导数与微分 习题

(2)
1 2x 1 2x sec sec 1 2+ 4 2 = csc x + y′ = 2 . 1 x x x tan 1 + tan 2 1 + 3cos 2 2 4 2 2
(3) (4)
y ′ = n sin n −1 xcox cos nx − n sin n x sin nx = n sin n −1 x cos(n + 1) x .
= 16 (3 3 − π ) . 9
解
(1)
f ′(ϕ ) = 3cos 3ϕ +
(2)
y′ = −
1 2 arcsin 2 x + , y′ x= 2 x x 1 − 4 x2
3 4
(3)
y ′ = −15e −5 x + 5, y ′ x =−1 = 5(1 − 3e5 ) .
3
7. (1) (3)
y = arcsin
1
(5)
y=
(6)
y=
x + 1 + x2
(7) (9)
解
y = x arcsin
y = ln chx +
(8) (10)
2 y = sh ch3x ; x y = a a + xa + a x .
x a a
(1)
y′ =
sec x tan x + sec2 x = sec x . sec x + tan x
切线方程为: y = x − e , 法线方程为:
y = −x + e .
5.
(1)
求下列函数的导数:
y = e−3 x ;
2
(2) (4)
(6) (8)
2 2
y = cos(4 − 3x 2 ) ; y = (arcsin x) 2 ;
y = cos 2 (tan 3 x) ; y=xx.
(3) (5)
s′ = (1 + sin t )′(1 + cos t ) − (1 + sin t )(1 + cos t )′ 1 + cos t + sin t = . (1 + cos t )2 (1 + cos t ) 2 1 2 gt , 2
以初速 v0 竖直上抛的物体, 其上升高度 s 与时间 t 的关系是 s = v0 t −
第二节
导数的运算法则
习题 2-2
1. (1) 解 (2) 2. (1) (3) (5)
推导余切函数及余割函数的导数公式: (cot x)′ = − csc2 x ; (1)
(cot x)′ = (
(2)
(csc x)′ = − csc x cot x .
cos x (cos x)′ sin x − (sin x)′ cos x 1 )′ = = − 2 = − csc2 x . 2 sin x sin x sin x
2 arcsin x 1 − x2
2
(4)
y ′ = 2(arcsin x)(arcsin x)′ =
.
(5) (6)
y ′ = a tan x ln a(tan x 2 )′ = a tan x ln a sec2 x 2 ( x 2 )′ = 2ln a ⋅ x sec2 x 2 a tan x . y ′ = 2cos(tan 3 x)[cos(tan 3 x)]′ = −2cos(tan 3 x)sin(tan 3 x)(tan 3 x)′ = −3sin(2 tan 3 x) tan 2 x(tan x)′ = −3tan 2 x sec2 x sin(2 tan 3 x) .
(2)
y ′ = f ′(sin 2 x)2sin x cos x − f ′(cos 2 x)2sin x cos x = sin 2 x[ f ′(sin 2 x) − f ′(cos 2 x)] . f ′( x ) 1 f ( x) 2 x g ′( x 2 )2 x f ′( x ) 2 xg ′( x 2 ) = + . 1 + g 2 ( x2 ) 2 x f ( x ) 1 + g 2 ( x2 )
x a
y ′ = a a ln a (a x )′ + a a x a
−1
+ a x ln a ( x a )′
4
a
= ln 2 aa x a a + a a x a
x
a
−1
+ a ln ax a −1a x .
a
8. (1) (3)
解
设 f ( x) 和 g ( x) 都可导, 求下列函数 y 的导数
y′ = −(1 + x) − (1 − x) 1 . =− 2 (1 + x) 1− x 1− x (1 + x) 2 x(1 − x) 1− 2 1+ x 1+ x 1 1
(5)
y′ =
2 x cos x 2 sin 2 x − sin x 2 2sin x cos x 2 x cos x 2 sin x − 2sin x 2 cos x = . sin 4 x sin 3 x 1 ( x + 1 + x 2 )2 (1 + 2x 2 1 + x2 )=− 1 1 + x2 ( x + 1 + x2 ) .
(1) 该物体的速度 v(t ) ; (2) 该物体到达最高点的时刻.
解
(1) v(t ) = s ′ = v0 − gt .
v0 . g
(2) 物体到达最高点时, v(t ) = 0 , 即 v0 − gt = 0 , 从而 t = 4.
解
求曲线 y = x(ln x − 1) 上横坐标为 x = e 的点处的切线方程和法线方程. 该点为 (e, 0) , 所求切线的斜率为 y ′ x =e = (ln x − 1 + 1) x =e = ln x x =e = 1 , 从而
(6)
y′ = −
(7)
y ′ = arcsin
x 1 +x 2 2
1 1− x 4
2
+
x = arcsin . 2 2 4−x
2
−2 x
(8) (9) (10)
2 2 2 2 2 2 y ′ = ch (− 2 )ch3x + 3sh sh3x = − 2 ch ch3 x + 3sh sh3x . x x x x x x y′ = shx shx 1 −2 = thx(1 − 2 ) = th 3 x . 3 chx 2ch x ch x
(3)
y′ =
+
2 f ( x) f ′( x) + (4) y′ =
2
g ′( x) 2 g ( x)
2 f ( x) + g ( x)
=
4 f ( x) f ′( x) g ( x) + g ′( x) 4 f 2 ( x ) + g ( x ) g ( x)
.
9. 设 f ( x) 在 (−l , l ) 内可导, 证明: 如果 f ( x) 是偶函数, 则 f ′( x) 是奇函数; 如 果 f ( x) 是奇函数, 则 f ′( x) 是偶函数. 证 如 果 f ( x) 是 偶 函 数 , 则 有 f (− x) = f ( x) , 对 等 式 两 边 对 x 求 导 , ′ 有, − f (− x) = f ′( x) , 从而 f ′(− x) = − f ′( x) , 即 f ′( x) 是奇函数. 如 果 f ( x ) 是 奇 函 数 , 则 有 f ( − x) = − f ( x) , 对 等 式 两 边 对 x 求 导 , 有, − f ′(− x) = − f ′( x) , 从而 f ′(− x) = f ′( x) , 即 f ′( x) 是偶函数.
5
(7)
解
y = arctan(e x ) ;
y = a tan x (a > 0) ;
y=2
sin 2 1 x
2
;
(1)
y ′ = e−3 x (−3x 2 )′ = −6 xe−3 x .
2
(2) (3)
y ′ = − sin(4 − 3x 2 )(4 − 3 x 2 )′ = 6 x sin(4 − 3x 2 ) . y′ = (e x )′ ex = . 1 + e2 x 1 + e2 x
sin 2 1 x sin 2 1 1 1 ln 2(sin 2 )′ = 2 x 2 ln 2sin (sin )′ x x x 1
2
2
(7)
y′ = 2
=2
sin 2
1 x 2ln 2sin
1 1 1 ln 2 sin 2 2 cos ( )′ = − 2 2 x sin . x x x x x 1 1 1 1 − ln x x ( ln x)′ = x x (− 2 ln x + 2 ) = x. x x x x2
(5)
y ′ = ( x3 )′ log 2x + x3 (log 2 x )′ = 3x 2 log 2x + x3
1 1 = x 2 (3log 2x + ). x ln 2 ln 2
1
(6) (7)
y′ =
(ln x)′ x − ln x 1 − ln x . = x2 x2
y ′ = ( x 2 )′ ln x cos x + x 2 (ln x)′ cos x + x 2 ln x(cos x)′
1
1
(8) 6. (1) (2) (3)