微積分33復(fù)合函數(shù)求導(dǎo)法則-資料下載頁
2025-01-20 05:44本頁面
【正文】 n)3( 22 ?? xy。)21()2( 10xy ?? )。0(e)1( 為常數(shù)?? axy ax.e)1( axax?? )e()1( ??? axxy axax ax ee ??)21()21(10 9 ???? xx.32 1a r c s i n 2xxx ??? ??2)1(1)1(1a r c s i n2??????xxx)1( ar c s i n1ar c s i n2 ???? xx])1[ ( ar c s i n)3( 2 ???? xy])1[ l n(21)1(21)4( 22 ???????? xxxxy1211212222?????xxx1)1(2112)1(12122222??????????????????xxxxxxxx11121112122222?????????????????xxxxxxx.122?? xx??????? ???xxx x )21l n ()21( 1.)21( )21l n ()21(2)21( 21 ????????????xxxxxx x])21[ l n ()21()5( 11 ????? xx xxy???????????????2)21l n (212)21(1xxxxx x證明 ),()( 00 xgxxgu ?????設(shè))()( 00 ufuuf ????0)()( 00 ?????? xgxxguxxgxxguufuufxy??????????? )()()()( 0000,0 xx ?一個改變量給uu ??? 0則這時uxgxxg ????? )()( 00)]([)]([ 00 xgfxxgfy ?????,}{\)(,0)( 000 xxOxg ?存在去心鄰域時當(dāng) ??,}{\)( 000 時使得 xxOxx ????,)( 0 點可導(dǎo)在由于 xxg,0)()(,0 00 ???????? xgxxgux 時則點可導(dǎo)可得在從而由 0)( uuf)()( 00 xguf ???xxgxxguufuufxyxux ?????????????????)()(l i m)()(l i ml i m 0000000即 式成立 . )43( ?)0()()()( 0 ??????? xxoxouf因此由導(dǎo)數(shù)定義式可得 .43 ）式仍成立即（ ?))(()()( 0 xooxouf ?????)0()()()( 00 ??????? xxoxgxxg）式可得由（時當(dāng) 23,0)( 0 ??? xg0))]([( 0 ?? ? xxxgf)()]([)]([)]([ 0000 ufxoufxgfxxgf ???????可得利用處可導(dǎo)在根據(jù) )23(,)()( 00 ?? xguuf例 2 .0)()(ln,)( 處的導(dǎo)數(shù)在求可導(dǎo)設(shè) ?? xfxfyxf解 )()(ln )( xfu xfu ???? ?,0)( 時當(dāng) ?xf))(( l n ??? xfy.)( )(xf xf ?? )(1)(xfuxfu????上式可推得公式 )73())() ( l n()( ???? xfxfxf—— 取對數(shù)求導(dǎo)法
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