(完整版)高中导数的概念与计算练习题带答案
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导数概念与计算
1.若函数42()f x ax bx c =++,满足'(1)2f =,则'(1)f -=( )
A .1-
B .2-
C .2
D .0
2.已知点P 在曲线4()f x x x =-上,曲线在点P 处的切线平行于直线30x y -=,则点P 的坐标为( )
A .(0,0)
B .(1,1)
C .(0,1)
D .(1,0)
3.已知()ln f x x x =,若0'()2f x =,则0x =( )
A .2e
B .e
C .
ln 2
2
D .ln2
4.曲线x y e =在点(0,1)A 处的切线斜率为( ) A .1
B .2
C .e
D .1e
5.设0()sin f x x =,10()'()f x f x =,21()'()f x f x =,…,1()'()n n f x f x +=,n N ∈,则2013()f x =
等于( )
A .sin x
B .sin x -
C .cos x
D .cos x -
6.已知函数()f x 的导函数为'()f x ,且满足()2'(1)ln f x xf x =+,则'(1)f =( )
A .e -
B .1-
C .1
D .e
7.曲线ln y x =在与x 轴交点的切线方程为________________.
8.过原点作曲线x y e =的切线,则切点的坐标为________,切线的斜率为____________. 9.求下列函数的导数,并尽量把导数变形为因式的积或商的形式: (1)1
()2ln f x ax x x
=--
(2)2
()1x
e f x ax =+
(3)21
()ln(1)2
f x x ax x =--+
(4)cos sin y x x x =-
(5)1cos x
y xe
-=
(6)1
1
x x e y e +=-
10.已知函数()ln(1)f x x x =+-. (Ⅰ)求()f x 的单调区间; (Ⅱ)求证:当1x >-时,1
1ln(1)1
x x x -≤+≤+.
11.设函数()b
f x ax x =-,曲线()y f x =在点(2,(2))f 处的切线方程为74120x y --=.
(Ⅰ)求()f x 的解析式;
(Ⅱ)证明:曲线()y f x =上任一点处的切线与直线0x =和直线y x =所围成的三角形
面积为定值,并求此定值.
12.设函数2()x x f x x e xe =+-. (Ⅰ)求()f x 的单调区间;
(Ⅱ)若当[2,2]x ∈-时,不等式()f x m >恒成立,求实数m 的取值范围.
导数作业1答案——导数概念与计算
1.若函数42()f x ax bx c =++,满足'(1)2f =,则'(1)f -=( )
A .1-
B .2-
C .2
D .0
选B .
2.已知点P 在曲线4()f x x x =-上,曲线在点P 处的切线平行于直线30x y -=,则点P 的坐标为( )
A .(0,0)
B .(1,1)
C .(0,1)
D .(1,0)
解:由题意知,函数f (x )=x 4-x 在点P 处的切线的斜率等于3,即f ′(x 0)=4x 30-1=3,∴x 0=1,将其代入f (x )中可得P (1,0). 选D .
3.已知()ln f x x x =,若0'()2f x =,则0x =( )
A .2e
B .e
C .
ln 2
2
D .ln2
解:f (x )的定义域为(0,+∞), f ′(x )=ln x +1,由f ′(x 0)=2, 即ln x 0+1=2,解得x 0=e. 选B .
4.曲线x y e =在点(0,1)A 处的切线斜率为( ) A .1
B .2
C .e
D .1
e
解:∵y ′=e x ,故所求切线斜率k =e x |x =0=e 0=1. 选A .
5.设0()sin f x x =,10()'()f x f x =,21()'()f x f x =,…,1()'()n n f x f x +=,n N ∈,则2013()f x =
等于( )
A .sin x
B .sin x -
C .cos x
D .cos x -
解:∵f 0(x )=sin x ,f 1(x )=cos x ,
f 2(x )=-sin x ,f 3(x )=-cos x ,f 4(x )=sin x ,… ∴f n (x )=f n +4(x ),故f 2 012(x )=f 0(x )=sin x , ∴f 2 013(x )=f ′2 012(x )=cos x . 选C .
6.已知函数()f x 的导函数为'()f x ,且满足()2'(1)ln f x xf x =+,则'(1)f =( )
A .e -
B .1-
C .1
D .e
解:由f (x )=2xf ′(1)+ln x ,得f ′(x )=2f ′(1)+1x
,