【微積分】求導法則(參考版)
2025-04-26 03:39本頁面
【正文】 ()(])()([)1(2??????????????????xvxvxvxuxvxuxvxuxvxuxvxuxvxuxvxuxvxu證 (3) ),0)((,)( )()( ?? xvxv xuxf設hxfhxfxfh)()(lim)(0?????hxvhxvhxvxuxvhxuh )()()()()()(lim0 ??????hxvxuhxvhxuh)()()()(l i m0?????證 (1)、 (2)略 . hxvhxvxvhxvxuxvxuhxuh )()()]()()[()()]()([lim0 ????????)()()()()()()()(lim0 xvhxvhxvhxvxuxvhxuhxuh ??????????2)]([)()()()(xvxvxuxvxu ????.)( 處可導在 xxf?推論 vCuCvCuC ?????? 2121 ][)2wuvwvuvwuu v w ???????)()3uCuC ???)()1 ( C為常數(shù) ) 例 1 .2lnc o s3 的導數(shù)求 ??? xxy解 23 xy ?? xx s i n3 2 ??例 2 .lns i n2 的導數(shù)求 xxxy ??解 xxxy lns i n2?? xxx lnc o s2?xxx1s i n2 ??xsin? 0?xxxxxxxx s i nlnc o slns i n2 2 ???例 3 .tan 的導數(shù)求 xy ?解 )co ss i n()( t a n ??? xxx xxxxx2c os)( c oss inc os)( s in ????xxx222coss i ncos ?? xx22 s e cc os1 ??.se c)( t a n 2 xx ?? .cs c)( cot 2 xx ???類似求得 即 例 4 .s e c 的導數(shù)求 xy ?解 )c o s1()( s e c ??? xx xx2c o s)( c o s ???xx2cossin? .t a ns e c xx?即 .ta ns e c)(s e c xxx ??類似求得 .c o tc s c)( c s c xxx ???其中 )(x? 在 ax ?因 )()()()( xaxxxf ?? ?????故 )()( aaf ????axafxfafax ?????)()(lim)(xxaax ????)()(lim ?)(lim
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