复习题二

(8) 设y = f ( ln x ) e
f ( x)
2
, 且f 可微,求dy.
1 f ( x) ′ f ( x) 解 dy = f ( ln x ) ⋅ e + f ( ln x ) e f ′ ( x ) dx x 1 f ( x) = e f ′ ( ln x ) ⋅ + f ( ln x ) f ′ ( x ) dx. x
(4) 设f ( x) 可导, F ( x) = f ( x) (1+ sin x ) ,则f ( 0) = 0是F ( x) 在x = 0处可导的( C) A.充分非必要条件 C.充分且必要条件 B.必要非充分条件 D.既非充分也非必要条件.
A. 16!
B.
15!
C. 14!
D. 0
(Q F ( x ) = f ( x ) (1 + sin x ) ,
F ( 0 + ∆x ) − F ( 0) f ( ∆x )(1 + sin ∆x ) − f ( 0) = lim+ = lim+ ∆x→0 ∆x→0 ∆x ∆x F ( ∆x ) f ( 0 ) = 0, ⇔ lim ∃ ∆x → 0 ∆x (5) 若f ( x ) 可导,且y = f ( e x ) , 则dy = ( B ) .
(1 + x ) (1 + x ) 1 f ′′ ( x ) = −2 ⋅ ( −2 ) , 3 (1 + x ) 1
2
f ′( x) =
− (1 + x ) − (1 − x )
( −1) 2 ⋅ n ! . n +1 (1 + x )
n
=
−2
2
,
LLL
f
(n)
(1 + x )
4
,
( −1) 2 ⋅ n ! . ( x) = n +1 (1 + x )
x = cos t + cos2 t, π (2)曲线 上在对应t = 点处的法线斜率为( A ) 4 y = 1+ sin t A.1+ 2 B. 2 −1 C. −1− 2 D.1− 2. 1 dy cos t 1 2 = π = π = , k法 = 1 + 2. dx t = 4 − sin t − sin 2t t = 4 −1 − 1 −1 − 2 2 16 (3) 设f ( x ) = x15 + 4 x5 − 2 x + 1, 则f ( ) (1) = ( D )
f ( e x ) ′ dx D. f ′ ( e x ) e x C.
(1) 设曲线f ( x) = xn在点(1,1) 处切线交x轴于点( xn ,0) ,求lim f ( xn ) ;
3.解答题 解答题
n −1 ∴ lim f ( xn ) = lim n →∞ n →∞ n − n −1 1 = lim 1 − = e −1. n →∞ n
d 1 1 1 (3) 设 f 2 = , 则f ′ = dx x x 2
1 −1 f ′ = = −1. 2 2⋅ 1 2
1− x (n) (4) 设f ( x ) = , 则f ( x ) = 1+ x
f ′′′ ( x ) = −2 ⋅ ( −2 ) ⋅ ( −3)
2.选择题 (1) 设函数f ( x ) 在x = 0处连续, 则下列命题错误的是(
=−πdx.
D
)
f ( x ) − f ( 0 ) − [ f ( − x ) − f ( 0 )] D中原式 = lim x →0 x
f ( x ) − f ( 0) f ( − x ) − f ( 0) = lim + lim = 2 f ′( x). x →0 x →0 x −x
f ( 0 + ∆x ) − f ( 0 ) = lim− 1 + e = 1, ∆x → 0 解 f −′ ( 0 ) = lim− ∆x 0 + ∆x ∆x → 0 ∆x 1 f ( 0 + ∆x ) − f ( 0 ) 1 + e 0+∆x = 0. f +′ ( 0 ) = lim+ = lim ∆x → 0 ∆x → 0 ∆x ∆x
∴ f ′ ( 0 ) 不存在. x = 1+ t2, d 2 y (6) 设 求 2; y = cos t , dx
d y (7) 设y = f ( x + y ) , f 具有二阶导数,且f ′ ≠ 1, 求 2 . dx f′ 解 y′ = f ′ ( x + y ) ⋅ (1 + y′ ) , ⇒ y′ = , 1− f ′ 2 y′′ = f ′′ ( x + y )(1 + y′ ) + f ′ ( x + y ) y′′, 2 f ′′ f ′′ f′ f ′′ 2 . y′′ = (1 + y′ ) = 1 + 1 − f ′ = 3 1− f ′ 1− f ′ (1 − f ′ )
将x = 0, y = 1代入方程 :1 − y′ 1 x
1 x
, x ≠ 0, x = 0.
f ( x ) 在x = 0处的可导性. 0 + ∆x
1 0 +∆x
求f −′ ( 0 ) 及f +′ ( 0 ) , 并确定
x =0 y =1
−1 = 1, ⇒ y′
x =0 y =1
=1
1 1 = + sin x ln tan x − = sin x ln tan x. sin x sin x
1 方程两边对x求导得 :cos(xy) ⋅ ( y + xy′) − ⋅ ( y′ −1) =1, y−x 1
(5) 设f ( x ) = 1 + e 0,
(4) 设方程sin ( xy) − ln( y − x) = x确定隐函数y = y ( x) , 求y′( 0) ; 解 当x = 0, 代入方程得:sin 0 − ln y = 0, y = 1,
+
dy ( cos t )′ − sin t 解 = = , dx 1 + t 2 ′ ( ) 2t dy − sin t ′ d 2 d y dx = 2t = 2 dx dx 2 ′ (1 + t )
−2t cos t + 2sin t sin t − t cos t = = . 3 3 8t 4t
e + 1+ e
x
2x
(e +
x
1+ e
) , 求y′ ′;
2x
)
x y′ = (ln tan )′ − (cos x ⋅ ln tan x)′ 2 2 ′ sec x 1 x = tan − − sin x ln tan x + cos x x 2 tan x tan 2 1 x ⋅ cos 1 2 2 x 2 cos cos 2 x ⋅ cos x 2 = + sin x ln tan x − cos x x sin x sin 2
dy (2) 设y = arctan e − ln 2 x , 则 e +1 dx
x
e
2x
e −1 x =1 = 2 1+ e
= arctan e − [2 x − ln ( e + 1)] 2
x 2x
e 1 e −1 = − 2 = . 2 2 1+ e e +1 1+ e
d 1 1 −2 1 f 2 = f ′ 2 ⋅ 3 = x x x dx x 2 −1 −1 1 −x f ′ 2 = = , ⇒ f ′ (u ) = 1 2 2u x 2⋅ 2 x
F ( 0 + ∆x ) − F ( 0 ) f ( ∆x )(1 − sin ∆x ) − f ( 0 ) = lim− lim− ∆x →0 ∆x →0 ∆x ∆x
A. f ′ ( e ) dx B. f ′ ( e ) de
x x
x
f ( e x ) = f ′ ( e x ) de x . Qd
1.填空题
复习题二
x (1) 设f ′ ( x0 ) = −1, 则 lim = x →0 f ( x − 2 x ) − f ( x − x ) 0 0
1
1 (原式 = lim x →0 f x − 2 x − f x ( 0 ) ( 0 ) − f ( x0 − x ) − f ( x0 ) x 1 = f ( x0 − 2 x ) − f ( x0 ) f ( x0 − x ) − f ( x0 ) (−2) lim + lim x →0 x →0 −2 x −x
4.证明题证明曲线 x + y = a ( a为常数) 上任一点的切线在 : 两坐标轴上的截距之和为常数. 证明 设切点为( x0 , y0 ) ,
对方程两边关于x求导得 + ⋅ y′ = 0, 2 x 2 y y0 y0 切线斜率:k = − , 切线方程为y = − ( x − x0 ) + y0 , x0 x0 y0 令− ( x − x0 ) + y0 = 0,得切线横截距x = x0 y0 + x0 , x0
1 = = 1. ( −2 ) ⋅ ( −1) + ( −1)
1 x 2x 2x y = arctan e − [ln ( e ) − ln ( e + 1)] 2 1