福州大学高等数学-第六章多元函数微分学习题

§6.1多元函数的基本概念p.13一.填空题p1322241.ln(1)xyzxy的定义域是.222(,)(1,)xyyfxyfxyx2.设,则.3(1),1,,zyfxyzx3.设且当时则222(,)01,4yxyxyx2220xyxxy(),.fxz3(1)x-11yx222.5..2yxzyxp13.一函数在间断22yx(,)(0,0)(,)(0,0)24sin4.lim,lim,xyxyxyxyxyx14(,)(0,0)11lim(sinsin).xyxyyx00二.计算题p131.ln[ln()],zxyx求的定义域并画出定义域草图.ln()0,xyx解{(,)|0,1}{(,)|0,01}.xyxyxxyxyx定义域为1yx0yxxy11O3313.2.(,),(,).pfxyxyxyfxy设求2,,2uvxuxyvxyuvy解令3322(,)()()(3)224uvuvvfuvuv23223(,)(3).44yxyyfxyxy.14..3.:p二求下列极限(,)(1,0)ln(1)(1)limxyxyy(,)(1,0)(,)(1,0)ln(1)limlim1.xyxyxyxxxy解原式22(,)(0,0)lim.||||xyxyxy(2)222(||||),xyxy解222(||||)0||||,||||||||xyxyxyxyxy(,)(0,0)lim(||||)0,xyxy22(,)(0,0)lim0.||||xyxyxy三.证明下列极限不存在p14三.证明下列极限不存在:220()1lim,xxxkxkxk原式(,)(0,0)1.limxyxyxy;2(0),yxkxk证明取则,k对不同的取不同的极限值,.原极限不存在224(,)(0,0)2.lim,xyxyxy2(0),xkyk证明取则42420lim,(1)1ykykkyk原式,k对不同的取不同的极限值,.原极限不存在222222(,)(0,0)1cos()14.3.lim,()xyxypxyxy2222(,)(0,0)lim2xyxyxy证明原式22(,)(0,0)111lim(),2xyyx.原极限不存在22222)(21~)cos(1:yxyx利用注§6.2偏导数p.15一.填空题p151.(,)(,)fxyab设在处偏导数存在,则arctan,;.xyzzzxyxy2.设则3.(1),,yzzxyx设则0(,)(,)lim.xfaxbfaxbx.zy2(,)xfab22yxy22xxy21(1)yyxy(1)[ln(1)]1yxyxyxyxy22224.ln,0,0.xxyyzxyxyzz设则当时221()(2,4,5)44zxyMy5.曲线在点处的切线6.(,)(,)xyyxfxyfxy若与均连续,则恒有(,)(,)xyyxfxyfxy..x与轴正向所成的倾角为4二.计算题p15.二计算题;解2sin()(,),.zzxaxbyabxy1.设为常数求sin()cos(),zaxbyaxaxbyx22cos()sin().zbaxbyabxaxbyx2.,,.xxxyyxyzyzzz设求和解2ln,(ln),xxxxxzyyzyy12,(1),xxyyyzxyzxxy111ln(1ln).xxxxyzxyyyyxyy2222223.,(0)uxyzxyz设222,xxuxyz解2232222,()xxyzuxyz2232222,()yyxzuxyz2232222,()zzxyuxyz22222.xxyyzzuuuuxyz+.xxyyzzuuu求4.,.xxyyzzxuzarctanuuuy设求解222222,,()xxxyzxyzuuxyxy222222,,()yyyxzxyzuuxyxyarctan,0,zzzxuuy0.xxyyzzuuu三.计算题p162221.,0urxyzr三设233513,,xxxxxuurrr解2235351313,,yyzzyzuurrrr0.xxyyzzuuu0.xxyyzzuuu求证§6.3.全微分p.17一.填空题p17221.(,),)(),zfxyxy若可微且则002.(,)(,)zfxyxy函数在点处连续和存在偏导数是00(,)xy它在处可微的条件.0lim.zdz3.(,)(,),(,)xyfxyfxyfxy函数的偏导数连续是函数(,).fxy可微的条件0必要充分4.,1,1,0.15,0.1xyzexyxy设则当时,.dzz225..yzdzxy设,则16.(,,)(),(1,1,1).zxfxyzdfy设则2227.,.xdxydyzdzduuxyz若则1.265ee32222()()xyxdyxydxdxdy222xyzc4e二.计算题p171.(,)||(,),(,)(0,0)fxyxyxyxy设其中在的某,),(0,0),xxyf个邻域内连续,问:(满足什么条件(0,0)?yf存在0(,0)(0,0)(0,0)limxxfxffx解0||(,0)lim(0,0),xxxx(0,0)0,(0,0);xf当时存在,(0,0)0,(0,0)yf同理当时存在.2218.2.(,),(1,2).ypfxyxyf设求2,(1,2)4.yyfyf解21,(1,)1,(1,)2,yxfxyfyy解法2.令得(1,2)4.yf21sin()03.(,),(0,1).00xxyxyxyfxyfxy设求0(0,1)(0,1)(0,1)limxxfxffx解201sin()limxxxx220sin()lim1.()xxx344.ln(1.030.981).利用全微分计算的近似值34(,)ln(1),fxyxy解设232433441134(,),(,),11xyxyfxyfxyxyxy11(1,1),(1,1),34xyff3411ln(1.030.981)0.030.020.005.34222222221()sin0(,)00xyxyxyfxyxy三.设,问(,)(0,0)??fxy(1)在处是否可微为什么(,),(,)(0,0)xyfxyfxy(2)在处是否连续?为什么?解(1)可微.22001sin(,0)(0,0)limlim0,xxxfxfxxx22001sin(0,)(0,0)limlim0,yyyfyfyyy(0,0)(0,0)0,xyff0(0,0)(0,0)limxyyzfxfy(2)(,),(,)(0,0).xyfxyfxy在处不连续220,xy当时222222121(,)2sincos,xxfxyxxyxyxy201limsin0,y22021,limcos.22xxyxxx取时不存在(,)(0,0),xfxy在处不连续(,)(0,0)yfxy同理在处不连续.§6.4复合函数的求导法则p.19一.填空题p19231.,sin,,xyzextyt设而.dzdt则(,,),(,),(),ufxyzzxyyx2.设,,,f其中均可微则(2)(1)[()],,zfxycc3.设则.xxyyxxzzzzdudx32sin2(cos6)tttte()(())xyzxyffxfx024.(,),(,2),(,2),xfxyfxxxfxxx设可微且则(,2).yfxx(,2)(,2)2(,2)1.xyfxxfxxfxx分析5.(,),()[,(,(,))],(1,1)1,fxyxfxfxfxxf设可微且(1,1),(1,1),(1).xyfafb则212x23aababb二.计算题p1922221.,,,,.vxyzzzuuevxyxx设求2233(),xyxyxyzee解33233,xyzxyex3324632(96)xyzexyxyx333333(32).xyxyexy2(2)2.(,,),,.yzzzfuxyCuxexxy设其中求12,yzfefx解2111132123()yyyzeffxeffxefxy2(2)20.3.(,)(),,,.xyzpzfxygfgCyxxy设其中求1221(),zyfyfgxyx解2111122(())zxfyfxfxyy221222211(())xffxfyyy2211yggxxx111222232311.xyfxyfffggyyxx三.证明题(2).(,),cos,sin,ufxyCxryr三设证明:22222222211uuuuuxyrrrrcossin,uuurxy证明1cossin,uuurrrxry22222cos(cossin)uuurxxy222sin(cossin),uuyxy2222222cossin2sin,uuuxxyy(sin)cosuuurrxy22222cossin((sin)cos)uuuurrrrxxxy222sincos((sin)cos)uuurrrryyxy222111cossinuuurrxry2222222222sinsin2cosuuurrrxxyy22cossinuuurrxy2222222sinsin2cosuuuxxyy22222222211.uuuuurrrrxy§6.5隐函数求导公式p.21一.填空题p211.(,),(,),(,)(,,)0xxyzyyzxzzxyFxyz设都是由.xyzyzx具有连续偏导数不为零的函数,则2.220,xyzxyz设则,.zzxy3.ln,.xzdzzy设则1yzxyzxyzxy2xzxyzxyzxy2()zzdxdyxyyxz21.4.2sin(23)23,pxyzxyz设则.zzxy(1)5.(,)(,)(-,-)0FuvCzzxyFcxazcybz设,是由,.zzabxy确定则1c二.计算题:p212221.1.,,.zzzpxyzexx设求(,,),zFxyzxyze解令(,,)1,xFxyz(,,)1,yFxyz(,,)1,zzFxyze1,1x