第一节 导数概念
1、填空题
2
(1)f (0) (2)4x 12x 9 (3)16 (4)y 2x 1 (5)e
2、选择题
(1)C (2)B (3)D (4)D (5)B 3、a 2,b 1
4、y f (x) nxn 1,k f'(1) n,切线方程为y 1 n(x 1) 由于切线过点( n,0),故0 1 n( n 1),解之得 n 1
limf( n) lim(1
n
1n
,从而f( n) (1
1n
),即
n
1n
n
) 1x1x
n
1e
1x
5、x 0,f (x) 2xsin
x 0,f (x) 2xcos
cos
2
1x (
1x
) 2xcos2
1x sin
1x
xsin
x 0,按左右导数来求
f(x) f(0)
x 0f(x) f(0)
x
xsin lim
x 0
22
1 0 1 0
f (0) lim
x 0
xxcos
f (0) lim
x 0
lim
x 0
x
11
2xsin cos,x 0 xx
x 0所以f(x) 0
11
2xcos sin,x 0
xx
6、f (0) lim
f( x) f(0)
x
1
( x)sin
lim
x 0
1
x 0
x
lim( x)
x 0
sin
1 x
所以 1时,f (0) 0
7、解:f (a) lim
f(x) f(a)
x a
x a
lim
(x a) (x)
x a
x a
(a)(因为 (x)在x a处连续)
g (a) lim
g(x) g(a)
x a
x a
lim
x a (x)x a
x a