基本初等函數(shù)的導(dǎo)數(shù)公式及導(dǎo)數(shù)運算法則綜合測試題(附答案)

編輯: 逍遙路 關(guān)鍵詞: 高二 來源: 高中學(xué)習(xí)網(wǎng)
選修2-2 1.2.2 第2課時 基本初等函數(shù)的導(dǎo)數(shù)公式及導(dǎo)數(shù)運算法則
一、
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.若對任意x∈R,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)}(n∈N*)的前n項和是(  )
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)}(n∈N*)的前n項和為:
Sn=11×2+12×3+13×4+…+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)的圖象過原點,且它的導(dǎo)函數(shù)y=f′(x)的圖象是過第一、二、三象限的一條直線,則函數(shù)y=f(x)的圖象的頂點在(  )
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)的圖象是過第一、二、三象限的一條直線,故2a>0,b>0,則f(x)=ax+b2a2-b24a,
頂點-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,求導(dǎo)運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-π4 B.cos2x-sin2x
C.sin2x+cos2x D.22cos2x+π4
[答案] A
[解析] y′=(sin2x-cos2x)′=(sin2x)′-(cos2x)′
=2cos2x+2sin2x=22cos2x-π4.
9.(2010?高二濰坊檢測)已知曲線y=x24-3lnx的一條切線的斜率為12,則切點的橫坐標(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y=f(x)在x=5處的切線的斜率為(  )
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、題
11.若f(x)=x,φ(x)=1+sin2x,則f[φ(x)]=_______,φ[f(x)]=________.
[答案] 2sinx+π4,1+sin2x
[解析] f[φ(x)]=1+sin2x=(sinx+cosx)2
=sinx+cosx=2sinx+π4.
φ[f(x)]=1+sin2x.
12.設(shè)函數(shù)f(x)=cos(3x+φ)(0<φ<π),若f(x)+f′(x)是奇函數(shù),則φ=________.
[答案] π6
[解析] f′(x)=-3sin(3x+φ),
f(x)+f′(x)=cos(3x+φ)-3sin(3x+φ)
=2sin3x+φ+5π6.
若f(x)+f′(x)為奇函數(shù),則f(0)+f′(0)=0,
即0=2sinφ+5π6,∴φ+5π6=kπ(k∈Z).
又∵φ∈(0,π),∴φ=π6.
13.函數(shù)y=(1+2x2)8的導(dǎo)數(shù)為________.
[答案] 32x(1+2x2)7
[解析] 令u=1+2x2,則y=u8,
∴y′x=y(tǒng)′u?u′x=8u7?4x=8(1+2x2)7?4x
=32x(1+2x2)7.
14.函數(shù)y=x1+x2的導(dǎo)數(shù)為________.
[答案] (1+2x2)1+x21+x2
[解析] y′=(x1+x2)′=x′1+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+x?2sinx?(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=cosx?sin3x;
(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′=(cosx?sin3x)′=(cosx)′sin3x+cosx(sin3x)′
=-sinxsin3x+3cosxcos3x=3cosxcos3x-sinxsin3x.
(3)y′=loga(x2+x-1)+x?1x2+x-1logae(x2+x-1)′=loga(x2+x-1)+2x2+xx2+x-1logae.
(4)y′=x+1x-1x-1x+1′log2e=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)2?g(x),求g(x).
[解析] ∵f′(x)=2cosx(1+x2)-2sinx?2x(1+x2)2
=2(1+x2)2[(1+x2)cosx-2x?sinx],
又f′(x)=2(1+x2)2?g(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y′x=y(tǒng)′u?u′x=f′(u)?-1x2=-1x2f′1x.
解法2:y′=f1x′=f′1x?1x′=-1x2f′1x.


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