第十五講第三章導(dǎo)數(shù)與微分3.4高階導(dǎo)數(shù)的概念

1、22),(),(,)(,)()(dxydxyxfxxfxxfxf或或或或記記作作的的二二階階導(dǎo)導(dǎo)數(shù)數(shù)在在稱稱為為函函數(shù)數(shù)的的導(dǎo)導(dǎo)數(shù)數(shù)在在的的導(dǎo)導(dǎo)函函數(shù)數(shù)函函數(shù)數(shù) 二、高階導(dǎo)數(shù)二、高階導(dǎo)數(shù)xxfxxfxfx )()(lim)(0 即即（一）高階導(dǎo)數(shù)定義（一）高階導(dǎo)數(shù)定義33),(),(,)(,)()(dxydxyxfxxfxxfxf或或或或記記作作的的三三階階導(dǎo)導(dǎo)數(shù)數(shù)在在稱稱為為函函數(shù)數(shù)導(dǎo)導(dǎo)數(shù)數(shù)的的在在的的二二階階導(dǎo)導(dǎo)函函數(shù)數(shù)函函數(shù)數(shù) xxfxxfxfnnxn )()(lim)()1()1(0)( 即即nnnndxydxyxfnxxfxnxf或或或或記記作作階階導(dǎo)導(dǎo)數(shù)數(shù)的的在在稱稱為為函函數(shù)數(shù)的

2、的導(dǎo)導(dǎo)數(shù)數(shù)階階導(dǎo)導(dǎo)函函數(shù)數(shù)在在的的函函數(shù)數(shù)),(),(,)(,)1()()()( )(tss 變變速速直直線線運(yùn)運(yùn)動(dòng)動(dòng)：瞬瞬時(shí)時(shí)速速度度一一階階導(dǎo)導(dǎo)數(shù)數(shù)：)()(tvts 瞬瞬時(shí)時(shí)加加速速度度二二階階導(dǎo)導(dǎo)數(shù)數(shù)：)()(tats 二階導(dǎo)數(shù)的物理意義二階導(dǎo)數(shù)的物理意義)(), 2, 1(1nnynxy求求例例 1 nnxy2)1( nxnny!)(nyn 解解用數(shù)學(xué)歸納法可以證明用數(shù)學(xué)歸納法可以證明aayxln )(),1, 0(2nxyaaay求求例例 2)(lnaayx nxnaay)(ln)( xnxee )()(特特例例：用數(shù)學(xué)歸納法可以證明用數(shù)學(xué)歸納法可以證明解解.sin3階階導(dǎo)導(dǎo)數(shù)數(shù)的

3、的求求例例nx)2sin()(sin)( nxxn)2sin(cos)(sin xxxxxx )2sin()(sin )22sin()2cos( xx解解xxx )22sin()(sin )23sin()22cos( xx用數(shù)學(xué)歸納法用數(shù)學(xué)歸納法).0(, 3/)0(,01cos22)(4yyyexfyx 求求且且確確定定由由方方程程函函數(shù)數(shù)例例 求求導(dǎo)導(dǎo)方方程程兩兩邊邊對(duì)對(duì) x) 1 (0sin22 yyex得得代代入入將將),1(, 3/)0(, 0 yx32)0( y解解得得求求導(dǎo)導(dǎo)式式兩兩邊邊再再對(duì)對(duì),)1(x得得代代入入將將),2(,3/2)0(, 3/)0(, 0 yyx 9310

4、)0( y)2(0sincos2 yyyyex).(,cossincosln)(5xytttytxxfy 求求確確定定由由參參數(shù)數(shù)方方程程設(shè)設(shè)函函數(shù)數(shù)例例tttttttttxtyxycoscos/sinsincoscos)()()( 解解xxydxdxy)()( ttxytxxycos)(cosln)(由由參參數(shù)數(shù)方方程程確確定定xxyxy )()( ttttt)cos(ln)cos( tttxy)cos()(: 注注意意tttttcos/sinsincos tttttsincossincos2 )( )(txxyt 則則階階導(dǎo)導(dǎo)數(shù)數(shù)有有設(shè)設(shè)函函數(shù)數(shù),)(),(nxvxu)()()()()1(

5、nnnvuvu )()()()2(nnucuc )()(0)()()3(kknnkknnvuCvu ),()0()0(vvuu 其其中中式式稱稱為為萊萊布布尼尼茲茲公公式式)3(（二）高階導(dǎo)數(shù)性質(zhì)（二）高階導(dǎo)數(shù)性質(zhì))(32,15nxyexy求求設(shè)設(shè)例例 則則令令,23xveux , 2,2 vxv由由萊萊布布尼尼茲茲公公式式得得), 2, 1(33)(nkeuxkk 0)()4( nvvv)()(! 2) 1()()()()(2)2(32)1(32)(3)(23)( xennxenxexeynxnxnxnxn)1(693232 nnnxxexn解解vuvuvu )(xuxuyy 乘乘、除除四四則則計(jì)計(jì)算算法法則則特特別別注注意意)1(復(fù)復(fù)合合求求導(dǎo)導(dǎo)法法則則)2(2)(vvuvuvu 函函數(shù)數(shù)關(guān)關(guān)系系注注意意分分析析清清楚楚小結(jié)小結(jié)1 導(dǎo)數(shù)計(jì)算導(dǎo)數(shù)計(jì)算要要求求反反函函數(shù)數(shù)求求導(dǎo)導(dǎo)公公式式)( 1)()3(1xfyf 0)( xf)( )() )()(txxyxyxytx 求求二二階階導(dǎo)導(dǎo)數(shù)數(shù)注注意意怎怎樣樣參參數(shù)數(shù)方方程程求求導(dǎo)導(dǎo)時(shí)時(shí)要要特特別別)5(txyxy) )()( .,)4(復(fù)復(fù)合合求求導(dǎo)導(dǎo)問問題題有有兩兩邊邊求求導(dǎo)導(dǎo)時(shí)時(shí)隱隱函函數(shù)數(shù)求求導(dǎo)導(dǎo)法法則則數(shù)數(shù)或或冪冪指指函函數(shù)數(shù)。子子乘乘