+ =++ =+ =+,
(4)
11 1
()
2
y
ax a x a
=?
+
,
3
() () ()
11
1 1 1 1 (1) ! (1) !
[( ) ( ) ] [ ]
22() ()
nn
nnn
y
axa xa axa xa
+ +

=?=?
+?+
11
(1) ! 1 1
[]
2 () ()
n
nn
n
a xa xa
+ +
=?
+
,
(5)
11
[ln(1 ) ln(1 )]
11
yxx
x x
′′=+=?
+?
,
11
() ( 1)
11 (1)(1)!(1)(1)!
()
11 (1) (1)
nn
nn
y
xx xx


=? =?
+? +?
1
11
(1) ( 1)![ ]
(1) (1)
n
nn
n
xx
=
+?
,
7,求下列函数的指定阶的导数,
(1) e cos
x
yx=,求
(4)
y ; (2)
2
s2yxinx=,求
(50)
y,
解 (1) e cos e sin e (cos sin )
xxx
yxxxx′=?=?,
e (cos sin ) e (sin cos ) 2e sin
xx x
yxxxx x′′ =+=?,
2e cos 2e sin 2e (cos sin )
xx x
yxxxx′′′ = =? +,
(4)
2e (cos sin ) 2e ( sin cos ) 4e cos
xx x
yxxxx=? + + =?,
(2)
(50) 2 (50) 2 (50) (49) (48)
50 49
( s 2 ) (sin 2 ) 50 2 (sin 2 ) 2(sin 2 )
2
yxinxxx xx x
==+?+
250 49 48
49
2sin(2 25π) 100 2 sin(2 π) 2450 2 sin(2 24π)
2
xx xx x=++ ++?+
50 2
1225
2 ( sin 2 50 s 2 sin 2 )
2
x xxcox x=? + +,
8,求下列方程所确定的隐函数 y 的二阶导数
2
2
d
d
y
x
,
(1) tan( )yxy=+; (2) e
x y
xy
+
=,
解 (1) 对方程两边对 x 求导,得
2
sec ( )(1 )yxyy′′=++
,
2
csc ( )yxy′=?+
,
对
2
csc ( )yxy′=? + 两边对 x 求导,得
4
223
2csc ( )cot( )(1 ) 2csc ( )cot ( )yxyxyy xyxy′′ ′= +++=?+ +,
(2) 对方程两边对 x 求导,得
e
e(1 ),
e
xy
xy
xy
yxyy
yxy y y
x xyx
+
+
+

′′′+= + = =

,
对 e(1 )
xy
yxy y
+
′′+= +两边对 x 继续求导,
2
2e(1)e
xy xy
yxy y y
++
′′′ ′ ′+= ++,
2222
23
e(1 )2 (1 )2 [(1)(1)]
xy
xy
yyxyyyyx y
y
xxyxx
+
+
′′ ′′+? ++?
′′ ===

,
9,求下列参数方程所确定的函数 y 的二阶导数
2
2
d
d
y
x
,
(1)
cos,
sin ;
x at
yb t
=?
=
(2)
(),
() ().
xft
ytft ft
′=
′=?
其中 ()f t′′ 存在且不为零,
解 (1)
dcos
cot
dsin
ybt b
t
xata
==?
,
2
2
2 23
dd d d 1
( cot ) ( cot ) csc
dd sin
yb btb b
ttt
xa ta xa atx at
=? =? = =?
,
(2)
2
2
d()dd1
,
d() d()d
ytft y t
t
x ft x ftx
′′
== ==
′ ′
,