2009年高考數(shù)學(xué)難點(diǎn)突破專題輔導(dǎo)三十四

難點(diǎn)34  導(dǎo)數(shù)的運(yùn)算法則及基本公式應(yīng)用

導(dǎo)數(shù)是中學(xué)限選內(nèi)容中較為重要的知識(shí),本節(jié)內(nèi)容主要是在導(dǎo)數(shù)的定義,常用求等公式.四則運(yùn)算求導(dǎo)法則和復(fù)合函數(shù)求導(dǎo)法則等問(wèn)題上對(duì)考生進(jìn)行訓(xùn)練與指導(dǎo).

●難點(diǎn)磁場(chǎng)

(★★★★★)已知曲線Cy=x3-3x2+2x,直線l:y=kx,且lC切于點(diǎn)(x0,y0)(x0≠0),求直線l的方程及切點(diǎn)坐標(biāo).

●案例探究

[例1]求函數(shù)的導(dǎo)數(shù):

6ec8aac122bd4f6e

命題意圖:本題3個(gè)小題分別考查了導(dǎo)數(shù)的四則運(yùn)算法則,復(fù)合函數(shù)求導(dǎo)的方法,以及抽象函數(shù)求導(dǎo)的思想方法.這是導(dǎo)數(shù)中比較典型的求導(dǎo)類型,屬于★★★★級(jí)題目.

知識(shí)依托:解答本題的閃光點(diǎn)是要分析函數(shù)的結(jié)構(gòu)和特征,挖掘量的隱含條件,將問(wèn)題轉(zhuǎn)化為基本函數(shù)的導(dǎo)數(shù).

錯(cuò)解分析:本題難點(diǎn)在求導(dǎo)過(guò)程中符號(hào)判斷不清,復(fù)合函數(shù)的結(jié)構(gòu)分解為基本函數(shù)出差錯(cuò).

技巧與方法:先分析函數(shù)式結(jié)構(gòu),找準(zhǔn)復(fù)合函數(shù)的式子特征,按照求導(dǎo)法則進(jìn)行求導(dǎo).

6ec8aac122bd4f6e

 

(2)解:y=μ3,μ=axbsin2ωx,μ=avby

v=x,y=sinγ  γx

y′=(μ3)′=3μ2?μ′=3μ2(avby)′

=3μ2(av′-by′)=3μ2(av′-byγ′)

=3(axbsin2ωx)2(abωsin2ωx)

(3)解法一:設(shè)y=f(μ),μ=6ec8aac122bd4f6e,v=x2+1,則

yx=yμμv?vx=f′(μ)?6ec8aac122bd4f6ev6ec8aac122bd4f6e?2x

=f′(6ec8aac122bd4f6e)?6ec8aac122bd4f6e6ec8aac122bd4f6e?2x

=6ec8aac122bd4f6e

解法二:y′=[f(6ec8aac122bd4f6e)]′=f′(6ec8aac122bd4f6e)?(6ec8aac122bd4f6e)′

=f′(6ec8aac122bd4f6e)?6ec8aac122bd4f6e(x2+1)6ec8aac122bd4f6e?(x2+1)′

=f′(6ec8aac122bd4f6e)?6ec8aac122bd4f6e(x2+1) 6ec8aac122bd4f6e?2x

=6ec8aac122bd4f6ef′(6ec8aac122bd4f6e)

[例2]利用導(dǎo)數(shù)求和

(1)Sn=1+2x+3x2+…+nxn1(x≠0,nN*)

(2)Sn=C6ec8aac122bd4f6e+2C6ec8aac122bd4f6e+3C6ec8aac122bd4f6e+…+nC6ec8aac122bd4f6e,(nN*)

命題意圖:培養(yǎng)考生的思維的靈活性以及在建立知識(shí)體系中知識(shí)點(diǎn)靈活融合的能力.屬

★★★★級(jí)題目.

知識(shí)依托:通過(guò)對(duì)數(shù)列的通項(xiàng)進(jìn)行聯(lián)想,合理運(yùn)用逆向思維.由求導(dǎo)公式(xn)′=nxn1,可聯(lián)想到它們是另外一個(gè)和式的導(dǎo)數(shù).關(guān)鍵要抓住數(shù)列通項(xiàng)的形式結(jié)構(gòu).

錯(cuò)解分析:本題難點(diǎn)是考生易犯思維定勢(shì)的錯(cuò)誤,受此影響而不善于聯(lián)想.

技巧與方法:第(1)題要分x=1和x≠1討論,等式兩邊都求導(dǎo).

解:(1)當(dāng)x=1時(shí)

Sn=1+2+3+…+n=6ec8aac122bd4f6en(n+1);

當(dāng)x≠1時(shí),

x+x2+x3+…+xn=6ec8aac122bd4f6e,

兩邊都是關(guān)于x的函數(shù),求導(dǎo)得

(x+x2+x3+…+xn)′=(6ec8aac122bd4f6e)′

Sn=1+2x+3x2+…+nxn1=6ec8aac122bd4f6e

(2)∵(1+x)n=1+C6ec8aac122bd4f6ex+C6ec8aac122bd4f6ex2+…+C6ec8aac122bd4f6exn,

兩邊都是關(guān)于x的可導(dǎo)函數(shù),求導(dǎo)得

n(1+x)n1=C6ec8aac122bd4f6e+2C6ec8aac122bd4f6ex+3C6ec8aac122bd4f6ex2+…+nC6ec8aac122bd4f6exn1,

x=1得,n?2n1=C6ec8aac122bd4f6e+2C6ec8aac122bd4f6e+3C6ec8aac122bd4f6e+…+nC6ec8aac122bd4f6e,

Sn=C6ec8aac122bd4f6e+2C6ec8aac122bd4f6e+…+nC6ec8aac122bd4f6e=n?2n1?

●錦囊妙計(jì)

1.深刻理解導(dǎo)數(shù)的概念,了解用定義求簡(jiǎn)單的導(dǎo)數(shù).

6ec8aac122bd4f6e表示函數(shù)的平均改變量,它是Δx的函數(shù),而f′(x0)表示一個(gè)數(shù)值,即f′(x)=6ec8aac122bd4f6e,知道導(dǎo)數(shù)的等價(jià)形式:6ec8aac122bd4f6e.?

2.求導(dǎo)其本質(zhì)是求極限,在求極限的過(guò)程中,力求使所求極限的結(jié)構(gòu)形式轉(zhuǎn)化為已知極限的形式,即導(dǎo)數(shù)的定義,這是順利求導(dǎo)的關(guān)鍵.

3.對(duì)于函數(shù)求導(dǎo),一般要遵循先化簡(jiǎn),再求導(dǎo)的基本原則,求導(dǎo)時(shí),不但要重視求導(dǎo)法則的應(yīng)用,而且要特別注意求導(dǎo)法則對(duì)求導(dǎo)的制約作用,在實(shí)施化簡(jiǎn)時(shí),首先必須注意變換的等價(jià)性,避免不必要的運(yùn)算失誤.

4.復(fù)合函數(shù)求導(dǎo)法則,像鏈條一樣,必須一環(huán)一環(huán)套下去,而不能丟掉其中的一環(huán).必須正確分析復(fù)合函數(shù)是由哪些基本函數(shù)經(jīng)過(guò)怎樣的順序復(fù)合而成的,分清其間的復(fù)合關(guān)系.

●殲滅難點(diǎn)訓(xùn)練

一、選擇題

1.(★★★★)y=esinxcos(sinx),則y′(0)等于(    )

試題詳情

A.0                              B.1                              C.-1                           D.2

試題詳情

2.(★★★★)經(jīng)過(guò)原點(diǎn)且與曲線y=6ec8aac122bd4f6e相切的方程是(    )

試題詳情

A.x+y=0或6ec8aac122bd4f6e+y=0                                         B.xy=0或6ec8aac122bd4f6e+y=0

試題詳情

C.x+y=0或6ec8aac122bd4f6ey=0                                        D.xy=0或6ec8aac122bd4f6ey=0

試題詳情

二、填空題

3.(★★★★)若f′(x0)=2,6ec8aac122bd4f6e =_________.

試題詳情

4.(★★★★)設(shè)f(x)=x(x+1)(x+2)…(x+n),則f′(0)=_________.

試題詳情

三、解答題

5.(★★★★)已知曲線C1:y=x2C2:y=-(x-2)2,直線lC1、C2都相切,求直線l的方程.

試題詳情

6.(★★★★)求函數(shù)的導(dǎo)數(shù)

(1)y=(x2-2x+3)e2x;

試題詳情

(2)y=6ec8aac122bd4f6e.

試題詳情

7.(★★★★)有一個(gè)長(zhǎng)度為5 m的梯子貼靠在筆直的墻上,假設(shè)其下端沿地板以3 m/s?的速度離開(kāi)墻腳滑動(dòng),求當(dāng)其下端離開(kāi)墻腳1.4 m時(shí),梯子上端下滑的速度.

試題詳情

8.(★★★★)求和Sn=12+22x+32x2+…+n2xn1?,(x≠0,nN*).

 

試題詳情

難點(diǎn)磁場(chǎng)

解:由l過(guò)原點(diǎn),知k=6ec8aac122bd4f6e(x0≠0),點(diǎn)(x0,y0)在曲線C上,y0=x03-3x02+2x0,

6ec8aac122bd4f6e=x02-3x0+2

y′=3x2-6x+2,k=3x02-6x0+2

k=6ec8aac122bd4f6e,∴3x02-6x0+2=x02-3x0+2

2x02-3x0=0,∴x0=0或x0=6ec8aac122bd4f6e

x≠0,知x0=6ec8aac122bd4f6e

y0=(6ec8aac122bd4f6e)3-3(6ec8aac122bd4f6e)2+2?6ec8aac122bd4f6e=-6ec8aac122bd4f6e

k=6ec8aac122bd4f6e=-6ec8aac122bd4f6e

l方程y=-6ec8aac122bd4f6ex 切點(diǎn)(6ec8aac122bd4f6e,-6ec8aac122bd4f6e)

殲滅難點(diǎn)訓(xùn)練

一、1.解析:y′=esinx[cosxcos(sinx)-cosxsin(sinx)],y′(0)=e0(1-0)=1

答案:B

2.解析:設(shè)切點(diǎn)為(x0,y0),則切線的斜率為k=6ec8aac122bd4f6e,另一方面,y′=(6ec8aac122bd4f6e)′=6ec8aac122bd4f6e,故

y′(x0)=k,即6ec8aac122bd4f6ex02+18x0+45=0得x0(1)=-3,y0(2)=-15,對(duì)應(yīng)有y0(1)=3,y0(2)=6ec8aac122bd4f6e,因此得兩個(gè)切點(diǎn)A(-3,3)或B(-15,6ec8aac122bd4f6e),從而得y′(A)=6ec8aac122bd4f6e =-1及y′(B)= 6ec8aac122bd4f6e ,由于切線過(guò)原點(diǎn),故得切線:lA:y=-xlB:y=-6ec8aac122bd4f6e.

答案:A

二、3.解析:根據(jù)導(dǎo)數(shù)的定義:f′(x0)=6ec8aac122bd4f6e(這時(shí)6ec8aac122bd4f6e)

6ec8aac122bd4f6e

答案:-1

4.解析:設(shè)g(x)=(x+1)(x+2)……(x+n),則f(x)=xg(x),于是f′(x)=g(x)+xg′(x),f′(0)=g(0)+0?g′(0)=g(0)=1?2?…n=n!

答案:n!

三、5.解:設(shè)lC1相切于點(diǎn)P(x1,x12),與C2相切于Q(x2,-(x2-2)2)

對(duì)于C1y′=2x,則與C1相切于點(diǎn)P的切線方程為

yx12=2x1(xx1),即y=2x1xx12                                                                                                                                             

對(duì)于C2y′=-2(x-2),與C2相切于點(diǎn)Q的切線方程為y+(x2-2)2=-2(x2-2)(xx2),即y=-2(x2-2)x+x22-4                                                                                             ②

∵兩切線重合,∴2x1=-2(x2-2)且-x12=x22-4,解得x1=0,x2=2或x1=2,x2=0

∴直線l方程為y=0或y=4x-4

6.解:(1)注意到y>0,兩端取對(duì)數(shù),得

lny=ln(x2-2x+3)+lne2x=ln(x2-2x+3)+2x

6ec8aac122bd4f6e

 (2)兩端取對(duì)數(shù),得

ln|y|=6ec8aac122bd4f6e(ln|x|-ln|1-x|),

兩邊解x求導(dǎo),得

6ec8aac122bd4f6e

7.解:設(shè)經(jīng)時(shí)間t秒梯子上端下滑s米,則s=5-6ec8aac122bd4f6e,當(dāng)下端移開(kāi)1.4 m時(shí),t0=6ec8aac122bd4f6e,又s′=-6ec8aac122bd4f6e (25-9t2)6ec8aac122bd4f6e?(-9?2t)=9t6ec8aac122bd4f6e,所以s′(t0)=9×6ec8aac122bd4f6e=0.875(m/s)

8.解:(1)當(dāng)x=1時(shí),Sn=12+22+32+…+n2=6ec8aac122bd4f6en(n+1)(2n+1),當(dāng)x≠1時(shí),1+2x+3x2+…+nxn-1?=6ec8aac122bd4f6e,兩邊同乘以x,得

x+2x2+3x2+…+nxn=6ec8aac122bd4f6e兩邊對(duì)x求導(dǎo),得

Sn=12+22x2+32x2+…+n2xn-1?

=6ec8aac122bd4f6e

 


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