蘇教版高中數(shù)學(xué)選擇性必修第一冊(cè)第5章導(dǎo)數(shù)及其應(yīng)用5-2導(dǎo)數(shù)的運(yùn)算課件

5.2

導(dǎo)數(shù)的運(yùn)算知識(shí)點(diǎn)1

基本初等函數(shù)的求導(dǎo)公式原函數(shù)導(dǎo)函數(shù)f(x)=xα(α為常數(shù))f'(x)=αxα-1f(x)=ax(a>0,且a≠1)f'(x)=axlnaf(x)=exf

'(x)=exf(x)=logax(a>0,且a≠1)f'(x)=

原函數(shù)導(dǎo)函數(shù)f(x)=lnxf'(x)=

f(x)=sinxf'(x)=cosxf(x)=cosxf'(x)=-sinx特別提醒

幾個(gè)常用函數(shù)的導(dǎo)數(shù)(1)若f(x)=C(C為常數(shù)),則f'(x)=0.(2)若f(x)=

,則f'(x)=-

.(3)若f(x)=

,則f'(x)=

.設(shè)函數(shù)f(x),g(x)均可導(dǎo),且其導(dǎo)數(shù)分別為f'(x),g'(x),則知識(shí)點(diǎn)2

函數(shù)的和、差、積、商的求導(dǎo)法則和的導(dǎo)數(shù)[f(x)+g(x)]'=f'(x)+g'(x)差的導(dǎo)數(shù)[f(x)-g(x)]'=f'(x)-g'(x)積的導(dǎo)數(shù)[Cf(x)]'=Cf'(x)(C為常數(shù)),[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)商的導(dǎo)數(shù)

'=

(g(x)≠0)

一般地,對(duì)于由函數(shù)y=f(u)和u=g(x)復(fù)合而成的函數(shù)y=f(g(x)),它的導(dǎo)數(shù)與函數(shù)y=f(u),u=g(x)的導(dǎo)數(shù)間的關(guān)系為y'x=y'u·u'x.知識(shí)點(diǎn)4

簡(jiǎn)單復(fù)合函數(shù)的導(dǎo)數(shù)知識(shí)辨析1.[f(x0)]'=f'(x0),對(duì)嗎?2.(ax)'=xax-1(a>0,且a≠1),對(duì)嗎?3.若f'(x)=1,則f'(x)的原函數(shù)一定是f(x)=x嗎?4.已知函數(shù)f(x)=x-

x2-lnx,則f'(-1)=3,正確嗎?一語(yǔ)破的1.不對(duì).f(x0)是一個(gè)常數(shù),所以[f(x0)]'=0,而f'(x0)是當(dāng)x=x0時(shí)f'(x)的函數(shù)值,不一定為0.2.不對(duì).(ax)'=axlna(a>0,且a≠1),而(xa)'=axa-1(a是常數(shù)).求導(dǎo)時(shí)不要混淆指數(shù)函數(shù)和冪函數(shù)的求

導(dǎo)公式.3.不一定.若f'(x)=1,則f(x)=x+c(c為常數(shù)).4.不正確.函數(shù)f(x)的定義域?yàn)閧x|x>0},所以f'(-1)的值不存在.利用導(dǎo)數(shù)的四則運(yùn)算法則求導(dǎo)的策略(1)若待求導(dǎo)的函數(shù)是兩個(gè)函數(shù)商的形式,則可先對(duì)函數(shù)進(jìn)行適當(dāng)變形,再求導(dǎo).(2)對(duì)于多個(gè)整式乘積形式的函數(shù),可以考慮展開,化為和、差形式,再求導(dǎo).(3)對(duì)于三角函數(shù),可考慮先進(jìn)行恒等變形,再求導(dǎo).定點(diǎn)1利用導(dǎo)數(shù)的四則運(yùn)算法則求導(dǎo)?關(guān)鍵能力定點(diǎn)破典例求下列函數(shù)的導(dǎo)數(shù).(1)y=lnx+

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

cos

;(4)y=

.解析

(1)y'=

'=(lnx)'+

'=

-

.(2)因?yàn)閥=(2x2-1)(3x+1)=6x3+2x2-3x-1,所以y'=(6x3+2x2-3x-1)'=18x2+4x-3.(3)因?yàn)閥=x-sin

cos

=x-

sinx,所以y'=

'=x'-

'=1-

cosx.(4)y'=

'=

=-

.1.復(fù)合函數(shù)求導(dǎo)的步驟定點(diǎn)2復(fù)合函數(shù)的導(dǎo)數(shù)(1)通常是將復(fù)合函數(shù)分解為基本初等函數(shù);(2)求導(dǎo)時(shí)分清是對(duì)哪個(gè)變量求導(dǎo);(3)計(jì)算結(jié)果盡量簡(jiǎn)單.2.求復(fù)合函數(shù)的導(dǎo)數(shù)的注意點(diǎn)典例求下列函數(shù)的導(dǎo)數(shù):(1)y=

;(2)y=(1-2x)3;(3)y=sin

;(4)y=22x+1.解析

(1)函數(shù)y=

可以看作函數(shù)y=

和u=3x+1的復(fù)合函數(shù),∴y'x=y'u·u'x=

'·(3x+1)'=

'·3=-3

=-3(3x+1

.(2)函數(shù)y=(1-2x)3可以看作函數(shù)y=u3和u=1-2x的復(fù)合函數(shù),∴y'x=y'u·u'x=(u3)'·(1-2x)'=-6u2=-6(1-2x)2.(3)函數(shù)y=sin

可以看作函數(shù)y=sinu和u=

-3x的復(fù)合函數(shù),∴y'x=y'u·u'x=(sinu)'·

'=-3cosu=-3cos

=3sin3x.(4)函數(shù)y=22x+1可以看作函數(shù)y=2u和u=2x+1的復(fù)合函數(shù),∴y'x=y'u·u'x=(2u)'·(2x+1)'=2·2uln2=2·22x+1ln2=4x+1ln2.切線問(wèn)題的處理思路(1)對(duì)函數(shù)進(jìn)行求導(dǎo);(2)若已知切點(diǎn),則直接求出切線斜率、切線方程;(3)若切點(diǎn)未知,則先設(shè)出切點(diǎn),用切點(diǎn)表示切線斜率,再根據(jù)條件求出切點(diǎn)坐標(biāo).在解決此類問(wèn)題時(shí),求函數(shù)的導(dǎo)數(shù)是基礎(chǔ),找出切點(diǎn)是關(guān)鍵.定點(diǎn)3利用導(dǎo)數(shù)運(yùn)算解決切線問(wèn)題典例(1)若直線l:y=kx+b

與曲線f(x)=ex-1和g(x)=ln(x+1)均相切,則直線l的方程為

;(2)若點(diǎn)P是曲線y=x2-lnx-1上任意一點(diǎn),則點(diǎn)P到直線y=x-3距離的最小值為

.y=x

解析

(1)設(shè)直線l與曲線f(x),g(x)分別相切于點(diǎn)A(x1,

),B(x2,ln(x2+1)),由f'(x)=ex-1,g'(x)=

,可得k=

=

,故曲線f(x)在點(diǎn)A處的切線方程為y-

=

(x-x1),即y=

x+

(1-x1),曲線g(x)在點(diǎn)B處的切線方程為y-ln(x2+1)=

(x-x2),即y=

x+ln(x2+1)-

,由

得

ln(1+x2)=ln(1+x2)-

,故

=

ln(1+x2),故x2=0或ln(1+x2)=1,若ln(1+x2)=1,則x2+1=e,則

=

<

,不合題意,舍去,故x2=0,此時(shí)直線l的方程為y=x.(2)由題意可得,當(dāng)點(diǎn)P到直線y=x-3的距離最小時(shí),曲線y=x2-lnx-1在點(diǎn)P處的切線平行于直線y

=x-3,設(shè)