(完整版)导数与微分21(选择题及答案)VIP免费

高等数学一、选择题（共40小题，）1、设则在处，，，不可导fxxxfxxAfBfCfD()()sin()()()()()()()()200200012、)(34)()(4)(3)()0(,43sin)0()(lim20答，，，等于则设DRCBAfxxfxfx3、设，其中，则等于答lim()cos()()()()()()()()xxfxxffABCD012100001244、设为可导函数且满足则曲线在点，处的切线斜率为答fxfafaxxyfxafaABCDx()lim()()()(())()()()()()02121125、)()(3)()32)((1)()()22()()(1)()()()32)((1)()()()(,0)(,)(22答等于则且可导在设afDaaafCafaBafAaxxafafafxfxafaxxf6、)(0)0()0()(0)0()0()(0)0()(0)0()(,0)()sin1)(()(,)(答则必有处可导在，若可导设ffDffCfBfAxxFxxfxFxf7、)()()(0)0()(1)0()(2)0()()(0),sin(cos)(答不可导处有则在设xfDfCfBfAxxxxxf8、设，则．．．．答yxxxyAxxxxxBxxxxxCxxxxxDxxxxxxtanlncos.cossinlncossinlnseccossinlnsectancossinln()111111112229、)(lnsincos1sincossincoscossincossinlncos答．．．．，则设axxxDaxxxCaxxxBeaxxxAyeaxxyxxxx10、设，则．．．．答yxxxxeyAxxxxxxxxBxxxxxxxxxeCxxxxxxxxxeDxxxxxxxxxsinlnsectansincoslnsectansincoslnsecsectansecsincoslnsecsectansincoslnsecsectan()32322322322333311311、)(1sectansecln21sectansec1sectansec21sectansec)0(lnsec22222答．．．．则设xchxshxxchxxDaaxchxshxxchxxCxchxshxxchxxBaxchxshxxchxxAyaathxchxxy12、)(112110lntansec2110lnsec2110lnsec2110lntanlog2222222102答．．．．，则设xxxeDxxxxeCxxxeBxxxeAyxxxey13、)(ln1ln1cosln1cos)0(sin)ln(答．．．．，则设aaxDaaaxCeaaxBeaxaAyaeaaxyxxxxx14、)(2ln1112cossin22ln1sec2cossin22ln1112cossin22ln1sec2cossin2logtancos22222222答．．．．，则设xxxxxxxDxxxxxxxCexxxxxxxBexxxxxxxAyexxxxy15、设，则．．．．yeaxxxchxyAxeaaxeaxxxxshxBxeaxeaxxxxshxCxeaxeaxxxxshxDxeaxexxxxxxxxxlnsincoscotcos(lncos)(lnsin)sincoscsccos(lncos)(lnsin)sincoscsccos(lncos)(lnsin)sincoscsccos(lncos)(1222222lnsin)sincoscsc()axxxxshx22答16、设，则．．．．yexxxxyAxexxexxxxxBxexxexxxxxCxexxexxxxxDxexxexxxxxxxxxxxxxlncostancoshcos(ln)lnsincossecsinhcos(ln)lncoscossecsinhcos(ln)lncoscossinhcos(ln)lncoscossec11111122222222sinh()x答17、设具有连续的一阶导数已知则．．．．答yfxfffffffffxABCDx(),(),()(),(),(),(),(),()()()112302111221301121121201118、设具有连续的一阶导数已知则．．．．答yfxfffffffffxABCDx(),(),(),(),(),(),(),(),()()()000212112121333121312111119、设，，则在处．可导．连续但不可导．左可导而右不可导．右可导而左不可导答fxxexxxfxxABCDx()()()300020、设，，，，则在处．可导．连续但不可导．不连续．左可导而右不可导答fxxxxxfxxABCD()sin()()1000021、)(0)(00sin)(答．左导不等于右导．不连续．连续但不可导．可导处在则，，，，设DCBAxxfxxxxxf22、设，，，为使在处可导则系数．，．，．，．，答fxxxaxbxfxxAabBabCabDab(),(),()21111221211223、)(3)(2)(1)(0)()0(,3)()(23答．．．．为：存在的最高阶数则使设DCBAnfxxxxfn24、关于函数在点处可导及可微三者的关系连续是可微的充分条件可导是可微的...