偏导数和全微分习题课

偏导数和全微分习题课相关概念复习偏导数和全微分作业题讲解课堂练习探索思考题偏导数(partialderivative)000000000,,,,limxxfxxyfxyfxyfxyxx000,,fxyPxyx在对的偏导数：000000000,,,,limyyfxyyfxyfxyfxyyy000,,fxyPxyy在对的偏导数：,,,,,.xyfffxyfxyxy偏导函数：00(,)xxdfxydx偏导数的几何意义00(,):xfxy空间曲线0(,)yyzfxy0000(,,(,))xyfxyx在点处的切线关于轴的斜率。00(,):yfxy曲线0(,)xxzfxy0000(,,(,))xyfxy在点处的切线关于y轴的斜率。全微分(totaldifferential)000022,,zfxxyyfxyAxByoxyAxByo若0000,,zfxxyyfxyAxByxy或,0,0xy其中、是时的无穷小量,000,,fxyPxy则称在可微。0000,,PfxyPxydzAxBy在的全微分：全微分00000,,.Pxydzfxyxfxyy且0000,,,fxyPxyfxyP若在可微,则在可导，000,,fxyPxy在可微0000,,1xyzfxyxfxyyo0000,,2xyzfxyxfxyyo00000,,lim03xyzfxyxfxyy判断函数在一点可微的方法(1),zAxByo0000,,,zfxxyyfxy1、计算判断是否有.zAxByxy或000,zxyPxy如判断在处是否可微。0000zxxyyxy00yxxyxy22,0,0lim0xyxyxy22xyoxyo判断函数在一点可微的方法(2)0000,,,;xyfxyfxy1、求偏导数0000,,;xyzfxyxfxyy2、计算000000000,,,,,lim0.xyxyzfxyxfxyyozfxyxfxyy3、判断是否有即224,00fxyxy、证明函数在，连续但偏导数不存在。证明：22,0,0,0,00lim,limlim00,0xyxyfxyxyf,0,0fxy在连续。200,00,00,0limlim,xxxfxfxfxx200limlim0,0xxxxxfxx而不存在，所以不存在，0,0yf同理可得也不存在。2222221sin,0,0,000xyxyxyfxyxy5、考察函数在，处的可微性。解：00,00,0000,0limlim0,xxxfxffxx0,00,0,0,00xyzfxfyfxyf0,0yf同理可得=0.222222221sin1sin,xyxyxyxyxyxy2222221sin,0,0,000xyxyxyfxyxy5、考察函数在，处的可微性。22220,00,01sin,xyzfxfyxyxyxy2222,0,01lim0,sin1,xyxyxyxy由于,0,00,00,0lim0,xyxyzfxfy所以0,00,0,0,0xyzfxfyof即在处可微。2222222,0,0,000xyxyxyfxyxy6、证明函数在，处连续且偏导数存在，但在此点不可微。222,0,0,0,0222200lim,limcossinlimlimcossin00,0,xyxyxyfxyxyf解：,0,0fxy所以在连续。00,00,0000,0limlim0,xxxfxffxx0,0yf同理可得=0.2222222,0,0,000xyxyxyfxyxy6、证明函数在，处连续且偏导数存在，但在此点不可微。3221,;2222xxyFxyxx当时，有222220,00,0,,xyzfxfyxyFxyxyxy0,0,0;xyFxy当时，有000,00,0limlim,xyzfxfyFxy所以不存在，0,0f即在处不可微.222222221sin,0,0,00000xyxyxyfxyxyf7、证明函数在，处连续且偏导数存在，但偏导数在此点不连续，而在，处可微。00,00,010,0limlimsin0,xxxfxffxxx2222222222112sincos,0,,0,0xxxxyxyxyxyfxyxy22,0,01lim2sin0,xyxxy2222,0,01limcosxyxxyxy不存在。1231(5)PEX,,,.uuufxyxyxy设求,sxytxy解：设，则1212,.uufyffxfxy或,uusutuuyxsxtxst.uusutuuxysytyst1234PEX,fxy设可微，cossin,sincos,xuvyuv,cossin,sincos,guvfuvuv2222.xyuvffgg求证cossin,sincos,uxyvxygffgff222222cossinsincos.uvxyxyxyggffffff1235PEX,232,fuFxtfxtfxt设是可微函数，0,00,0.xtFF试求和解：,2232322332,xFxtfxtxtfxtxtxxfxtfxt,22323222232,tFxtfxtxtfxtxtttfxtfxt0,0030=40,0,00.xtFfffF所以可微的几何意义00(,)(,)fxyxy在可微0000,(,,(,))zfxyxyfxy空间曲面在点z存在不平行于轴的切平面.0000000(,)()(,)()xyzzfxyxxfxyyy法线方程：0000000.,,1xyxxyyzzfxyfxy课堂练习1,,0,0fxyxy、设考察该函数在处是否连续，是否可微，是否存在偏导数。3(,),(,,),(,,),,.ufxyvgxyuwhxuvwwxy、设求44,.xyzdzxy、设求22223273,1,1xyz、求曲面在处的切平面与法线方程。探索思考题00001,,xfxyfxyx、存在能保证在点连续吗？01,yy由此能否进一步断言，对充分接近的10,fxyx在连续?探索思考题2,,,,00fxyxyxyxy、设其中在，,xy点的某个邻域上有定义，要求给加上适当的条件，使得1,0,0fxy在点连续；2,0,0fxy在点存在偏导数；3,0,0fxy在可微。探索思考题3,,,,zfxyabcddzfxyD、设在开集上可微，且全微分恒为零，问在内是否应取常数值？证明你的结论。