[研究生入學(xué)考試]清華大學(xué)微積分課件全x-資料下載頁(yè)
2024-10-19 01:15本頁(yè)面
【正文】 函數(shù)稱為的二元函數(shù)上在則各偏導(dǎo)數(shù)仍然是定義中處處有偏導(dǎo)數(shù)在區(qū)域若二元函數(shù)yxfDyyxfxyxfDyxfz?????.),(,),(),(的二階偏導(dǎo)數(shù)則稱它們是存在的偏導(dǎo)數(shù)也和如果偏導(dǎo)函數(shù)yxfyyxfxyxf????三、高階偏導(dǎo)數(shù) 2021/11/10 33 ),()( 22yxfx fxfx xx????????),()( 22yxfyfyfy yy????????),()(2yxfxyfxfy yx?????????),()(2yxfyxfyfx xy?????????的二階偏導(dǎo)數(shù)關(guān)于 xf的二階偏導(dǎo)數(shù)關(guān)于 yf的二階混合偏導(dǎo)數(shù)后對(duì)先對(duì) yxf的二階混合偏導(dǎo)數(shù)后對(duì)先對(duì) xyf2021/11/10 34 的二階偏導(dǎo)數(shù)求例 )2c o s (]1[ yxxyz ???][解 )2s i n ( yxyxz ?????由 )2s i n (2 yxxyz ?????)2c o s (22yxxz ????? )2c o s (422yxyz ?????得到)]2s i n ([2yxyyxyz ???????? )2c o s (21 yx ???)]2s i n (2[2yxxxyxz ???????? )2c o s (21 yx ???xyzyxz??????? 22可見(jiàn)2021/11/10 35 ??????????)0,0(),(0)0,0(),(),(]2[ 2222yxyxyxyxxyyxf設(shè)例yxfxyf?????? )0,0()0,0( 22 和求][解xyfy??? ),0(,0 計(jì)算對(duì)于)( 2222yxyxxyx ????22222222222)()(2)(2yxyxxyxxxyyxyxy????????222244)(]4[)(yxxyxyyxy????2021/11/10 36 xf?? )0,0(, 計(jì)算利用偏導(dǎo)數(shù)的定義0?xfxfxfx ???)0,0()0,(lim)0,0(0????????????0,00,),0(yyyyf x這樣就求得同理可求得??????000)0,(xxxxf y得到再令 ,0?x yxyf ???? ),0( )0( ?y2021/11/10 37 因而1??1?yfyfxyf xxy ???)0,0(),0(l i m)0,0(02 ??????xfxfyxf yyx ???)0,0()0,(lim)0,0(02 ??????xyfyxf??????? )0,0()0,0( 222021/11/10 38 那麼就有連續(xù)并且在點(diǎn)的鄰近存在在點(diǎn)和偏導(dǎo)數(shù)的兩個(gè)二階混合如果函數(shù)混合偏導(dǎo)數(shù)定理定理,),(,),(),(),(),()(:000022yxyxyxyxfxyyxfyxf??????yxyxfxyyxf??????? ),(),( 002021
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