习题221推导余切函数及余割函数的导数公式(cotx)csc2x(cscx)cscxcotx解xxxxxxxx2sincoscossinsin)sincos()(cotxxxxx22222cscsin1sincossinxxxxxxcotcscsincos)sin1()(csc22求下列函数的导数(1)1227445xxxy(2)y5x32x3ex(3)y2tanxsecx1(4)ysinxcosx(5)yx2lnx(6)y3excosx(7)xxyln(8)3ln2xeyx(9)yx2lnxcosx(10)ttscos1sin1解(1))12274()12274(14545xxxxxxy2562562282022820xxxxxx(2)y(5x32x3ex)15x22xln23ex(3)y(2tanxsecx1)2sec2xsecxtanxsecx(2secxtanx)(4)y(sinxcosx)(sinx)cosxsinx(cosx)cosxcosxsinx(sinx)cos2x(5)y(x2lnx)2xlnxx2x1x(2lnx1)(6)y(3excosx)3excosx3ex(sinx)3ex(cosxsinx)(7)22ln1ln1)ln(xxxxxxxxy(8)3422)2(2)3ln(xxexxexexeyxxxx(9)y(x2lnxcosx)2xlnxcosxx2x1cosxx2lnx(sinx)2xlnxcosxxcosxx2lnxsinx(10)22)cos1(cossin1)cos1()sin)(sin1()cos1(cos)cos1sin1(tttttttttts3求下列函数在给定点处的导数(1)ysinxcosx求6xy和4xy(2)cos21sin求4dd(3)553)(2xxxf求f(0)和f(2)解(1)ycosxsinx21321236sin6cos6xy222224sin4cos4xy(2)cossin21sin21cossindd)21(4222422214cos44sin214dd(3)xxxf52)5(3)(2253)0(f1517)2(f4以初速v0竖直上抛的物体其上升高度s与时间t的关系是2021gttvs求(1)该物体的速度v(t)(2)该物体达到最高点的时刻解(1)v(t)s(t)v0gt(2)令v(t)0即v0gt0得gvt0这就是物体达到最高点的时刻5求曲线y2sinxx2上横坐标为x0的点处的切线方程和法线方程解因为y2cosx2xy|x02又当x0时y0所以所求的切线方程为y2x所求的法线方程为xy21即x2y06求下列函数的导数(1)y(2x5)4(2)ycos(43x)(3)23xey(4)yln(1x2)(5)ysin2x(6)22xay(7)ytan(x2)(8)yarctan(ex)(9)y(arcsinx)2(10)ylncosx解(1)y4(2x5)41(2x5)4(2x5)328(2x5)3(2)ysin(43x)(43x)sin(43x)(3)3sin(43x)(3)22233236)6()3(xxxxexexey(4)222212211)1(11xxxxxxy(5)y2sinx(sinx)2sinxcosxsin2x(6))()(21])[(22121222122xaxaxay222122)2()(21xaxxxa(7)ysec2(x2)(x2)2xsec2(x2)(8)xxxxeeeey221)()(11(9)y21arcsin2)(arcsinarcsin2xxxx(10)xxxxxytan)sin(cos1)(coscos17求下列函数的导数(1)yarcsin(12x)(2)211xy(3)xeyx3cos2(4)xy1arccos(5)xxyln1ln1(6)xxy2sin(7)xyarcsin(8))ln(22xaxy(9)yln(secxtanx)(10)yln(cscxcotx)解(1)2221)21(12)21()21(11xxxxxy(2))1()1(21])1[(21212212xxxy222321)1()2()1(21xxxxx(3))3)(3sin(3cos)2()3(cos3cos)(2222xxexxexexeyxxxx)3sin63(cos213sin33cos21222xxexexexxx(4)1||)1()1(11)1()1(1122222xxxxxxxy(5)22)ln1(2)ln1(1)ln1()ln1(1xxxxxxxy(6)222sin2cos212sin22cosxxxxxxxxy(7)2222121)(11)()(11xxxxxxy(8)])(211[1)(12222222222xaxaxaxxaxxaxy2222221)]2(211[1xaxxaxax(9)xxxxxxxxxxysectansecsectansec)tan(sectansec12(10)xxxxxxxxxxycsccotcsccsccotcsc)cot(csccotcsc128求下列函数的导数(1)2)2(arcsinxy(2)2tanlnxy(3)xy2ln1(4)xeyarctan(5)ysinnxcosnx(6)11arctanxxy(7)xxyarccosarcsin(8)y=ln[ln(lnx)](9)xxxxy1111(10)xxy11arcsin解(1))2(arcsin)2(arcsin2xxy)2()2(11)2(arcsin22xxx21)2(11)2(arcsin22xx242arcsin2xx(2))2(2sec2tan1)2(tan2tan12xxxxxyxxxcsc212sec2tan12(3))ln1(ln121ln1222xxxy)(lnln2ln1212xxxxxx1ln2ln1212xxx2ln1ln(4))(arctanarctanxeyx)()(112arctanxxex)1(221)(11arctan2arctanxxexxexx(5)ynsinn1x(sinx)cosnxsinnx(sinnx)(nx)nsinn1xcosxcosnxsinnx(sinnx)nnsinn1x(cosxcosnxsinxsinnx)nsinn1xcos(n1)x(6)222211)1()1()1()11(11)11()11(11xxxxxxxxxxy(7)222)(arccosarcsin11arccos11xxxxxy22)(arccosarcsinarccos11xxxx22)(arccos12xx(8))(lnln1)ln(ln1])[ln(ln)ln(ln1xxxxxy)ln(lnln11ln1)ln(ln1xxxxxx(9)2)11()121121)(11()11)(121121(xxxxxxxxxxy22111xx(10)2)1()1()1(1111)11(1111xxxxxxxxxy)1(2)1(1xxx9.设函数f(x)和g(x)可导且f2(x)g2(x)0试求函数)()(22xgxfy的导数解])()([)()(212222xgxfxgxfy)]()(2)()(2[)()(2122xgxgxfxfxgxf)()()()()()(22xgxfxgxgxfxf10设f(x)可导求下列函数y的导数dxdy(1)yf(x2)(2)yf(sin2x)f(cos2x)解(1)yf(x2)(x2)f(x2)2x2xf(x2)(2)yf(sin2x)(sin2x)f(cos2x)(cos2x)f(sin2x)2sinxcosxf(cos2x)2cosx(sinx)sin2x[f(sin2x)f(cos2x)]11求下列函数的导数(1)ych(shx)(2)yshxechx(3)yth(lnx)(4)ysh3xch2x(5)yth(1x2)(6)yarch(x21)(7)yarch(e2x)(8)yarctan(thx)(9)xxy2ch21chln(10))11(ch2xxy解(1)ysh(shx)(shx)sh(shx)chx(2)ychxechxshxechxshxechx(chxsh2x)(3))(lnch1)(ln)(lnch122xxxxy(4)y3sh2xchx2chxshxshxchx(3shx2)(5))1(ch2)1()1(ch122222xxxxy(6)222)1()1(112422xxxxxy(7)12)(1)(142222xxxxeeeey(8)xxxxxxxy222222ch1chsh11ch1th11)th()th(11xxx222sh211shch1(9))ch(ch21)ch(ch124xxxxyxxxxxshch2ch21chsh4xxxxxxxx323chshchshchshchshxxxxxx33332thchshch)1ch(sh(10))11()11(sh)11(ch2])11(ch[)11(ch2xxxxxxxxxxy)112(sh)1(2)1()1()1()112(sh22xxxxxxxx12求下列函数的导数(1)yex(x22x3)(2)ysin2xsin(x2)(3)2)2(arctanxy(4)nxxyln(