文档介绍：1/21上页下页返回结束导数与微分第三章习题课一、用导数定义求导(可导充要条件)二、用求导法则求导四、函数的微分三、高阶导数求法2/21上页下页返回结束一、????????)()(lim)(0000hxfhxfh)()(lim000????00)()(lim0xxxfxfxx????点导数xxfxxfxfx????????)()(lim)(0导函数hxfhxfh)()(lim0????xtxftfxt????)()(lim)(0xf?)(xf?xyxx????0lim3/21上页下页返回结束【例1】).0()1(),100()1()(ffxxxxf?????和求设?【解Ⅰ】—用导数定义1)1()(lim)1(1?????xfxffx)100()2(lim1????xxxx?!99??【解Ⅱ】—用求导法则先求导函数])100()1([)(?????xxxxf?])100()2()[1()]100()2([)1(??????????xxxxxxxx??])100()2()[1()100()2(????????xxxxxxx??故!99)99()2)(1(1)1(?????????f同理可求f?(0)(自己练习)4/21上页下页返回结束【例2】已知可导函数f (x)表示的曲线在xxfx1)(lim20??【分析】切线斜率点导数导数定义极限【解】]1)([1)(lim1)(lim020??????xfxxfxxfxx]1)0([)0(???ff]1)([0)0()(lim0?????xfxfxfx1)11(21???点(0,1)处的切线的斜率为1/2 ,求5/21上页下页返回结束二、(注意适用类型)【复习】幂指函数的导数求法方法Ⅰ:化为)()(xvxuy?)(ln)(xuxvey?复合函数链式法则方法Ⅱ:)()(xvxuy?)(ln)(xuxve?6/21上页下页返回结束【例7】求导数:【解】)1(ln2xxeey?????????xuuyy则【分析】复合函数链式法则)1(2???xxee??)122(22xxxeee??112????xxee【关键】搞清每一部分的复合结构——用相应的导数公式,1),(2xxeeuufy????令112???xxee7/21上页下页返回结束.,)0,0()(22dxydyxxyxfyyx求所确定由方程设函数????【例9】【解】两边取对数,ln1ln1xyyx?,lnlnxxyy?即,1ln1ln????????xyyyyy,1ln1ln????yxy即2)1(ln1)1(ln)1(ln1?????????yyyxyxy322)1(ln)1(ln)1(ln?????yxyxxyy【分析】隐函数求导(幂指函数)——对数求导法8/21上页下页返回结束.,)(sincosyxxyx???求设【例10】【解】等式两边取对数有)sincossinlnsin1()(sin2cosxxxxxxxyx?????即【分析】含有幂指函数——对数求导法有取导数两边对xxxxxxyxsinlncosln])(sinln[lncos????xxxxxxyycossin1cossinlnsin1??????9/21上页下页返回结束【例11】设)(,)()()(tftftftytfx??????????存在且不为零,求22;dxyddxdy【分析】参数方程的求导,特别注意高阶导数每次都要用参数方程求导公式.【解】ttxydxdy???)()()()(tftftfttf?????????ttftft??????)()()(22dxdydxddxyd?dxdtdxdydtd??)(dtdxtdtd1)(??)(11tf????)(1tf???ty)(??高阶导数10/21上页下页返回结束三、高阶导数求法①直接法;②归纳法;③四则运算法;④间接法;【常用n 阶导数公式】nnxnx??????????)1()1()()4()(?nnnxnx)!1()1()(ln)5(1)(????)2sin()(sin)2()(????nkxkkxnn)2cos()(cos)3()(????nkxkkxnn)0(ln)()1()(???aaaanxnxxnxee??)()()1(1)(!)1()1()6(???nnnxnx