精選基本初等函數(shù)的導(dǎo)數(shù)公式及導(dǎo)數(shù)運(yùn)算法則測(cè)試題

思而思學(xué)網(wǎng)

一、選擇題

1.函數(shù)y=(x+1)2(x-1)在x=1處的導(dǎo)數(shù)等于()

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.若對(duì)任意xR,f(x)=4x3,f(1)=-1,則f(x)=()

A.x4 B.x4-2

C.4x3-5 D.x4+2

[答案] B

[解析] ∵f(x)=4x3.f(x)=x4+c,又f(1)=-1

1+c=-1,c=-2,f(x)=x4-2.

3.設(shè)函數(shù)f(x)=xm+ax的導(dǎo)數(shù)為f(x)=2x+1,則數(shù)列{1f(n)}(nN)的前n項(xiàng)和是()

A.nn+1 B.n+2n+1

C.nn-1 D.n+1n

[答案] A

[解析] ∵f(x)=xm+ax的導(dǎo)數(shù)為f(x)=2x+1,

m=2,a=1,f(x)=x2+x,

即f(n)=n2+n=n(n+1),

數(shù)列{1f(n)}(nN)的前n項(xiàng)和為:

Sn=112+123+134+…+1n(n+1)

=1-12+12-13+…+1n-1n+1

=1-1n+1=nn+1,

故選A.

4.二次函數(shù)y=f(x)的圖象過(guò)原點(diǎn),且它的導(dǎo)函數(shù)y=f(x)的圖象是過(guò)第一、二、三象限的一條直線(xiàn),則函數(shù)y=f(x)的圖象的頂點(diǎn)在()

A.第一象限 B.第二象限

C.第三象限 D.第四象限

[答案] C

[解析] 由題意可設(shè)f(x)=ax2+bx,f(x)=2ax+b,由于f(x)的圖象是過(guò)第一、二、三象限的一條直線(xiàn),故2a0,b0,則f(x)=ax+b2a2-b24a,

頂點(diǎn)-b2a,-b24a在第三象限,故選C.

5.函數(shù)y=(2+x3)2的導(dǎo)數(shù)為()

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)若函數(shù)f(x)=ax4+bx2+c滿(mǎn)足f(1)=2,則f(-1)=()

A.-1 B.-2

C.2 D.0

[答案] B

[解析] 本題考查函數(shù)知識(shí),求導(dǎo)運(yùn)算及整體代換的思想,f(x)=4ax3+2bx,f(-1)=-4a-2b=-(4a+2b),f(1)=4a+2b,f(-1)=-f(1)=-2

要善于觀察,故選B.

7.設(shè)函數(shù)f(x)=(1-2x3)10,則f(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.函數(shù)y=sin2x-cos2x的導(dǎo)數(shù)是()

A.22cos2x- B.cos2x-sin2x

C.sin2x+cos2x D.22cos2x+4

[答案] A

[解析] y=(sin2x-cos2x)=(sin2x)-(cos2x)

=2cos2x+2sin2x=22cos2x-4.

9.(2010高二濰坊檢測(cè))已知曲線(xiàn)y=x24-3lnx的一條切線(xiàn)的斜率為12,則切點(diǎn)的橫坐標(biāo)為()

A.3 B.2

C.1 D.12

[答案] A

[解析] 由f(x)=x2-3x=12得x=3.

10.設(shè)函數(shù)f(x)是R上以5為周期的可導(dǎo)偶函數(shù),則曲線(xiàn)y=f(x)在x=5處的切線(xiàn)的斜率為()

A.-15 B.0

C.15 D.5

[答案] B

[解析] 由題設(shè)可知f(x+5)=f(x)

f(x+5)=f(x),f(5)=f(0)

又f(-x)=f(x),f(-x)(-1)=f(x)

即f(-x)=-f(x),f(0)=0

故f(5)=f(0)=0.故應(yīng)選B.

二、填空題

11.若f(x)=x,(x)=1+sin2x,則f[(x)]=_______,[f(x)]=________.

[答案] 2sinx+4,1+sin2x

[解析] f[(x)]=1+sin2x=(sinx+cosx)2

=|sinx+cosx|=2sinx+4.

[f(x)]=1+sin2x.

12.設(shè)函數(shù)f(x)=cos(3x+)(0<),若f(x)+f(x)是奇函數(shù),則=________.

[答案] 6

[解析] f(x)=-3sin(3x+),

f(x)+f(x)=cos(3x+)-3sin(3x+)

=2sin3x++56.

若f(x)+f(x)為奇函數(shù),則f(0)+f(0)=0,

即0=2sin+56,+56=kZ).

又∵(0,),6.

13.函數(shù)y=(1+2x2)8的導(dǎo)數(shù)為_(kāi)_______.

[答案] 32x(1+2x2)7

[解析] 令u=1+2x2,則y=u8,

yx=y(tǒng)uux=8u74x=8(1+2x2)74x

=32x(1+2x2)7.

14.函數(shù)y=x1+x2的導(dǎo)數(shù)為_(kāi)_______.

[答案] (1+2x2)1+x21+x2

[解析] y=(x1+x2)=x1+x2+x(1+x2)=1+x2+x21+x2=(1+2x2)1+x21+x2.

三、解答題

15.求下列函數(shù)的導(dǎo)數(shù):

(1)y=xsin2x;(2)y=ln(x+1+x2);

(3)y=ex+1ex-1;(4)y=x+cosxx+sinx.

[解析] (1)y=(x)sin2x+x(sin2x)

=sin2x+x2sinx(sinx)=sin2x+xsin2x.

(2)y=1x+1+x2(x+1+x2)

=1x+1+x2(1+x1+x2)=11+x2 .

(3)y=(ex+1)(ex-1)-(ex+1)(ex-1)(ex-1)2=-2ex(ex-1)2 .

(4)y=(x+cosx)(x+sinx)-(x+cosx)(x+sinx)(x+sinx)2

=(1-sinx)(x+sinx)-(x+cosx)(1+cosx)(x+sinx)2

=-xcosx-xsinx+sinx-cosx-1(x+sinx)2.

16.求下列函數(shù)的導(dǎo)數(shù):

(1)y=cos2(x2-x); (2)y=cosxsin3x;

(3)y=xloga(x2+x-1); (4)y=log2x-1x+1.

[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=(cosxsin3x)=(cosx)sin3x+cosx(sin3x)

=-sinxsin3x+3cosxcos3x=3cosxcos3x-sinxsin3x.

(3)y=loga(x2+x-1)+x1x2+x-1logae(x2+x-1)=loga(x2+x-1)+2x2+xx2+x-1logae.

(4)y=x+1x-1x-1x+1log2e=x+1x-1log2ex+1-x+1(x+1)2

=2log2ex2-1.

17.設(shè)f(x)=2sinx1+x2,如果f(x)=2(1+x2)2g(x),求g(x).

[解析] ∵f(x)=2cosx(1+x2)-2sinx2x(1+x2)2

=2(1+x2)2[(1+x2)cosx-2xsinx],

又f(x)=2(1+x2)2g(x).

g(x)=(1+x2)cosx-2xsinx.

18.求下列函數(shù)的導(dǎo)數(shù):(其中f(x)是可導(dǎo)函數(shù))

(1)y=f1x;(2)y=f(x2+1).

[解析] (1)解法1:設(shè)y=f(u),u=1x,則yx=y(tǒng)uux=f(u)-1x2=-1x2f1x.

解法2:y=f1x=f1x1x=-1x2f1x.

(2)解法1:設(shè)y=f(u),u=v,v=x2+1,

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