大一高數(shù)課件第二章-wenkub
2023-05-26 01:58:40 本頁(yè)面
【正文】 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 xy ??? 112)1(1xy ????? 3)1(!2xy ?????4)4()1(!3xy ??? ??)1!0,1()1( )!1()1( 1)( ?????? ? nxny nnn例 4 .,si n )( nyxy 求設(shè) ?解 xy c os?? )2s in( ??? 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 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)()()3( kknnkknn vuCvu ????? 萊布尼茲公式 vu ???? 3)( ?vu vuvu ????)( ??vu )( ????? vuvu vuvu ?????? 2 vu ???)( ???vu vu ???? vu ???? 3 vu ????用數(shù)學(xué)歸納法可證 萊布尼茲公式 成立 .
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