高等数学练习答案8-2

习题821求下列函数的偏导数(1)zx3yy3x解323yyxxz233xyxyz(2)uvvus22解21)(uvvuvvuuus21)(vuuuvvuvvs(3))ln(xyz解xyxyxxxz1lnln121)lnln()ln(21xyx同理)ln(21xyyyz(4)zsin(xy)cos2(xy)解yxyxyyxyxz)]sin([)cos(2)cos()]2sin()[cos(xyxyy根据对称性可知)]2sin()[cos(xyxyxyz(5)yxztanln解yxyyyxyxxz2csc21sectan12yxyxyxyxyxyz2csc2sectan1222(6)z(1xy)y解121)1()1(yyxyyyxyyxz]1)1[ln()1ln()1ln(xyxyxyeeyyzxyyxyy]1)1[ln()1(xyxyxyxyy(7)zyxu解)1(zyxzyxuxxzzxxyuzyzyln11lnxxzyzyxxzuzyzyln)(ln22(8)uarctan(xy)z解zzyxyxzxu21)(1)(zzyxyxzyu21)(1)(zzyxyxyxzu2)(1)ln()(2设glT2试证0gTglTl解因为lglT1ggglgT1)21(223所以0glglgTglTl3设)11(yxez求证zyzyxzx222解因为2)11(1xexzyx2)11(1yeyzyx所以zeeyzyxzxyxyx2)11()11(224设yxyxyxfarcsin)1(),(求)1,(xfx解因为xxxxf1arcsin)11()1,(所以1)1,()1,(xfdxdxfx5曲线4422yyxz在点(245)处的切线与正向x轴所成的倾角是多少？解因为242xxxztan1)5,4,2(xz故46求下列函数的22xz22yzyxz2(1)zx4y44x2y2解2384xyxxz2222812yxxzyxyyz23842222812xyyzxyyxyyyxz16)84(232(2)xyzarctan解22222)(11yxyxyxyxz22222)(2yxxyxz2222)1(11yxxxxyyz22222)(2yxxyyz22222222222222)()(2)()(yxxyyxyyxyxyyyxz(3)zyx解yyxzxlnyyxzx222ln1xxyyz222)1(xyxxyz)1ln(1ln)ln(112yxyyyyxyyyyyxzxxxx7设f(xyz)xy2yz2zx2求fxx(001)fxz(102)fyz(010)及fzzx(201)解因为fxy22xzfxx2zfxz2xfy2xyz2fyz2zfz2yzx2fzz2yfzzx0所以fxx(001)2fxz(102)2fyz(010)0fzzx(201)08设zxln(xy)求yxz23及23yxz解1)ln()ln(xyxyyxxyxzxxyyxz122023yxzyxyxyxz122231yyxz9验证(1)nxeytknsin2满足22xykty证明因为nxeknknnxetytkntknsin)(sin2222nxnexytkncos2nxenxytknsin2222nxeknxyktknsin2222所以22xykty(2)222zyxr满足rzryrxr2222222证明rxzyxxxr222322222rxrrxrxrxr由对称性知32222ryryr32222rzrzr因此322322322222222rzrryrrxrzryrxrrrrrrzyxr23)(332232222