高中數(shù)學導(dǎo)數(shù)與微分教學課件(編輯修改稿)
2025-06-29 21:38 本頁面
【文章內(nèi)容簡介】 yyyyyyyxys i n111)s i n1(0s i n10s i n1)????????????????????解：（導(dǎo)數(shù)與微分 exeeyyxxeeyexeyyxeeyyxeyxyyyyyyyyyy?????????????????????????01|1)0(101)1(0)0(12)時解：求（導(dǎo)數(shù)與微分 25|1)2(21|1)2()4,2(),0,2(,4,0,44421,22222)2(223)420221222????????????????????????????????????????yxyxyxyxyyxyxyyyyyxyxyxyyyyxyxyxyxyx及解得代入原方程：將解：求（導(dǎo)數(shù)與微分 yxyxyxyxyxyxyxexyeyyeexyyeyxyexyexy?????????????????????????)()1()()(4)解：（導(dǎo)數(shù)與微分 )]()()()(ln)([)]([)()()()(ln)(1)(ln)(ln:)]([.4)()(xfxfxgxfxgxfyxfxfxgxfxgyyxfxgyxfyxgxg?????????????取對數(shù)化成隱函數(shù)數(shù)皆為變量）稱為冪指函數(shù)（底和指冪指函數(shù)求導(dǎo)法則導(dǎo)數(shù)與微分 )s i nln( c oss i nlnc os1lns i nlnln)2().ln1(1ln1,lnln1)7s i ns i ns i nxxxxxyxxxxyyxxxyxyxxyxxxyyxxyxyxxxxx???????????????????（：求下列函數(shù)的導(dǎo)數(shù)例導(dǎo)數(shù)與微分 )ln()ln(ln)( l nlnlnlnln,lnln3)xxyxyyxyyxyyyxxyyyxyxyxyyxxyyxyxxyxy??????????????????（導(dǎo)數(shù)與微分 ? 注：對一些較復(fù)雜的乘積，商或根式函數(shù)求導(dǎo)時，可利用先取對數(shù)后求導(dǎo)的方法計算 62333623232333333333121112)1313(311)]1l n ()1[ l n (3111ln31)11l n (ln11)4(31xxxxyxxxxxxyyxxxxxxyxxy???????????????????????????解：導(dǎo)數(shù)與微分 2111111))1( l n ()()()()()1l n (1)8)()()()()(:2tttar c t gttxtyxyar c t gtytxtxtyxyttyytxxtt??????????????????????????????（的函數(shù)的導(dǎo)數(shù)：求下列參數(shù)方程給出例求導(dǎo)公式：由參數(shù)方程給出的函數(shù) ??導(dǎo)數(shù)與微分 1)0(111,001|c oss i ns i nc osc oss i ns i nc oss i nc os)()(0c oss i n)3(2)c os1(s i n)s i n()c os1()()()c os1()s i n()2(0???????????????????????????????????????????????????xyxyyxtyktttt
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