§3導(dǎo)數(shù)的計(jì)算課后訓(xùn)練鞏固提升1.(多選題)下列各式中,正確的是().A.(ex)'=ex B.(lnx)'=1C.(e2x)'=e2x D.(lnx)'=1解析:根據(jù)求導(dǎo)公式可知A,B正確.(e2x)'=[(e2)x]'=(e2)xlne2=2e2x,故C錯(cuò)誤.(lnx)'=1x·(x)'=1x·答案:ABD2.已知點(diǎn)P(x0,y0)是拋物線y=f(x)=3x2+6x+1上一點(diǎn),且f'(x0)=0,則點(diǎn)P的坐標(biāo)為().A.(1,10) B.(-1,-2)C.(1,-2) D.(-1,10)解析:因?yàn)閒'(x)=6x+6,所以f'(x0)=6x0+6=0,解得x0=-1.因?yàn)辄c(diǎn)P在拋物線上,所以y0=3×(-1)2+6×(-1)+1=-2.所以點(diǎn)P的坐標(biāo)為(-1,-2).答案:B3.下列點(diǎn)中,在曲線y=x2上,且曲線在此點(diǎn)處的切線的傾斜角為π4的是()A.(0,0) B.(2,4)C.14,1解析:y'=2x.∵切線的傾斜角為π4∴切線的斜率為1.設(shè)切點(diǎn)坐標(biāo)為(x0,y0).根據(jù)導(dǎo)數(shù)的幾何意義,得2x0=1,解得x0=12,從而y0=14.故選答案:D4.若曲線y=f(x)=x2+ax+b在點(diǎn)(0,b)處的切線方程是x-y+1=0,則().A.a=1,b=1 B.a=-1,b=1C.a=1,b=-1 D.a=-1,b=-1解析:∵ΔyΔx=∴f'(0)=limΔx由曲線在點(diǎn)(0,b)處的切線方程可知f'(0)=a=1.∵(0,b)在切線上,∴0-b+1=0,∴b=1.答案:A5.已知函數(shù)f'(x)=0,且f(0)=-1,則f(x)=.

解析:∵f'(x)=0,∴f(x)=c(c為常數(shù)).又f(0)=-1,∴c=-1,∴f(x)=-1.答案:-16.已知點(diǎn)P在曲線y=f(x)=x2+1上,若曲線y=x2+1在點(diǎn)P處的切線與曲線y=g(x)=-2x2-1相切,求點(diǎn)P的坐標(biāo).解:設(shè)點(diǎn)P的坐標(biāo)為(x0,y0),易知曲線y=f(x)=x2+1在點(diǎn)P處的切線的斜率存在,設(shè)為k,則k=f'(x0)=limΔx→0ΔyΔ又點(diǎn)P在曲線y=f(x)=x2+1上,所以y0=x02+1.所以切線方程為y-y0=2x0(x-x即y=2x0x+1-x0由題意知此直線與曲線y=g(x)=-2x2-1相切.由y=2x0x+1-x02,y=-2Δ=4x02-8(2-x0解得x0=±233,此時(shí)y0=所以點(diǎn)P的坐標(biāo)為237.已知曲線y=f(x)=1x.求(1)曲線在點(diǎn)P(1,1)處的切線方程;(2)曲線過(guò)點(diǎn)Q(1,0)的切線方程.解:因?yàn)閥=1x,所以y'=-1(1)切線的斜率為函數(shù)y=1x在x=1處的導(dǎo)數(shù)即k=f'(1)=-1.所以曲線在點(diǎn)P(1,1)處的切線方程為y-1=-(x-1),即y=-x+2.(2)因?yàn)辄c(diǎn)Q(1,0)不在曲線y=1x上,所以可設(shè)過(guò)該點(diǎn)的切線與曲線相切于點(diǎn)Aa,1a,則該切線的斜率k=f'(a于是切線方