1
第二章 导数与微分
第一节 导数的概念
习题 2-1
1,当物体的温度高于周围介质的温度时,物体就不断冷却,若物体的温度T
与时间t的函数关系为()TTt=,应怎样确定该物体在时刻t的冷却速度?
解 该物体在时刻t的冷却速度为
0
d()()
lim
d
t
TTttTt
tt
Δ→
+Δ?
=
Δ
,
2,设对1g质量的物体加热,使他的温度从0℃升高到t℃,这物体吸收的热量为()qqt=,求物体在温度
0
t℃时的比热(比热是1g物体温度升高1℃所需的热量),
解 物体在温度
0
t℃时的比热为
0
00
0
()()d
lim
d
t
tt
qt t qtq
Δ→
=
+Δ?
=
Δ
,
3,自由落体的运动规律为
2
1
2
s gt=,其中g是重力加速度,求
(1) 物体在3s到4s这一时段的平均速度;
(2) 物体在3s时的瞬时速度,
解 (1) 物体在3s到4s这一时段的平均速度为
22
11
43
()() 7
22
43 2
gg
st t st
vg
t
+Δ?
===
Δ?
( ms),
(2) 物体在3s时的瞬时速度为
22
0
11
(3 ) 3
22
lim 3
t
gt g
vg
t
Δ→
+Δ?
==
Δ
( ms),
4,证明 (cos ) sinx x′=?,
2
证
00
2
2sin( )sin
cos( ) cos
22
(cos ) lim lim
xx
x xx
xx x
x
Δ→ Δ→
+ΔΔ
+Δ?
′==
ΔΔ
0
sin
2
2
lim sin( ) sin
2
2
x
x
xx
x
xΔ→
Δ
+Δ
=? =?
Δ
,
5,假设
0
()f x′存在,按照导数的定义求下列极限,
(1)
00
0
()()
lim
x
f xxfx
x
Δ→
Δ?
Δ; (2)
00
0
(3)()
lim
h
f xhfx
h
→
+?;
(3)
00
0
()()
lim
h
f xhfxh
h
→
+; (4)
00
1
lim [ ( ) ( )]
2
n
nfx fx
n
→∞
+?,
解 (1)
00
0
()()
lim
x
f xxfx
x
Δ→
Δ?
Δ
=
00
0
0
()()
lim ( )
x
fx x fx
f x
x
Δ→
Δ?
′?=?
Δ
,
(2)
00 00
0
(3)() (3)()
lim 3lim 3 ( )
3
hh
fx h fx fx h fx
f x
hh
→→
+? +?
′==,
(3)
00 0000
()() ()()()()
lim lim
f x h fx h fx h fx fx fx h+ +? +
=
00 00
0
( ) () ( ) ()
lim lim 2 ( )
hh
fx h fx fx h fx
f x
hh
→→
+
′=+=
,
(4)
00
00 0
1
()()
11 1
2
lim [ ( ) ( )] lim ( )
1
22 2
2
nn
fx fx
n
nfx fx f x
n
n
→∞ →∞
+?
′+? = =,
6,设(0) 0f =,且(0)f ′存在,求
0
()
lim
x
f x
x
→
,
解
00
() () (0)
lim lim (0)
0
xx
fx fx f
f
xx
→→
′==
,
7,求下列函数的导数,
(1)
3 2
yx= ; (2)
3 5
yx x=? ; (3)
2 3
5
x x
y
x
= ; (4)
2
e
x
y =,
解 (1)
21
33
2
()
3
yx x
′′==; (2)
16 11
55
16
()
5
yx x′′==;
3
(3)
17
66
1
()
6
yx x

′′==?; (4)
2222
[(e ) ] (e ) ln e 2e
x xx
y′′== =,
8,如果()f x为偶函数,且(0)f ′存在,证明(0) 0f ′ =,
证 因为
00
() (0) ()
(0) (0) lim lim
0
xx
f xf fx
ff
x x
++
+
→→
′′== =
,
() (0) ()
(0) (0) lim lim
0
f xf fx
ff
x x

′′== =?

,
所以(0) (0)ff′′=?,即(0) 0f ′ =,
9,求曲线e
x
y =在点(0,1)处的切线方程,
解 曲线e
x
y =在点(0,1)处的切线的斜率为
0
0
e1
x
x
x
y
=
=
′ = =,切线的方程为
1( 0)yx?=?,即10xy?+=,
10,在抛物线
2
yx=上取横坐标为
1
1x =及
2
3x =的两点,作过这两点的割线,
问该抛物线上哪一点的切线平行于这条割线?
解 割线的斜率为
22
31
4
31
k
==
,设所求的点为
2
00
(,)x x,由题意知
0
0
24
xx
yx
=
′ ==,故而所求的点为(2,4),
11,讨论下列函数在点0x =处的连续性与可导性,
(1) sinyx= ; (2)
2
1
sin,0,
0,0.
xx
y x
x
≠
=
=
解 (1) sinyx=在0x =处连续,因为
00
sin sin 0
sin
(0) lim lim 1
0
xx
x
x
f
xx
++
+
→→
′ ===
,
00
sin sin 0
sin
(0) lim lim 1 (0)
0
xx
x
x
ff
xx
+
→→
′′==?=?≠
,
所以sinyx=在0x =处不可导,
(2) 由于
2
0
1
lim sin 0
x
x
x
→
=,所以函数在0x =处连续,另外
2
00
1
sin 0
1
(0) lim lim sin 0
0
xx
x
x
fx
xx
→→
′ ===
,
4
故而()f x在0x =处可导,
12,设函数
2
,1,
()
,1.
xx
fx
ax b x
≤?
=
+ >?
为了使函数在点1x =处连续且可导,常数,ab应取什么值?
解 因为
1
(1 0) lim
x
f ax b a b
+
→
+= +=+,
2
1
(1 0) lim 1
x
fx
→
==,
所以由函数在点1x =处连续条件知,1ab+ =,即1ba? =?,另外
00
1
(1) lim lim
11
xx
ax b ax a
f a
xx
++
+
→→
+
′ ===

,
2
1
1
(1) lim 2
1
x
x
f
x
→
′ = =
,
所以由函数在点1x =处可导的条件知,2,1ab= =?,
13,已知
2
,0,
()
,0,
xx
fx
xx
≥?
=
<?
求(0)f
+
′及(0)f
′,又(0)f ′是否存在?
解 由于
2
00
(0) lim 0,(0) lim 1
xx
ff
+?
+?
→→

′′====?,所以(0)f ′不存在,
14,已知
sin,0,
()
,0,
xx
fx
xx
<?
=
≥
求()f x′,
解 当0x <时,( ) (sin ) cosf xxx′′==; 当0x >时,() () 1fx x′ ′= = ; 当0x =时,
00
0sin0
(0) lim 1,(0) lim 1
xx
ff
+?
+?
→→

′′=== =,从而(0) 1f ′ =,
故而
cos,0,
()
1,0.
xx
fx
x
<?
′ =
≥