选择题
1.函数y=(x+1)2(x-1)x=1处导数等( )
A.1 B.2
C.3 D.4
[答案] D
[解析] y′=[(x+1)2]′(x-1)+(x+1)2(x-1)′
=2(x+1)·(x-1)+(x+1)2=3x2+2x-1
∴y′|x=1=4
2.意x∈Rf′(x)=4x3f(1)=-1f(x)=( )
A.x4 B.x4-2
C.4x3-5 D.x4+2
[答案] B
[解析] ∵f′(x)=4x3∴f(x)=x4+cf(1)=-1
∴1+c=-1∴c=-2∴f(x)=x4-2
3.设函数f(x)=xm+ax导数f′(x)=2x+1数列{}(n∈N*)前n项( )
A B
C D
[答案] A
[解析] ∵f(x)=xm+ax导数f′(x)=2x+1
∴m=2a=1∴f(x)=x2+x
f(n)=n2+n=n(n+1)
∴数列{}(n∈N*)前n项:
Sn=+++…+
=++…+
=1-=
选A
4.二次函数y=f(x)图象原点导函数y=f′(x)图象第二三象限条直线函数y=f(x)图象顶点( )
A.第象限 B.第二象限
C.第三象限 D.第四象限
[答案] C
[解析] 题意设f(x)=ax2+bxf′(x)=2ax+bf′(x)图象第二三象限条直线2a>0b>0f(x)=a2-
顶点第三象限选C
5.函数y=(2+x3)2导数( )
A.6x5+12x2 B.4+2x3
C.2(2+x3)2 D.2(2+x3)·3x
[答案] A
[解析] ∵y=(2+x3)2=4+4x3+x6
∴y′=6x5+12x2
6.(2010·江西文4)函数f(x)=ax4+bx2+c满足f′(1)=2f′(-1)=( )
A.-1 B.-2
C.2 D.0
[答案] B
[解析] 题考查函数知识求导运算整体代换思想f′(x)=4ax3+2bxf′(-1)=-4a-2b=-(4a+2b)f′(1)=4a+2b∴f′(-1)=-f′(1)=-2
善观察选B
7.设函数f(x)=(1-2x3)10f′(1)=( )
A.0 B.-1
C.-60 D.60
[答案] D
[解析] ∵f′(x)=10(1-2x3)9(1-2x3)′=10(1-2x3)9·(-6x2)=-60x2(1-2x3)9∴f′(1)=60
8.函数y=sin2x-cos2x导数( )
A.2cos B.cos2x-sin2x
C.sin2x+cos2x D.2cos
[答案] A
[解析] y′=(sin2x-cos2x)′=(sin2x)′-(cos2x)′
=2cos2x+2sin2x=2cos
9.(2010·高二潍坊检测)已知曲线y=-3lnx条切线斜率切点横坐标( )
A.3 B.2
C.1 D
[答案] A
[解析] f′(x)=-=x=3
10.设函数f(x)R5周期导偶函数曲线y=f(x)x=5处切线斜率( )
A.- B.0
C D.5
[答案] B
[解析] 题设知f(x+5)=f(x)
∴f′(x+5)=f′(x)∴f′(5)=f′(0)
f(-x)=f(x)∴f′(-x)(-1)=f′(x)
f′(-x)=-f′(x)∴f′(0)=0
f′(5)=f′(0)=0应选B
二填空题
11.f(x)=φ(x)=1+sin2xf[φ(x)]=_______φ[f(x)]=________
[答案] 1+sin2
[解析] f[φ(x)]==
=|sinx+cosx|=
φ[f(x)]=1+sin2
12.设函数f(x)=cos(x+φ)(0<φ<π)f(x)+f′(x)奇函数φ=________
[答案]
[解析] f′(x)=-sin(x+φ)
f(x)+f′(x)=cos(x+φ)-sin(x+φ)
=2sin
f(x)+f′(x)奇函数f(0)+f′(0)=0
0=2sin∴φ+=kπ(k∈Z).
∵φ∈(0π)∴φ=
13.函数y=(1+2x2)8导数________.
[答案] 32x(1+2x2)7
[解析] 令u=1+2x2y=u8
∴y′x=y′u·u′x=8u7·4x=8(1+2x2)7·4x
=32x(1+2x2)7
14.函数y=x导数________.
[答案]
[解析] y′=(x)′=x′+x()′=+=
三解答题
15.求列函数导数:
(1)y=xsin2x (2)y=ln(x+)
(3)y= (4)y=
[解析] (1)y′=(x)′sin2x+x(sin2x)′
=sin2x+x·2sinx·(sinx)′=sin2x+xsin2x
(2)y′=·(x+)′
=(1+)=
(3)y′==
(4)y′=
=
=
16.求列函数导数:
(1)y=cos2(x2-x) (2)y=cosx·sin3x
(3)y=xloga(x2+x-1) (4)y=log2
[解析] (1)y′=[cos2(x2-x)]′
=2cos(x2-x)[cos(x2-x)]′
=2cos(x2-x)[-sin(x2-x)](x2-x)′
=2cos(x2-x)[-sin(x2-x)](2x-1)
=(1-2x)sin2(x2-x).
(2)y′=(cosx·sin3x)′=(cosx)′sin3x+cosx(sin3x)′
=-sinxsin3x+3cosxcos3x=3cosxcos3x-sinxsin3x
(3)y′=loga(x2+x-1)+x·logae(x2+x-1)′=loga(x2+x-1)+logae
(4)y′=′log2e=log2e
=
17.设f(x)=果f′(x)=·g(x)求g(x).
[解析] ∵f′(x)=
=[(1+x2)cosx-2x·sinx]
f′(x)=·g(x).
∴g(x)=(1+x2)cosx-2xsinx
18.求列函数导数:(中f(x)导函数)
(1)y=f(2)y=f().
[解析] (1)解法1:设y=f(u)u=y′x=y′u·u′x=f′(u)·=-f′
解法2:y′=′=f′·′=-f′
(2)解法1:设y=f(u)u=v=x2+1
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