高数第二章自测题答案

二、1. y = x2 , 求dy. 1+ x
dy
=
2x + x2 (1 + x)2
dx.
( ) 2. y = tan e−2x + 1 + cos π ,求y′. 4
y′ = −2e−2 x ⋅ sec2 (e−2 x + 1).
导数为零
3. y = x a2 − x2 + a2 arcsin x (a > 0),求dy.
lim
x→0−
e ax
−1−
b x
+
a
+
2
= lim eax − 1 = lim ax = a,
x x→0−
x x→0−
f +′ (0)
=
lim
x→0+
b(1 +
sin
x)+
a+ x
2
−
(b
+
a
+
2)
=
lim
x→0+
b sin x
x
=
b,
Q可导, ∴a = b
(2)
由(1)(2)得 a = b = −1.
⎤ f (t)⎥
⎥
a h→0 ⎢
h
⎢⎣
a
−h a
⎥ ⎥⎦
= 1 [ f ′(t) + f ′(t)] = 2 f ′(t)
a
a
5.曲线y = xe− x在x = 1处的切线方程是
法线方程是 x = 1
y − 1 = 0, e
6. xe x的n阶导数是
( ) ( ) ( ) xe x (n) = e x (n) x + n e x (n−1) ⋅1 = ( x + n)e x
Q lim x→0+
f ′(x) =
lim
x→0+
⎡ ⎢⎣
2
x
sin
1 x
− cos
1 x
⎥⎦⎤,
不存在,
∴ f+′(0)不存在.
f
(x) =
⎧ ⎪
x
2
⎨ห้องสมุดไป่ตู้
sin
1 x
,
x>0
f+′(0) =
⎪⎩ x 3 , f
lim
x→0+
( x) − x
x≤0 f (0) = lim
x→0+
x2
sin 1 x
x
或求1~3阶导数,找规律.
[ ] 7.设f (x)可导， F (x) = f f 2 (sin x) ,则
[ ] dF = f ′ f 2 (sin x) ⋅ 2 f (sin x)⋅ f ′(sin x)⋅ cos x
dx
8.设f
(x)
=
⎧b(1
⎨ ⎩e
ax
+ −
sin 1,
x)+
a
+
2,
x x
≥ <
⎪⎩3 x2 ,
x>0 .
x≤0
由于可导必连续 , 所以 f ( x)在(−∞ ,+∞ )内连续 .
x→1 x − 1
七、研究函数
f
(x)
=
⎧ ⎪
x
2
⎨
sin
1 x
,
x>0 ,
⎪⎩ x 3 ,
x≤0
在(− ∞,+∞ )内的连续性与可导性， 并求出 f ′(x).
x
>
0时,
f
′(x)
=
2x
sin
1 x
+
x2
cos
1 x
⋅ ⎜⎛ ⎝
− x
1
2
⎟⎞, ⎠
x < 0时, f ′(x) = 3x2 ,
x = 0时,
(1)式整理, 得
注：此题不用解出
y′ = e xy (1 + xy + x2 y′)
y′ = L,
两边对x求导, 得
y′′ = L.
y′′ = e xy ( y + xy′)(1 + xy + x2 y′) + e xy ( y + xy′ + 2xy′ + x2 y′′)
将x = 0, y = 1, y′ = 1代入上式,得
1 y′x − y = 1 2x + 2 yy′
1 + ⎜⎛ y ⎟⎞2 x2
2 x2 + y2
⎝ x⎠
⇒ y′x − y = x + yy′, (1)
⇒ dy = x + y , dx x − y
(1)式两边对x求导,得y′′x + y′ − y′ = 1 + ( y′)2 + yy′′
( ) ⇒
或 ln y = sin x ln cos x y′ = cos x ln cos x − sin x tan x y
y′ = (cos x)sin x (cos x ln cos x − sin x tan x)
5.
设 arctan y = ln x
x2
+
y2
,求
dy dx
,
d2y dx 2
.
( ) 将方程整理得 arctan y = 1 ln x 2 + y2 , x2 方程两边对 x求导, 得
高等数学第二章自测题解答
一、1.
y=
f
(
x
)在点
x
可微的定义是
0
Δy = A⋅ Δx + o(Δx) (Δx → 0)
2. 函数可导与连续的关系是
可导
连续
函数可导与可微的关系是 可导 ⇔ 可微
3. d ( 4 tan x + C ) = sec2 x dx
4
4
4. 设f
(u)可导, t, a与h无关,则lim h→0
f ′(−1) = −2 + a, g′(−1) = −2b,
⎧− 2 + a = −2b
则⎪⎨1 − a = 0 ⎪⎩b + c = 0
⇒ a = 1, b = 1 , c = − 1 ,
2
2
f ′(−1) = −1,
切线方程为 y = −( x + 1),即y + x + 1 = 0.
( ) 四、设⎨⎧x ( ) ⎩ y
0 处处可导, 则a 0
=
__,
b
=
__
.
( ) Q可导 ⇒ 连续, 而
f (0− ) = lim eax − 1 = 0, x→0−
f (0+ ) = lim [b(1 + sin x)+ a + 2] = b + a + 2, x→0+
∴b + a + 2 = 0,
(1)
( ) 又f−′(0)
=
−0
=
0,
f −′ (0)
=
lim
x→0−
f (x)− x
f (0)
= lim x3 − 0 = 0, x x→0−
f−′(0) = f+′(0) = 0, ∴ f ′(0) = 0,
故f ( x)在(−∞,+∞)内可导,且
f
′(
x)
=
⎪⎧2 ⎨
x
sin
1 x
+
x2
cos
1 x
⋅
⎜⎛ ⎝
−1 x2
⎟⎞, ⎠
= =
t 2 − arctan ln 1 + t 4
t2
, 确定x是y的函数，求
d2x dy 2
.
[ ( )] dx = [ ( )] dy
t 2 − arctan t 2 ln 1 + t 4 ′
′
=
2t
−
1
2t +t
4
4t 3
=
t2 2
,
1+ t4
[ ( )] d 2 x =
⎜⎜⎝⎛
t2 2
⎟⎟⎠⎞′
1 h
⎡ ⎢ ⎣
f
⎜⎛ t ⎝
+
h ⎟⎞ − a⎠
f
⎜⎛ t ⎝
−
h a
⎟⎠⎞⎥⎦⎤
=
f ⎜⎛ t + h ⎟⎞ − f (t) + f (t) − f ⎜⎛ t − h ⎟⎞
= lim ⎝ a ⎠
⎝ a⎠
h→0
h⋅a
a
=
⎡
1
⎢ lim⎢
f ⎜⎛ t + h ⎟⎞ − ⎝ a⎠
f (t) +
f ⎜⎛ t − h ⎟⎞ − ⎝ a⎠
2
2
a
dy = a2 − x2 dx.
4. y = (cos )x sin x , 求y′.
y = esin x lncos x ,
y′ = esin xlncos x ⎜⎛ cos x ln cos x + sin x − sin x ⎟⎞
⎝
cos x ⎠
= (cos x)sin x (cos x ln cos x − sin x tan x)
y′′ = 1 + 1 = 2. x=0
六、已知函数f (x)在x = 1处连续，且 lim f (x) = 2,求f ′(1).
x→1 x − 1
由于 f ′(1) = lim f (x)− f (1)，