3-復合函數(shù)-隱函數(shù)求導(編輯修改稿)
2025-08-20 06:27 本頁面
【文章內(nèi)容簡介】 0nk n!nk 121 knknkxknnnnxy)())(()( )()(?1nn)n()1x(!n)1()1x1(??????.y),x1ln(y:)2( )n(求設 ??xy ??? 11解 )1!0,1()1( )!1()1( 1)( ?????? ? nxny nnn)1n()n( )y(y ??? )1n()x11( ???1nn)n()1x(!n)1()1x1( ??????.y,xs iny )3( )n(求設 ?解 xy c os?? )2s i n ( ??? xxs iny ???? )22s in ( ???? xxc osy ????? )23s in ( ???? x??)2s i n()( ???? nxy n)2c o s ()( c o s )( ???? nxx n同理可得 xs iny )4( ? )24xs in ( ????xay )4( ? nx)n(x )a( l na)a( ?x)n(x e)e(: ?特別地由復合函數(shù)求導法則有 : )2nbaxs in (a))bax( s in()2( n)n( ??????)2nbaxc o s (a))bax( c o s ()3( n)n( ??????baxn)n(bax ea)e)(1( ?? ?n1nn)n( a)bax(!n)1()bax1)(4(?????2. 高階導數(shù)的運算法則 : 則階導數(shù)具有和設函數(shù) ,nvu)n()n()n( vu)vu()2( ???????)n()n( Cu)Cu()1( ?)k()kn(n0kkn)n(nn)k()kn(kn)2n(2n)1n(1n)n(0n)n(vuCuvCvuCvuCvuCvuC)vu()3(?????????????????
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