2024高中數(shù)學(xué)第三章變化率與導(dǎo)數(shù)3.4導(dǎo)數(shù)的四則運(yùn)算法則精練含解析北師大版選修1-1

PAGE6-§4導(dǎo)數(shù)的四則運(yùn)算法則A組1.若f(x)=,則f'(-1)=()A. B.- C. D.-解析:因?yàn)閒(x)=,所以f'(x)=-,所以f'(-1)=-×(-1=-.答案:D2.已知f(x)=ax3+3x2+2,若f'(-1)=4,則a的值是 ()A. B. C. D.解析:因?yàn)閒'(x)=3ax2+6x,所以f'(-1)=3a-6,所以3a-6=4,故a=.答案:D3.已知點(diǎn)P在曲線y=2sincos上,α為曲線在點(diǎn)P處的切線的傾斜角,則α的取值范圍是()A. B.C. D.解析:∵y=2sincos=sinx,∴y'=cosx.設(shè)P(x0,y0),由題意,知切線的斜率存在,則曲線在點(diǎn)P處的切線的斜率為tanα=cosx0,∴-1≤tanα≤1.∵0≤α<π,∴α∈,故選D.答案:D4.若函數(shù)f(x)=f'(1)x3-2x2+3,則f'(1)的值為()A.0 B.-1 C.1 D.2解析:因?yàn)閒(x)=f'(1)x3-2x2+3,所以f'(x)=3f'(1)x2-4x,所以f'(1)=3f'(1)-4,所以f'(1)=2.答案:D5.已知點(diǎn)P在曲線y=上,α為曲線在點(diǎn)P處的切線的傾斜角,則α的取值范圍是.

解析:∵y=,∴y'=<0.∵=-=-≥-=-1,當(dāng)且僅當(dāng)x=0時(shí)取等號(hào),∴-1≤y'<0.∴-1≤tanα<0,即≤α<π.答案:6.曲線y=在點(diǎn)(-1,-1)處的切線方程為.

解析:由y=,得y'=,所以所求切線的斜率為2,故所求切線方程為y-(-1)=2(x+1),即2x-y+1=0.答案:2x-y+1=07.若f(x)=cos2-sin2+tan,則f'=.

解析:∵f(x)=cosx+,∴f'(x)=-sinx,∴f'=-sin=-.答案:-8.求下列函數(shù)的導(dǎo)數(shù).(1)y=xcosx-sinx;(2)y=x-sincos;(3)y=.解(1)∵y=xcosx-sinx,∴y'=(xcosx)'-(sinx)'=cosx-xsinx-cosx=-xsinx.(2)∵y=x-sincos=x-sinx,∴y'='=x'-(sinx)'=1-cosx.(3)∵y=,∴y'='=.9.已知函數(shù)f(x)=ax3+bx2+cx過點(diǎn)(1,5),其導(dǎo)函數(shù)y=f'(x)的圖像如圖所示,求f(x)的解析式.解∵f'(x)=3ax2+2bx+c,且f'(1)=0,f'(2)=0,f(1)=5,∴解得∴函數(shù)y=f(x)的解析式為f(x)=2x3-9x2+12x.B組1.已知f(x)=x-5+3sinx,則f'(x)等于()A.-5x-6-3cosx B.x-6+3cosxC.-5x-6+3cosx D.x-6-3cosx解析:y'=-5x-6+3cosx.答案:C2.函數(shù)f(x)=的導(dǎo)數(shù)是()A.(x>0) B.C. D.解析:f(x)=,∴f'(x)=.答案:C3.函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且滿意f(x)=2xf'(e)+lnx,則f'(e)等于()A.e-1 B.-1 C.-e-1 D.-e解析:∵f(x)=2xf'(e)+lnx,∴f'(x)=2f'(e)+,∴f'(e)=2f'(e)+,解得f'(e)=-,故選C.答案:C4.若曲線f(x)=acosx與曲線g(x)=x2+bx+1在交點(diǎn)(0,m)處有公切線,則a+b=.

解析:f'(x)=-asinx,g'(x)=2x+b,∵曲線f(x)=acosx與曲線g(x)=x2+bx+1在交點(diǎn)(0,m)處有公切線,∴f(0)=a=g(0)=1,且f'(0)=0=g'(0)=b,∴a+b=1.答案:15.導(dǎo)學(xué)號(hào)01844037已知曲線C:f(x)=x3-ax+a,若過曲線C外一點(diǎn)A(1,0)引曲線C的兩條切線,它們的傾斜角互補(bǔ),則a的值為.

解析:函數(shù)f(x)的導(dǎo)數(shù)f'(x)=3x2-a.過直線外A(1,0)作曲線C的切線.設(shè)切點(diǎn)(x0,f(x0)),則切線方程為y=(3-a)(x-1),將(x0,f(x0))代入得f(x0)=-ax0+a,即2-3=0,解得x0=0或x0=.故滿意條件的切線有兩條,且它們的斜率分別為-a與-a.因?yàn)閮蓷l切線的傾斜角互補(bǔ),所以-a+-a=0,故a=.答案:6.求下列函數(shù)的導(dǎo)數(shù).(1)y=;(2)y=-sin.解(1)y==-2,∴y'='=.(2)y=-sin·cos=-sinx,∴y'='=-cosx.7.導(dǎo)學(xué)號(hào)01844038已知函數(shù)f(x)=x3+x-16.(1)求曲線y=f(x)在點(diǎn)(2,-6)處的切線的方程;(2)直線l為曲線y=f(x)的切線,且經(jīng)過原點(diǎn),求直線l的方程及切點(diǎn)坐標(biāo);(3)假如曲線y=f(x)的某一切線與直線y=-x+3垂直,求切點(diǎn)坐標(biāo)與切線的方程.解(1)∵f'(x)=3x2+1,∴f(x)在點(diǎn)(2,-6)處的切線的斜率為k=f'(2)=13.∴切線的方程為13x-y-32=0.(2)解法一:設(shè)切點(diǎn)為(x0,y0),則直線l的斜率為f'(x0)=3+1,∴直線l的方程為y=(3+1)(x-x0)++x0-16,又∵直線l過原點(diǎn)(0,0),∴0=(3+1)(-x0)++x0-16,整理得,=-8,∴x0=-2,∴y0=-26,k=13.∴直線l的方程為y=13x,切點(diǎn)坐標(biāo)為(-2,-26).解法二:設(shè)直線l的方程為y=kx,切點(diǎn)為(x0,y0),則k=,又∵k=f'(x0)=3+1,∴=3+1,解得x0=-2,∴y0=-26,k=13.∴直線l的方程為y=13x,切點(diǎn)坐標(biāo)為(-2,-26).(3)