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习题841设zu2v2而uxyvxy求xzyz解xvvzxuuzxz2u12v12(uv)4xyvvzyuuzyz2u12v(1)2(uv)4y2设zu2lnv而yxuv3x2y求xzyz解xvvzxuuzxz31ln22vuyvu222)23(3)23ln(2yyxxyxyxyvvzyuuzyz)2()(ln222vuyxvu2232)23(2)23ln(2yyxxyxyx3设zex2y而xsintyt3求dtdz解dtdyyzdtdxxzdtdz2223)2(costeteyxyx)6(cos)6(cos22sin223ttettettyx4设zarcsin(xy)而x3ty4t3求dtdz解dtdyyzdtdxxzdtdz22212)(113)(11tyxyx232)43(1)41(3ttt5设zarctan(xy)而yex求dxdz解dxdyyzxzdxdzxxxexxeeyxxyxy2222221)1(116设1)(2azyeuax而yasinxzcosx求dxdu解dxdzdzudxdyyuxudxdu)sin(1cos11)(222xaexaaeazyaeaxaxax)sincoscossin(122xxaxaxaaeaxxeaxsin7设yxzarctan而xuvyuv验证22vuvuvzuz证明)()(vyyzvxxzuyyzuxxzvzuz)()(111)(11222yxyxyyx)1()()(111)(11222yxyxyyx22222vuvuyxy8求下列函数的一阶偏导数(其中f具有一阶连续偏导数)(1)uf(x2y2exy)解将两个中间变量按顺序编为12号2122212)()(fyefxxefxyxfxuxyxy212)2212)((fxefyyefyyxfyuxyxy(2)),(zyyxfu解1211)()(fyzyxfyxxfxu)()(21zyyfyxyfyu2121fzfyx)()(21zyzfzxzfzu22fzy(3)uf(xxyxyz)解yzfyffxu3211321fyzfyf3232fxzfxxzfxfyu33fxyxyfzu9设zxyxF(u)而xyuF(u)为可导函数证明xyzyzyxzx证明yzyxzx])([])()([yuuFxxyxuuFxuFyx)]([)]()([uFxyuFxyuFyxxyxF(u)xyzxy10设)(22yxfyz其中f(u)为可导函数验证211yzyzyxzx证明uffxyufxfyxz2222uffyufufyfyufyz2222)(1)2()(所以)(11221122ufyuffyuffyyzyxzx211yzzyy11设zf(x2y2)其中f具有二阶导数求22xzyxz222yz解令ux2y2则zf(u)fxxuufxz2)(fyyuufyz2)(fxfxufxfxz2224222fxyyufxyxz422fyfyufyfyz42222212求下列函数的22xzyxz222yz(其中f具有二阶连续偏导数)(1)zf(xyy)解令uxyvy则zf(uv)ufyvfyufxvvfxuufxz0vfufxvfxufyvvfyuufyz1因为f(uv)是u和v的函数所以uf和vf也是u和v的函数从而uf和vf是以u和v为中间变量的x和y的函数)()()(22ufxyufyxxzxxz222222222)0()(ufyvufyufyxvvufxuufy)(1)()(2ufyyufufyyxzyyxz)(222yvvufyuufyufvufyufxyufvufxufyuf222222)1()()()()(22vfyufyxvfufxyyzyyzyvvfyuuvfyvvufyuufx222222)(1)1(222222vfxuvfvufxufx2222222vfvufxufx(2)),(yxxfz解令uxyxv则zf(uv)vfyufxvvfdxduufxz1vfyxdydvvfyz2因为f(uv)是u和v的函数所以uf和vf也是u和v的函数从而uf和vf是以u和v为中间变量的x和y的函数)(1)()1()(22vfxyufxvfyufxxzxxz)(1)(222222xvvfdxduuvfyxvvufdxduuf22222212vfyvufyuf)1()(2vfyufyxzyyxz)(1)1()(vfyyvfydydufyyvvfyvfyyvvuf222112232221vfyxvfyvufyx)()()(2222vfyyxvfyxyyzyyz22423222322vfyxvfyxyvvfyxvfyx(3)zf(xy2x2y)解zxf1y2f22xyy2f12xyf2zyf12xyf2x22xyf1x2f2zxxy2[f11y2f122xy]2yf22xy[f21y2f222xy]y4f112xy3f122yf22xy3f214x2y2f22y4f114xy3f122yf24x2y2f22zxy2yf1y2[f112xyf12x2]2xf22xy[f212xyf22x2]2yf12xy3f11x2y2f122xf24x2y2f212x3yf222yf12xy3f115x2y2f122xf22x3yf22zyy2xf12xy[f112xyf12x2]x2[f212xyf22x2]2xf14x2y2f112x3yf122x3yf21x4f222xf14x2y2f114x3yf12x4f22(4)zf(sinxcosyexy)解zxf1cosxf3exycosxf1exyf3zyf2(siny)f3exysinyf2exyf3zxxsinxf1cosx(f11cosxf13exy)exyf3exy(f31cosxf33exy)sinxf1cos2xf11exycosxf13exyf3exycosxf31e2(xy)f33sinxf1cos2xf112exycosxf13exyf3e2(xy)f33zxycosx[f12(siny)f13exy]exyf3exy[f32(siny)f33exy]sinycosxf12exycosxf13exyf3exysinyf32e2(xy)f33sinycosxf12exycosxf13exyf3exysinyf32e2(xy)f33zyycosyf2siny[f22(siny)f23exy]exyf3exy[f32(siny)f33exy]cosyf2sin2yf22exysinyf23exyf3exysinyf32f33e2(xy)cosyf2sin2yf222exysinyf23exyf3f33e2(xy)13设uf(xy)的所有二阶偏导数连续而23tsx23tsy证明2222)()()()(tusuyuxu及22222222tusuyuxu.证明因为yuxusyyusxxusu2321yuxutyyutxxutu2123所以2222)2123()2321()()(yuxuyuxutusu22)()(yuxu又因为)2321()(22yuxussussu)(23)(21222222syyusxxyusyyxusxxu)2321(23)2321(21222222yuxyuyxuxu22222432341yuyxuxu)2123()(22yuxuttuttu)(21)(23222222tyyutxxyutyyxutxxu)2123(21)2123(23222222yuxyuyxuxu22222412343yuyxuxu所以22222222yuxutusu
本文标题:高等数学练习答案8-4
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