§—多元函數(shù)的偏導(dǎo)數(shù)與全微分(復(fù)習(xí))

1、16.36.3—6.56.5多元函數(shù)的偏導(dǎo)數(shù)與全微分多元函數(shù)的偏導(dǎo)數(shù)與全微分偏導(dǎo)數(shù)偏導(dǎo)數(shù)在求多元函數(shù)對(duì)某個(gè)自變量的偏導(dǎo)數(shù)時(shí)，只需把其余自變量看作常數(shù)，然后直接利用一在求多元函數(shù)對(duì)某個(gè)自變量的偏導(dǎo)數(shù)時(shí)，只需把其余自變量看作常數(shù)，然后直接利用一元函數(shù)的求導(dǎo)公式及復(fù)合函數(shù)求導(dǎo)法則來(lái)計(jì)算之。元函數(shù)的求導(dǎo)公式及復(fù)合函數(shù)求導(dǎo)法則來(lái)計(jì)算之。13求下列函數(shù)的偏導(dǎo)數(shù)：（3）；(00)yxzxyxy????解：解：（把看作常數(shù)，對(duì)求導(dǎo)）??1lnyxyxx

2、xzxyyxyy???????yx（把看作常數(shù)，對(duì)求導(dǎo)）??1lnyxyxyxzxyxxxy???????xy（6）；2cossin()zxyxy??解：解：（把看作常數(shù)，對(duì)求導(dǎo)）??2cossin()xxzxyxy????yx??sin()2sin()cos()sin()2cos()1xyyxyxyyyxyxy????????（把看作常數(shù)，對(duì)求導(dǎo)）??2cossin()yyzxyxy????xy??sin()2sin()cos()si

3、n()2cos()1xyxxyxyxxxyxy????????高階偏導(dǎo)數(shù)高階偏導(dǎo)數(shù)例設(shè)求yxxyyxxz?????223334()()().xxxyyyfxyfxyfxy??????解()xfxy?1361222????yxyx()yfxy?1632???xyx，()xxfxy??[()]xxfxy???22(12631)xxxyy?????246xy??()yyfxy??[()]yyfxy???2(361)yxxy????6x??，(

4、)xyfxy??[()]xyfxy???22(12631)yxxyy?????66xy??()yxfxy??[()]yxfxy???2(361)xxxy????.66yx??17求下列函數(shù)的二階偏導(dǎo)數(shù)：（1）；cossinyxzxeye??3（2），求；22sinuvzeuxvy????，，zzxy????，解一：解一：，2sin2xyze??2222sin2sin2sin2sin2()cos()(4)xyxyxyxyxxyyzeexz

5、eey???????????????，解二：解二：由變量分析圖可得由變量分析圖可得222cos2()(sin)coscosuvuvxyuzzduexexexxudx?????????????2222cos2()()(2)24uvuvxyvzzveyeyyeyvy???????????????????（4），求。2sincos1axyzueyaxzxa?????，，dudx解一：解一：，2sincos1axaxxuea???222sinc

6、oscossinsincos()sin111axaxaxaxduaxxaxxaxxueeaexedxaaa??????????????解二：解二：由變量分析圖可得由變量分析圖可得222222222()()(sin)()(cos)11111cos(sin)111sincos11cos(sin)sin111axaxaxxyzaxaxaxaxaxaxaxyyzyzyzduuuuzeeaxexdxxyxzxaaayzaeeaxexaaaaxxa

7、eeaxexexaaa??????????????????????????????????????????????????抽象復(fù)合函數(shù)的偏導(dǎo)數(shù)或?qū)?shù)：抽象復(fù)合函數(shù)的偏導(dǎo)數(shù)或?qū)?shù)：例設(shè)，其中f具有二階連續(xù)偏導(dǎo)數(shù)，求。()yzfxyx?zzyy????，解：引進(jìn)中間變量，函數(shù)z可看作如下的復(fù)合函數(shù)而()zfuv?yuxyvx??由變量分析圖可得由變量分析圖可得2()uxvxuvyzfufvfyfxx?????????????????；1uy