清华名师：微积分课程各章节知识点题解：第3章 导数概念、性质与计算

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n
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E5?¨é?9
b

è 

p 1è

￥ò¨?
b
λ
λ
(
x
ey =
3
,,,
2
L
xx
eyey
λλ
λλ =
′′
=
′


xnn
ey
λ
λ=
)(
b

è 

p ￥ò¨?
b
xy sin=

3


+==
′
2
sincos
π
xxy






+=
+=
′′
2
2sin
2
cos
ππ
xxy





l
l
l
l
l




)
2
sin(
)(
π
nxy
n
+= b


?
?f
)(xf D )(xg eé ?ú¨?

H
è ¨?/
? ? L
!
)(xf
 )(xg (μ ¨?


?1 ??
DG
T

n
)()()]()([
)()(
0
)(
xgxfCxgxf
kknk
n
n
k
n?
=
∑
=

!
)1()1(
!)!(
!
k
knnn
kkn
n
C
k
n
+
=
=
L

i?

 b
)()(
)0(
xfxf = )()(
)0(
xgxg =
è 
￥ ¨?
^





xxy sin
2
= 

3
? ??
DG
T
?i? ￥ ¨?
(1
,
5μ

2
x 3≥n
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L 1101 è
 62781785
2005
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)99(2)100(2)100(
)(sin)(100)(sin xxxxy
′
+=
























)98(2
)(sin)(
!2
99100
xx
′′
×
+











++
+=
2
99
sin200
2
100
sin
2
ππ
xxxx



























+×+
2
98
sin99100
π
x










xxxxx sin9900cos200sin
2
= b

è 
p
22
1
ax
y
= ￥ ¨?
b
n

3


)()(
22
)(
11
2
11
nn
n
axaxa
ax
y
+
=
=
+
=
++ 11
)(
!)1(
)(
!)1(
2
1
n
n
n
n
ax
n
ax
n
a


+
=
++ 11
)(
1
)(
1
2
!
)1(
nn
n
axax
a
n



è 

)32ln()( xxf?= ￥ ¨?
^





 b

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L 1101 è
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"


10
10
)32(
!103
x?


#


10
10
)32(
!93
x?



$


10
10
)32(
!103
x?

%


10
10
)32(
!93
x?

b

3
s?1 %
bo??i? 
￥Q


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a¨e￥2T+ ￥Z
a' Vb





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p?E5D±sE

? ? 

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)();( ufyx
dx
du
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′
=? ?
? ))(( xuu?= ) 9 μ?
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du
dy
′
=
5 ˉf
)]([ xfy?=
? x)9μ?

O

dx
du
du
dy
dx
dy
= 
)()(})]([{ xufxf
′
′
=
′


è 
p
x
ay
1
arctan
= D

)1ln(
2
++= xxy ￥?
b

3

)(
1
1
ln)(
2
2
1
arctan
1
arctan
+
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x
x
aaa
xx


x
a
x
a
1
arctan
2
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ln
+
=

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L 1101 è
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L 1101 è
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)2
12
1
1(
1
1
])1[ln(
22
2
x
xxx
xx
+
+
++
=
′
++








1
1
2
+
=
x


è 
pf
￥?
b
)0(
sin
>= xxy
x
3
 Z E 


? ?f
?S
a·f
b
n5
H| 

¤?f




xxy lnsinln = b

??f
p?E¤

x
x
xxy
y
sin
lncos
1
+=
′



V7
]
sin
ln[cos
sin
x
x
xxxy
x
+=
′
b


??5 |
p ?￥ ZE? S| 
p? E b"
a¨ ?
a ·f
?
cμ?y
TMe"{eZa 7Z￥f
9
a¨b

)(
)(
xv
xuy =

ZE 


|
a·f
?1a
 ¨ˉf
p ?
E5#eE
T

xxx
exy
lnsinsin
==
è 



!

1
lnarctan
2
2
+
=
x
x
x
e
e
ey


5


1
1
2
1
+
=
=
e
e
dx
dy
x
b

3
|f
Vr
T?1

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L 1101 è
 62781785
2005
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L 1101 è
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)1ln(
2
1
arctan
2
++?=
xx
exey
5

1
1
1
2
2
2
+
+?
+
=
′
x
x
x
x
e
e
e
e
y


1
1
1
2
2
2
1
+
+?
+
=
=
e
e
e
e
dx
dy
x
1
1
2
+
=
e
e
b


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p?9
H
 31???WM

 üf
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 ¨ˉ
f
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 Ka1ü??￥?WM
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 V[????WM
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 o1T???μ

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 °?K?
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?u 11M

H
 )(ufy = ￥ ± s





uufdy d)(
′
= b

? u1?WM

 u
^M
 ￥ V±f
x )(xu?=
H
5 )]([ xfy?= ￥±s

uufxxufxxfy d)(d)()(d})]([{d
′
=
′′
=
′
= Vn
í
^1M

^?WM

f
u )(ufy = ￥±s?
T?Mb


?f
p?ED±sE


? )(xyy = ?B? ?f
Z ? 0),( =yxF ??
 5 V
j1
0))(,( ≡xyxF
°¤ ?¨ˉf
p?E5é?p?


3 ' Vb

x
y′
/
 è
a
üp?f
￥=¨?
￥ZEb

è 



!
p4
22
=++ yxyx y
′′
b

3

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22
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5¤?
T
4)()(
22
=++ xyxxyx
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L 1101 è
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2005
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Z?
H1? p?¤
x
022 =
′
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3¤
yx
yx
y
2
2
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x



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2
=
′′
+
′
+
′′
+
′
+
′
+ yyyyxyy
3

yx
yy
y
2
)(222
2
+
′
+
′
+
=
′′


| y
′
￥Vr
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i? ?¤

33
22
)2(
24
)2(
)(6
yxyx
yxyx
y
+
=
+
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′′
b

è 


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1ln =+ yxy
??
pwL )(xyy = 
)￥ELZ?b
1=x
3
 ?X?Z?
7
5 μ1=x 1ln =+ yy
A?μ 1=y b?X?Z?
HsY
1? xp?


0
1
=
′
++
′
xx
y
y
yyx


7 1,1 == yx
?N V¤? 012
1
=+
′
=xx
y

yN
2
1
1
=
′
=xx
y

[ )(xyy =  

)EL|
q 2=k bE LZ?1
)1(21?=? xy


12?= xy b



Qf
D?
Z?￥p?E5

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2005
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? ? 

Qf
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? )(yx?= 
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=??a V?
 O 0)( ≠
′
y?
5 
Qf
)(xfy = ?￥uW
=9 V?
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)(
1
)(
y
xf
′
=
′
b




?
? ú￥Qf
àμ?Me ?f
￥M
)(xfy =
:|b

è 

! 1??f

 1Qf

O)(xf )(xg
3
1
)1(,2)1(?=
′
= ff

1)1( =
′′
f





p )2(g
′′



p
x
fxf
x
2
)1()1(
lim
0

→
b

<3>

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 )(xg 1 )(xf ￥Qf
￥Hq?
Xü?M
M
:|


1 ?¨Qf
?

T
?| )(xg ^1 )(yg
? )(xfy = b

?Qf
?

T V¤?
1)()( =
′′
ygxf

H1? xQp?
 0)()()()( =
′′′′
+
′′′
x
yygxfygxf




0)()]([)()(
2
=
′′′
+
′′′
ygxfygxf

7
?μ1=x 2=y b?i?

3
)1(
1
)2(?=
′
=
′
f
g
yN¤?

33)2( =
′′
g b




6
3
)1(
2
1
2
)1()1(
lim
0
=
′
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→
f
x
fxf
x
b?



^¨ ?
?
?lb

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L 1101 è
 62781785
2005
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L 1101 è
 62781785


? ? 

?
Z???￥f
￥p?E5


? Tttyytxx ∈== ),(),( ?
V?
O 0)( ≠
′
t?
5??
Z?

=
=
),(
),(
tyy
txx




Tt∈

??￥f
9 V?
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)(
)(
tx
ty
x
y
dx
dy
t
t
′
′
=
′
′
= b


?+Y?i

)(
)(
2
2
tx
ty
dx
yd
′′
′′
≠
??
T1


)(
1
)]([
)()()()(
)(
22
2
tx
tx
txtytxty
x
y
dt
d
dx
yd
t
t
′
′
′′′
′′′
=
′
′
=
3
)]([
)()()()(
tx
txtytxty
′
′′′
′′′
=


è 

p?L



=
=
),cos1(
),sin(
tay
ttax

2
π
=t )￥MLZ?b

3
??

t
t
ta
ta
x
y
dx
dy
t
t
cos1
sin
)cos1(
sin
=
=
′
′
= )2( πkt ≠



[?L
2
π
=t )￥ML|
q1


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L 1101 è
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2005
￡
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TDn bê
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L 1101 è
 62781785
1
cos1
sin
22
=
=
==
ππ
tt
t
t
dx
dy


?L
??
2
π
=t ￥?
^
aa,1
2
π



#
p￥MLZ?1 axay
=? 1
2
π




'

0
2
2 =
+? ayx
π
b

è 


! tbytax sin,cos ==
p
x
y
′′
b


3













t
a
b
ta
tb
x
y
y
t
t
x
cot
sin
cos
=
=
′
′
=
′



)cos(
)cot(
)(
′
′
=
′
′′
=
′′
ta
t
a
b
x
y
y
t
tx
x


ta
b
ta
t
a
b
32
2
sin
1
sin
sin
1
=
=

è 

!

3
1
)(
+
=
x
x
xf
p
( )
)(1
10
f 

3

1ZL9

5 Sí) ??/

)(xf
3
4
1
3
43
3
1
)(
+
=
+
+
=
+
=
xx
x
x
x
xf


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T? www.tsinghuatutor.com - 21 - bvD ? S
L 1101 è
 62781785
2005
￡
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TDn bê
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L 1101 è
 62781785
11
11
)10(
)3(
!9)1(4
)(
+

=
x
xf
911
)10(
2
!9
2
!94
)5( =
=?f

è 
! 0>δ
 )(xf  ],[ δδ?
μ?l
 1)0( =f
O
@


0
1
)(sin)1ln(
lim
2
0
=
+?
→
x
x
e
xfxx






1

￡
üK
1
sin)(sin
lim
2
0

→
x
x
e
xxfx
i
ipNK′b



2

￡
üf
)(xf  ) V±
ip0=x )0(f
′
b





<3>






ZE 

n5


x
xf
e
xxfx
x
x
x
1)(
lim
1
sin)(sin
lim
00
2
=

→→
b

?X?

1
)(sin)1ln(
lim
2
0
+?
→
x
x
e
xfxx



2
0
)(sin)1ln(
lim
x
xxfxxx
x
++?
=
→


+
+?
=
→
2
0
)1ln(
lim
x
xx
x
2
0
)(sin
lim
x
xxfx
x

→


x
x
x
xf
x
x
x
x
xxx
sin
)(
lim
sin
lim
2
1
1
1
lim
000
+
+
=
→→→
0
sin
11)(
lim
22
1
lim
00
=
+?
+
=
→→
x
x
x
xf
x
xx


￡
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L 1101 è
 62781785
2005
￡
!·°
TDn bê
ê?7<v0 1006 bvD ? S
L 1101 è
 62781785
1
sin
1
lim
1)(
lim
22
1
lim
000
x
x
x
x
xf
x
xxx
+
+
=
→→→
xx
xx
x
xf
xx
sin
sin
lim
1)(
lim
2
1
00
+
+?=
→→


2
3
00
6
1
lim
1)(
lim
2
1
x
x
x
xf
xx →→
+
+?=

00
)0()(
lim
2
1
0
=+
+?=
→
x
fxf
x


¨K
E5
 VwK



x
fxf
x
)0()(
lim
0
→
i
O


2
1)0()(
lim
0
=
→
x
fxf
x
b

pSE


1
)(sin)1ln(
lim0
2
0
+?
=
→
x
x
e
xfxx



2
0
)(
lim
x
xfxx
x
+?
=
→


0
1)(
lim
0
=
=
→
x
xf
x



yN 0
1
sin)(sin
lim
2
0
=

→
x
x
e
xxfx
b

￡
!·° Iù
T? www.tsinghuatutor.com - 23 - bvD ? S
L 1101 è
 62781785
2005
￡
!·°
TDn bê
ê?7<v0 1006 bvD ? S
L 1101 è
 62781785


ZE 

?¨
ü à
T5μ








1
)(sin)1ln(
lim
2
0
+?
→
x
x
e
xfxx


2
22
0
)(sin)(
2
1
lim
x
xfxxoxx
x
++
=
→



??
?l

K

x
xf
x
1)(
lim
0
→
2
1)0()(
lim
0
=
=
→
x
fxf
x
i



?
^ )(xf  ) V±
iO

0=x
)0(f
′
2
1)0()(
lim
0
=
=
→
x
fxf
x
b


è 

! )(xyy =
^=¨è±sZ?

x
eqyypy
3
=+
′
+
′′
@
SHq 0)0()0( =
′
= yy

￥+3
5?
Hf
0→x
)(
)1ln(
2
xy
x+
￥K




 b

"

?ib








#
?? b

#

$

?? b




%

?? b

<3>

)(
lim
)(
)1ln(
lim
2
0
2
0
xy
x
xy
x
xx →→
=
+


2
1
2
)(
2
lim
)(
2
lim
00
==
′′
=
′
=
→→
xyxy
x
xx
b

ê $
b

￡
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L 1101 è
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