《高階導(dǎo)數(shù)數(shù)分教案》ppt課件-文庫(kù)吧
2024-12-31 17:38 本頁(yè)面
【正文】 )( nyRxy 求設(shè) ??? ?解 1????? xy)( 1 ????? ??xy 2)1( ????? x??3)2)(1( ???????? x))1(( 2 ??????? ??xy)1()1()1()( ???????? ?? nxny nn ?則為自然數(shù)若 ,n?)()( )( nnn xy ? ,!n? )!()1( ??? ny n .0?例 3 .),1l n ( )( nyxy 求設(shè) ??解 注意 : xy ??? 112)1(1xy ?????3)1(!2xy ????? 4)4()1(!3xy ?????)1!0,1()1( )!1()1( 1)( ?????? ? nxny nnn 求 n階導(dǎo)數(shù)時(shí) ,求出 13或 4階后 ,不要急于合并 ,分析結(jié)果的規(guī)律性 ,寫(xiě)出 n階導(dǎo)數(shù) .(數(shù)學(xué)歸納法證明 ) 例 4 .,s i n )( nyxy 求設(shè) ?解 xy c os?? )2s i n( ??? x)2c os ( ????? xy )22s i n( ????? x )22s i n( ???? x)22c o s ( ??????? xy )23s i n(???? x??)2s i n()( ???? nxy n)2c os ()( c os )( ???? nxx n同理可得 例 5 .),(s i n )( nax ybabxey 求為常數(shù)設(shè) ?解 bxbebxaey axax c oss i n ???)c oss i n( bxbbxae ax ??)ar c t an()s i n(22 abbxbae ax ???????)]c os ()s i n ([22 ?????????? bxbebxaebay axax)2s i n (2222 ??????? bxbaeba ax??)s i n()( 222)( ????? nbxebay axnn )a r c t a n( ab??2. 高階導(dǎo)數(shù)的運(yùn)算法則 : 則階導(dǎo)數(shù)具有和設(shè)函數(shù) ,nvu)()()()()1( nnn vuvu ???)()()()2( nn CuCu ?)()(0)()()()2()1()()(!)1()1(!2)1()()3(kknnkknnkknnnnnvuCuvvukknnnvunnvnuvuvu???????????????????????萊布尼茲公式 例 6 ., )20(22 yexy x 求設(shè) ?解 則由萊布尼茲公式知設(shè) , 22 xveu x ??0)()(!2)120(20)()(20)(2)18(22)19(22)20(2)20(???????????xexexeyxxx22!21920222022182192220????????xxxexexe)9520(2 2220 ??? xxe x : 常用高階導(dǎo)數(shù)公式 nn xnx ??? ??????? )1()1()()4( )( ?nnnxnx )!1()1()( l n)5( 1)( ??? ?)2s i n()( s i n)2( )( ???? nkxkkx nn)2c os ()( c os)3( )( ???
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