2025優(yōu)化設(shè)計(jì)一輪第2節(jié) 導(dǎo)數(shù)的運(yùn)算

第2節(jié)導(dǎo)數(shù)的運(yùn)算高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2025研考點(diǎn)精準(zhǔn)突破目錄索引

強(qiáng)基礎(chǔ)固本增分12強(qiáng)基礎(chǔ)固本增分知識(shí)梳理(1)基本初等函數(shù)的導(dǎo)數(shù)公式

原函數(shù)導(dǎo)數(shù)f(x)=c(c為常數(shù))f'(x)=_________f(x)=xα(α∈R,且α≠0)f'(x)=_________f(x)=sin

xf'(x)=_________f(x)=cos

xf'(x)=_________f(x)=ax(a>0,且a≠1)f'(x)=_________f(x)=exf'(x)=_________f(x)=logax(a>0,且a≠1)f'(x)=_________f(x)=ln

xf'(x)=_________0αxα-1cosx-sinxaxlnaex(2)導(dǎo)數(shù)的運(yùn)算法則①[f(x)±g(x)]'=__________.

②[f(x)g(x)]'=_________________________,特別地,[cf(x)]'=__________.

(3)復(fù)合函數(shù)的求導(dǎo)法則:復(fù)合函數(shù)y=f(g(x))的導(dǎo)數(shù)和函數(shù)y=f(u),u=g(x)的導(dǎo)數(shù)間的關(guān)系為y'x=__________,即y對(duì)x的導(dǎo)數(shù)等于y對(duì)u的導(dǎo)數(shù)與u對(duì)x的導(dǎo)數(shù)的乘積.

f'(x)±g'(x)f'(x)g(x)+f(x)g'(x)cf'(x)y'u·u'x常用結(jié)論

3.奇函數(shù)的導(dǎo)數(shù)是偶函數(shù),偶函數(shù)的導(dǎo)數(shù)是奇函數(shù),周期函數(shù)的導(dǎo)數(shù)還是周期函數(shù).自主診斷題組一

思考辨析(判斷下列結(jié)論是否正確,正確的畫(huà)“√”,錯(cuò)誤的畫(huà)“×”)1.f'(a)=[f(a)]'.(

)2.若函數(shù)f(x)定義域?yàn)镽,且f(-x)+f(x)=0,則f'(-x)=f'(x).(

)3.若f(x)=ln3,則f'(x)=.(

)4.若f(x)=sin2x,則f'(x)=cos2x.(

)×√××題組二

回源教材

題組三

連線高考

1研考點(diǎn)精準(zhǔn)突破考點(diǎn)一導(dǎo)數(shù)的運(yùn)算例1(1)(多選題)(2024·吉林長(zhǎng)春模擬)在下列求導(dǎo)運(yùn)算中,不正確的是(

)BCD(2)(2024·湖南長(zhǎng)沙模擬)已知函數(shù)f(x)=ex+x3f'(1),則f(1)=(

)C解析

設(shè)x∈(-2π,-π),則x+2π∈(0,π),所以f(x+2π)=(x+2π)sin(x+2π)=(x+2π)sin

x,又因?yàn)閒(x)=f(x+π)=f(x+2π),所以f(x)=(x+2π)sin

x,此時(shí)f'(x)=sin

x+(x+2π)·cos

x,B規(guī)律方法函數(shù)求導(dǎo)運(yùn)算的基本原則(1)熟記常見(jiàn)函數(shù)的導(dǎo)數(shù)公式和四則運(yùn)算法則,切勿記錯(cuò)記混;(2)求導(dǎo)前,應(yīng)利用代數(shù)、三角恒等變換對(duì)函數(shù)解析式進(jìn)行化簡(jiǎn),以便減少運(yùn)算量,減少差錯(cuò);(3)復(fù)合函數(shù)求導(dǎo),要正確分析函數(shù)的復(fù)合過(guò)程,分清內(nèi)外層函數(shù),按照法則進(jìn)行求導(dǎo);(4)求函數(shù)在某一點(diǎn)處的導(dǎo)數(shù)且解析式未知時(shí),應(yīng)先根據(jù)條件求出該點(diǎn)所在區(qū)間的解析式再求導(dǎo);(5)當(dāng)函數(shù)解析式中含有待定系數(shù)(如f'(x0)等)時(shí),應(yīng)將待定系數(shù)看成常數(shù)進(jìn)行求解.考點(diǎn)二導(dǎo)數(shù)運(yùn)算的應(yīng)用(多考向探究預(yù)測(cè))考向1與導(dǎo)數(shù)運(yùn)算有關(guān)的新定義問(wèn)題例2(2024·山東煙臺(tái)模擬)給出定義:若函數(shù)f(x)在D上可導(dǎo),即f'(x)存在,且導(dǎo)函數(shù)f'(x)在D上也可導(dǎo),則稱f(x)在D上存在二階導(dǎo)函數(shù),記f″(x)=(f'(x))'.若f″(x)<0在D上恒成立,則稱f(x)在D上為凸函數(shù),以下四個(gè)函數(shù)在(0,)內(nèi)不是凸函數(shù)的是(

)A.f(x)=cos

x+sin

x

B.f(x)=ln

x+3xC.f(x)=-x3+4x-8 D.f(x)=xexD規(guī)律方法導(dǎo)數(shù)新定義問(wèn)題的求解策略新定義題型的特點(diǎn)是:通過(guò)給出一個(gè)新概念,或約定一種新運(yùn)算,或給出幾個(gè)新模型來(lái)創(chuàng)設(shè)全新的問(wèn)題情境,要求考生在閱讀理解的基礎(chǔ)上,依據(jù)題目提供的信息,聯(lián)系所學(xué)的知識(shí)和方法,實(shí)現(xiàn)信息的遷移,達(dá)到靈活解題的目的.遇到新定義問(wèn)題,應(yīng)耐心讀題,分析新定義的特點(diǎn),弄清新定義的性質(zhì),按新定義的要求,“照章辦事”,逐條分析、驗(yàn)證、運(yùn)算,使問(wèn)題得以解決.[對(duì)點(diǎn)訓(xùn)練1](2024·浙江杭州模擬)曲率是衡量曲線彎曲程度的重要指標(biāo),其定義為:若f'(x)是f(x)的導(dǎo)函數(shù),f″(x)是f'(x)的導(dǎo)函數(shù),則曲線y=f(x)在點(diǎn)(x,f(x))處的曲率K=

.已知f(x)=cos(x-1)-ln

x,則曲線y=f(x)在點(diǎn)(1,f(1))處的曲率為_(kāi)_________.

0考向2導(dǎo)函數(shù)性質(zhì)的應(yīng)用例3(2024·山東德州模擬)已知函數(shù)f(x)及其導(dǎo)函數(shù)f'(x)的定義域均為R,且f(x-1)為奇函數(shù),f'(2-x)+f'(x)=2,f'(-1)=2,則

=(

)A.2025

B.2024 C.1013

D.1012B解析

由f'(2-x)+f'(x)=2,令x=1,得2f'(1)=2,所以f'(1)=1.由f(x-1)為奇函數(shù),得f(x-1)=-f(-x-1),所以f'(x-1)=f'(-x-1),故f'(x)=f'(-x-2),①又因?yàn)閒'(2-x)+f'(x)=2,②由①和②得f'(2-x)+f'(-x-2)=2,即f'(4-x-2)+f'(-x-2)=2,所以f'(x)+f'(x+4)=2,③令x=-1,得f'(-1)+f'(3)=2,得f'(3)=0;令x=1,得f'(1)+f'(5)=2,得f'(5)=1.又因?yàn)閒'(x+4)+f'(x+8)=2,④由③-④得f'(x)-f'(x+8)=0,即f'(x)=f'(x+8),所以函數(shù)f'(x)是以8為周期的周期函數(shù),故f'(7)=f'(-1)=2,所以f'(1)+f'(3)+f'(5)+f'(7)=1+0+1+2=4,=506[f'