變化率與導(dǎo)數(shù)【知識梳理】1.必會(huì)知識教材回扣填一填(1)函數(shù)y=f(x)在x=x0處的導(dǎo)數(shù):①定義:稱函數(shù)y=f(x)在x=x0處的瞬時(shí)變化率__________________=為y=f(x)在x=x0處的導(dǎo)數(shù)，記作f′(x0)或即f′(x0)==___________________.②幾何意義:函數(shù)f(x)在點(diǎn)x0處的導(dǎo)數(shù)f′(x0)的幾何意義是在曲線y=f(x)上點(diǎn)(x0,f(x0))處的___________.相應(yīng)地,切線方程為_____________________.(2)函數(shù)y=f(x)的導(dǎo)函數(shù):稱函數(shù)f′(x)=__________________為函數(shù)y=f(x)的導(dǎo)函數(shù),導(dǎo)函數(shù)有時(shí)也記作y′.切線的斜率y-f(x0)=f′(x0)(x-x0)(3)基本初等函數(shù)的導(dǎo)數(shù)公式:原函數(shù)導(dǎo)函數(shù)f(x)=c(c為常數(shù))f′(x)=__f(x)=xα(α∈Q*)f′(x)=______f(x)=sinxf′(x)=_____f(x)=cosxf′(x)=______0αxα-1cosx-sinx原函數(shù)導(dǎo)函數(shù)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)=lnxf′(x)=_____axlnaex(4)導(dǎo)數(shù)四則運(yùn)算法則:①[f(x)±g(x)]′=_______________.②[f(x)·g(x)]′=______________________.③＝__________________(g(x)≠0).f′(x)±g′(x)f′(x)g(x)+f(x)g′(x)2.必備結(jié)論教材提煉記一記(1)曲線y=f(x)在點(diǎn)P(x0,y0)處的切線是以點(diǎn)P(x0,y0)為切點(diǎn),以f′(x0)為斜率的直線,而曲線y=f(x)過點(diǎn)P(x0,y0)的切線,點(diǎn)P(x0,y0)不一定是切點(diǎn).(2)函數(shù)y=f(x)的導(dǎo)數(shù)f′(x)反映了函數(shù)f(x)的瞬時(shí)變化趨勢,其正負(fù)號反映了變化的方向,其大小|f′(x)|反映了變化的快慢,|f′(x)|越大,曲線在這點(diǎn)處的切線越“陡”.3.必用技法核心總結(jié)看一看(1)常用方法:利用導(dǎo)數(shù)求切線的方法.(2)數(shù)學(xué)思想:轉(zhuǎn)化與化歸、數(shù)形結(jié)合.(3)記憶口訣:導(dǎo)數(shù)概念要理清,專門刻畫變化量,放大放大再放大,逼近逼近再逼近.幾何意義在切線,物理應(yīng)用求速度.常見函數(shù)的導(dǎo)數(shù),定義證明會(huì)推導(dǎo).導(dǎo)數(shù)的四則運(yùn)算,記住法則計(jì)算巧.【小題快練】1.思考辨析靜心思考判一判(1)求f′(x0)時(shí),可先求f(x0)再求f′(x0).(

)(2)曲線的切線不一定與曲線只有一個(gè)公共點(diǎn).(

)(3)與曲線只有一個(gè)公共點(diǎn)的直線一定是曲線的切線.(

)(4)若f(x)=f′(a)x2+lnx(a>0),則f′(x)=2xf′(a)+.(

)【解析】(1)錯(cuò)誤.應(yīng)先求f′(x),再求f′(x0).(2)正確.如y=1是曲線y=sinx的切線,但其交點(diǎn)個(gè)數(shù)有無數(shù)個(gè).(3)錯(cuò)誤.如y=0與拋物線y2=x只有一個(gè)公共點(diǎn),但是y=0不是拋物線y2=x的切線.(4)正確.f′(x)=(f′(a)x2+lnx)′=(f′(a)x2)′+(lnx)′=2xf′(a)+.答案:(1)×

(2)√

(3)×

(4)√2.教材改編鏈接教材練一練(1)(選修1-1P80T1改編)在高臺跳水運(yùn)動(dòng)中,ts時(shí)運(yùn)動(dòng)員相對于水面的高度(單位:m)是h(t)=-4.9t2+6.5t+10.則運(yùn)動(dòng)員的速度v=

,加速度a=

.【解析】v=h′(t)=-9.8t+6.5,a=v′(t)=-9.8.答案:-9.8t+6.5

-9.8(2)(選修1-1P85T3改編)已知函數(shù)r(V)=,則r′()=_____.【解析】因?yàn)閞′(V)=所以r′()=答案：3.真題小試感悟考題試一試(1)(2014·廣東高考)曲線y=-5ex+3在點(diǎn)(0,-2)處的切線方程為

.【解析】因?yàn)閥′=-5ex,y′|x=0=-5,即在點(diǎn)(0,-2)處的切線斜率為-5,所以切線方程為y-(-2)=-5(x-0),5x+y+2=0.答案:5x+y+2=0(2)(2013·江西高考)若曲線y=xα+1(α∈R)在點(diǎn)(1,2)處的切線經(jīng)過坐標(biāo)原點(diǎn),則α=

.【解析】因?yàn)閥′=α·xα-1,所以在點(diǎn)(1,2)處的切線斜率k=α,則切線方程為y-2=α(x-1),又切線過原點(diǎn),故0-2=α(0-1),解得α=2.答案:2(3)(2015·陽泉模擬)直線