高中數(shù)學(xué)導(dǎo)數(shù)與微分教學(xué)課件-展示頁(yè)
2025-05-30 21:38本頁(yè)面
【正文】 處可導(dǎo)，則 2)()()()(vvuvuvuuccuvuvuvuvuvu????????????????????導(dǎo)數(shù)與微分 ? xxxxxxyxxxxyxc os12)s i n(s i n1)2122??????????????（xxxxxxxxxyxxyxln1ln)( l nln)()ln(ln)21????????????（導(dǎo)數(shù)與微分 222)1(2)1(11)1()1)(1()1()1(11113????????????????????????xxxxxxxxxxxyxxy）（）（導(dǎo)數(shù)與微分 !)1()()2)(1(0)0()0()()()()()()(y),()2)(1()(,2!)1()()2)(1(0)()2)(1(lim0)0()(lim)0(1)0(),()2)(1()4(00nnfyxfxxfxfxxfxyxxfnxxxxfnnxnxxxxxfxfyynxxxxynnxx??????????????????????????????????????????????????則：令解法：利用導(dǎo)數(shù)的定義計(jì)算解法求導(dǎo)數(shù)與微分 。求導(dǎo)自變量對(duì)乘中間變量求導(dǎo)對(duì)中間變量即函數(shù)點(diǎn)處可導(dǎo)，且在則點(diǎn)處可導(dǎo)，在相應(yīng)的點(diǎn)處可導(dǎo)，在若定理：設(shè)( x )xu( u )uy( x )( u )yx( x ) ]f[yu)(x)(),(),(??????????????ffufxxuufy導(dǎo)數(shù)與微分 注：復(fù)合函數(shù)求導(dǎo)法則的關(guān)鍵在于： （ 1) 將復(fù)合函數(shù)分解成若干個(gè)基本初等函數(shù)； （ 2) 分別求出這些函數(shù)的導(dǎo)數(shù)并相乘； （ 3) 將所設(shè)中間變量還原 導(dǎo)數(shù)與微分 ? ?3 2223 2222134)4()(3121)(21,:,21)2(s e cs e c1)()( l n,ln:,ln( 1)4323131xxxuyxuyxuuyxyxc t gxxut gxuyt gxuuyt gxy????????????????????????????????令令求下列函數(shù)的導(dǎo)數(shù)例導(dǎo)數(shù)與微分 xxxxxxxxxt geeeeeevuevuyevvuuyey??????????????????c oss i n)s i n(1)()( c os)( l n,c os,ln:,c osln)3( 令導(dǎo)數(shù)與微分 ?1)1()1(2121111)()()( l n,ln:)1(,ln)4(2?????????????????????yxar c t gxxxvuxar c t gvuyxvar c t gvuuyyxar c t gy令求導(dǎo)數(shù)與微分 xxuxuxuxxxvvyvvuyy1c o s2111c o ss i n22ln)()s i n(2ln2)(( c os)2(,c os,2:25)121????????????????????）令（導(dǎo)數(shù)與微分 xxxxxtvuxtvuyxttvvuuyxy2c os14s i n2c os12c os2s i n22)s i n(221)2()( c os1()(2,c os,1,:2c os16)22222???????????????????????????）令（導(dǎo)數(shù)與微分 )(ln2)(ln2)(()(,),(:)()(7)22222xxvvvxafaxaufaxaxaufyxvauufyafuf???????????????????）令的導(dǎo)數(shù)存在，求已知（導(dǎo)數(shù)與微分 ? 例 5：證明：偶函數(shù)的導(dǎo)數(shù)是奇函數(shù)。同理可證奇函數(shù)的導(dǎo)數(shù)是奇函數(shù)故 )()()()()(),()()(xfxfxfxfxfxfxxf????????????????
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