高等数学( 北大版)习题6.4

习题6.422222222222222222222332222223221111.:(1)ln().11,.()(2).1,.(3),lnln.1ln,(lyxyyyyzxxyxxyzxxxyxyyxyzyyxxyxyxxyxzxyzxyxxyxyxyzxyyxyzxzxxzyxxxzyxzx求下列函数的一阶偏导数1222222222n),1lnln,(ln).(4).,()().()()(5)arcsin().,.121(6).()(1yyyxyxyxyxyxxzzxxxzxxzyyxyzxyzxyxyyxxyxyzxyyxxyxyxyzxyyzzxxyxyyxyzxezexeyexyx2222),.(7).111,,.xyzxeyyzxuxyzuyuzuxxxzyxyzyz11(8)().(),(),().ln()zzzzuxyuuuyzxyxzxyxyxyxyz(0,1)(0,1)2(0,1)00(0,1)12arccos(1)(1)cos(1),.1sinsin(1)1sincos1,1sin(1sin)(1)(1sin(1))(1)cos(1)1sin(1)(1sxxyxyyxzzzxyxyzdxdxxxxdxxdxxzdydyyyydyydy求下列函数在指定点的偏导数:求及21(,1)(,1)22222(,1)(,1)221.in(1))2(2),.cos2sincos2cos,2(cos)(cos)(cos)2,0.(3)(,,)ln(),(2,1,0),(2,1,0),(2,1,0).(,,)yxyzxyyzzzyxxyzyxzyxyxxyxyyxyxzzxyfxyzxyzffffxyz求及求1,(,,),(,,).11(2,1,0),(2,1,0)1,(,,).22yzxyzyxfxyzfxyzxyzxyzxyzfffxyz222220,(,)(0,0),3.(,)||||0,(,)(0,0)(0,0),(0,0).(,)|||||0((,)(0,0)),||||(,)(0,0)0((,)(0,0)),(,)(0,0)|(0,0)limxxxxyxyfxyxyxyfxyfxyxyxyxyfxyfxyfxyxf证明函数在连续但是不存在在连续.证|0|lim.||xxxxx不存在24.sin.211sincos,cos,2sincoscossin.222yzzzzxxyxxyzyyyzyxxxxxxyxxxzzxyyyyyxyzxyxyxxxxxx设,证明为齐次函数根据关于齐次函数微分的一个定理立得结论直接计算如下证1/2,,..22322225.:(1)(,)ln(23).26,.23(23)(2)(,)sin.cos,cos.(3)(,)4ln(1).2112,2.1(4)(,)ln()lnln.lnln1xyxxyxxxxyxxyxffxyxyffxyxyfxyyxefyxefxfxyxxyxxxfyxfyxfxyxxyxxxyfyx求下列函数的二阶混合偏导数2232223322332222222221,.6.cos3,Laplace0.3sin3,9cos3,3cos3,9cos3,0.7.(,)4cos(33)xyyyyyyxctfyuuuexuuxyuuexexxxuuexexyyuuuxyuuuxtexctct设证明满足平面方程证明函数满足波动方程证222222222222.12sin(33),36cos(33),12sin(33),36cos(33),.8.(,)(,),.xctxctxctxctxuucecxctcecxctttuuexctexctxxuuctxuuxyvvxyDuuuvuvDxyyx故设及在内又连续的二阶偏导数,且满足方程组证明及在内证2222Laplace0,.uvuuxy满足平面方程其中222222222,(),0.0.uvvuvvvvvxxyxyyyxyxxyyxxyuv和连续故类似证证2222/19.(,)sin(0,)2sin.111sinsinln(1),11(0,)2sin,(,)sinln|1|2sin1(2)sinln|1|.10.:(1).yxzzxyyzyyyzxxyydxxyxyCxyyzyCyyzxyxyxyyyyxyyxyyzed已知函数满足以及.试求的表达式求下列函数的全微分解z=//22222222222222.()()()()(2)(2))(2)..()()(3)arctanarctanarctanarccot,0.2()2(()(4),yxyxyxdyydxzedexxxydxdyxyxydxdyydxxdyzdzxyxyxyyxyyzdzxyxxdxyzxdxydyzdzuxyzduxyzxy2.z334223433422343344234223223411.(,)(4103)(15125),(,).4103,15125.(4103)53(),1512()15125,()zxydzxxyydxxyxyydyfxyzzxxyyxyxyyxyzxxyydxxxyxyCyzxyxyCyxyxyyyCy已知函数的全微分求的表达式解454234522222222222222225,().(,)53.12.(,)(),(,).()11()()2211yCyyCfxyxxyxyyCzfxyyxdzxdxydyzxyxyxyyxdzxdxydyxyxyxdyydxxdxydyxyxdyydxydxxdxydxyyx已知函数的全微分求的表达式解22222221()arctan21()arctan.21()arctan.2ydxydxyxydxyxyzxyCx222000000000000013.(,):{()()}0,0.:(,).(,),(,)(,)[(,)(,)][(,)(,)](,)()(,)()0.(,)(,),(,).xyffzfxyDxxyyRfxyxyxyDfxyfxyfxyfxyfxyfxyfyxxfxyyfxyfxyxyD证明在区域上恒等于常数证222214.:(,)||(0,0),(0,0),(0,0),(,)(0,0).(,)|||0(0,0)((,)(0,0)),(,)||(0,0)(0,0)0,(0,0)0.(,)(0,0)()(0),(,)xyxyofxyxyfffxyfxyxyfxyfxyxyfffxyxyxyfxx证明函数在点处连续存在但在处不可微在点处连续.若在处可微,将有f(x,y)=特别应有证|||(2||)(0),.oxxx但此式显然不成立12222115.(,)(,)(,),,()..(,)(,).,.,,,(),(),,.PxydxQxydyDuxyPQCDPQyxuuduPxydxQxydyPQxyPuuQuuyyxyxxxyxyPQuuPQPQCDCDyxyxxyyx设在区域中是某个函数之全微分且证明由假设由得故即证22222(),(,)(0,0)16.(,)0(,)(0,0).(1)(0,)(0);(2)(0,0)0;(3)(0,0)1;(4)(,0),(0,0)1.[2((1)0,(,)xxxyyyxxxyxyxyfxyxyxyfyyfffxfxyxyfxy设函数计算根据偏导数定义证明在上述结果的基础上证明重复上述步骤于并证明设则证2222222225402222222222)]()2(),()(0,).(2)(,0)0,(0,0)0.(3)(0,0)()|1.[2()]()2()(4)0,(,),()(,0).(0,)0,(0,0)0.(0,0)xxxyyyyyyxyyxyxxyxyxyyfyyyfxffyxyxyxxyyxyxyxfxyxyfxxfyffx设则03332232322322|1.17.ln(),.11ln()ln()1,,,11,.xzzzxxyxxyzyzzxyxxyxxyxxxxzzxyyxyy设求解