《高等数学》第07章多元函数微分法及其应用习题详解

1、第七章 多元函数微分法及其应用习题详解第七章多元函数微分法及其应用习题7-11.判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并指出集合的边界(1) (x,y)x0,y0；(2) (x,y)1x2y24；，一、2(3)(x,y)yx；(4)(x,y)x2(y1)21且x2(y1)24.解(1)集合是开集，无界集；边界为(x,y)|x0或y0.(2)集合既非开集，又非闭集，是有界集；边界为(x,y)|x2y21J(x,y)|x2y24.(3)集合是开集，区域，无界集；边界为(x,y)yx2.(4)集合是闭集，有界集；边界为(x,y)x2(y1)21J(x,y)|x2(y1)242 .

2、已知函数f(u,v)uv,试求f(xy,xy).解f(xy,xy)xy(xy).3 .设f(x,y)Jx4y42xy,证明：f(tx,ty)t2f(x,y).解 f (tx, ty).tx4ty42t2xyt2_x4y42t2xyt2x4y42xyt2f(x,y).27(x0),求f(x).212y解由于fx -1 x2-,则f15.求下列各函数的定义域:(1)22xy.22'xyy(2)zln(yx)arcsin-；x(3)ln(xy)；(4)Z21x2(5)z(6)ux2y2(1)定义域为（x,y）y(2)定义域为(x,y)(3)定义域为(x,y)xy即第一、三象限（不含坐标轴）(

3、4)定义域为(x,y)2x2a2yb2(5)定义域为(x,y)0,y0,x2y.，(6)定义域为(x,y,z)0,x26.求下列各极限:(1)lim(x,y)(2,0)(2)(3)lim(x2(x,y)(0,0)(5)(x,y)m(0,1)(1解：（1）lim(x,y)(2,0)(2)(加0,0)(3)因为(x,y)(4)y2)sin1xy)x；2xxyxy(4)(6)f(2,0)1cosx2y2ln(x2y21)lum0叫,小y2)0(x,y)m(2,0)sin(xy)(x,yl)m(2,0)lim(x,y)(0,0)lim(x,y)(2,0)(x,y)limln(1u)1cosln(x2s

4、in(xy)(1ulim-2u0u且sinxysin(xy)xxy1有界，故212;2x-2y1)2xyy)e2lim(x(x,y)(0,0)'y2)sin0；xy(5)lim(1(x,y)(0,1)xy)xlim(x,y)1(0,1)(11Xyy1xy)ee；(6)当xN0,yN0时，有0/22(xy)exy(xy)2exy而(x.yjmexylimulimu2ulim按夹逼定理得lim（x,y）（/22.xy)(xy)e0.(x,y) (0,0) x(2)设 f (x,y)0,lim(x,y) (0,0)f (x, y).7.证明下列极限不存在:(1)lim0,证明(x)x kx

5、limk0x kxkx0,（1）当（x,y）沿直线ykx趋于（0,0）时极限与k有关，上述极限不存在（2）当（x, y）沿直线x和曲线2 .x趋于(0,0)有(x,y)m(0,0)2x y42x ylimx 0 y x2x xI 2xlimx 0y x(x,y)m(0,0)limx 02 y xlim4 x 2x4故函数f（x, y）在点（0,0）处二重极PM不存在.8.指出下列函数在何处间断:(1) z ln(x2y2)；1y2 2x解（1）函数在（0,0）处无定义，故该点为函数z ln(x2 y2)的间断点;（2）函数在抛物线y2 2x上无定义，故y2 12x上的点均为函数 z -的间断y

6、 2x点.9.用二重极限定义证明:证Jx2 y2 |OP|,于2 x2yxy lim0.(x,y) (0,0)、,2、,2xy日2 i 1-2P(x, y),其中0,0;当0时,xy_22x y成立，由二重极限定义知lim(x,y) (0,0)xy22x y0.10.设 f (x, y)2sinx ,证明f(x, y)是R上的连续函数.2证设 Po(xo,yo) R .0 ,由于sinx在X0处连续，故时,|sinxsinx01以上述作P0的邻域U(P0,)，则当P(x,y)U(P0,)时，显然|xXo|(P,Po)从而R2|f(x,y)f(x°,y°)|sinxsinx0

7、|即f(x,y)sinx在点P0(x0,y0)连续.由P0的任意性知，sinx作为x、y的二元函数在上连续.习题721.设zf(x,y)在(xo,y°)处的偏导数分别为fx(x0,y°)A,fy(x0,y°)B,问下列极限是什么?(1)f(%h,yO)f(%,yo)(3)limh0hf(xo,yo2h)f(xo,yo)f(xO,yo)f(x°,y°h)hf(x°h,y°)f(xo,y。)(4)limh0hf(xoh,y°)f(xoh,y°)(2)4(x°,y°)A；f(x0,y0)f(

8、%,y0h)f(x0,y0h)f(x,y。)Zy(x0,y。)B；(3)limh0f(xo,y02h)f(xo,y°)lim2h0f(Xo,yo2h)f(Xo,y°)2h2B;（4）iim）f(%h,y°)f(Xoh,y°)limh 0f(xoh,yo)f(xo,yo)f(xo,y°)f(xoh,y°)hf(%h,y°)f(x0,y°)f(x°h,y°)“飞)。)hf(%h,y°)f(x0,y°)limf(%h,y°)f(x0,y°)AA2A2.求下列函数

9、的一阶偏导数:(1)xy(2)xz ln tan ； y(3)exy;(4)2y； xy(5)22x ln( x2、y )；(6)z ,Jn(xy)；(9)sec(xy)；(8)z (1xy)y ；arctan(xy)z(10)解（1）(2)tan2xsec一y1,x2x一cotsec一(3)(4)tanexy2x2xsec一yxyyye，xyxexx-2cot-I"；2x.sec一；y/2xy(x2xy2、y)y2x*2y(x222xy2、y)y222,22、z2yxy(xy)x2xy(xy)x222yxyxy(5) 2xln(x2xy2)2x .2-22x 2xln(xx yy2

10、)2x32 2x y-22xy.22，xy(6)12 Jn(xy)1-y xy1z二/=,一2x ln(xy)y111x :2 .ln(xy) xy 2y ln(xy)tan(xy)sec(xy)yytan(xy)sec(xy)，xztan(xy)sec(xy)xxtan(xy)sec(xy),y(8)_zy(1xy)y1yy2(1xy)y1,xeyln(1xy)y(1xy)ln(1xy)xy-yy1xy-rrz(xy)z1;z,x1(xy)1(xy)11 (x y)2zz(xy)z1 ( 1)z(x y)z11 (x y)2z11 (x y)2z(xy)z ln(x y)(x y)zln(x

11、 y)1 (x y)2z(10)z 1x1zyyz 1ux一 z yyln3.设 f (x, y)1nx "fy(1,0)由于f(1,y)ln1,所以fy(1,y)解法二fx(x,y)1y2xv行，fy(x,y)2x1y2x12xfx(1,0)1-02,11，fy(1,0)0124.设f(x,y)(y1)arcsinJ,求yfx(x,1).解法一由于f(x,1)x(11)arcsinx,fx(x,1)(x)1.5.设f(x,y)1，fx(x,1)ydt，求fx(x,y)解法二fx(x,y)fy(x,y).解fx(x,y)ex2*4,f(x,y)_y6.设zxyxex证明zxxzy一y

12、xyz.解由于yexyxexyexyexyxex所以zy一yyexyexyyxyex(xy)xyyexxyyxexxyxy7. (1) z(2)z41xy,在点(1,1,J3)处的切线与y轴正向所成的倾角是多少？x1解(1)按偏导数的几何意义，zx(2,4)就是曲线在点(2,4,5)处的切线对于x轴正向所成倾角的斜率，而1.，Zx(2,4),Xx21,即ktan(2)按偏导数的几何意义,Zy(1,1)就是曲线在点(1,1,J3)处的切线对于y轴正向所成tan倾角的斜率，而zy(1,1)一=2y2J18.求下列函数的二阶偏函数：(1)已知z3.3.xsinyysinx,(2)已知z(3)已知zl

13、n(xx2),lnxy2十z,求；xy2z；xy(4)+y+2zzarctan求一2xx解(1)3x2sinyx3ycosx,3x2cosy3y2cosxylnxlny1xlnylnxyxlnxlnylnxInxyInx1(1Inxlny);(3)22xxy22、xy2x2x2,x2y2(x,x21x23，2y2)y2，2z2x2x2.x22z2y2.x2(4)x12yx121工xc2一22z2满足12xyy222x y 2y22 2x y22yX22 2，x y2222z x y 2xyxx2 y2 222yx22 2 .x y9.设 f (x, y, z)xy2yz2 zx2 ,求 fxx

14、(0,0,1),fxz(1,0,2) , fyz(0, 1,0)及fzzx(2,0,1).2斛因为fxy2xz,fxx2z,fxz2x,2fy2xyz,fyz2z,2fz2yzx,fzz2y,fzzx0,所以fxx(0,0,1)2,fxz(1,0,2)2,fyz(0,1,0)0,fzzx(2,0,1)0.10.验证:(1)kn2ty(2)sinnx满足t(1)kn2kn2tesinnxnekn2tcosnx,2y2xkn2tsinnxkn2t.sinnx(2)因为x2r由函数关于自变量的对称性,2r2z2z3r2r所以x2r2y2J2z2y3r2z3r习题2J“、st11uu-22-2；St(

15、2)(x22xyy)e(4) Z,x(3)zarcsin(y0)；yyz(6) u x ./L,/222、(5)uln(xyz)；(1)2s(s22_2t)2s(st2)t224st2TV2t(s2 t2)4s2t2t(s2t2)2st2(2)(3)(4)dux22xe4st2:y2xyt2ds4s2t2dtt24stt22(tdssdt)(x22)e22xyxy22xy(x222xyy2)y22xyxy2x44xy2，xy由函数关于自变量的对称性可得dzdzdz22xyxy2x44xy2xydx2y22xyexy2y2xyxydyxarcsin一y12x2y1一dxyxFdyyyy2ydxx

16、dydxdy(5)dudln(z2)2dz2xdx2ydy2zdz2,I,、22-2-222(xdxydyzdz);xyzxyz(6)dudxyzyzxyz1dxxyzzlnxdyxyzylnxdzxyz1yzdxxzlnxdyxylnxdz.2.求下列函数的全微分:(1)zln(1x2y2)在x2处的全微分;(2)z,xarctan21y1处的全微分.解（1）因为dzdln(1y2)11x2d(1yy2)(2xdx2ydy)y所以dz11(2dx264dy)3dx(2)因为dzdarctan-12d所以dzxy3.求函数z解因为dz所以当4.求函数z并求两者之差.解因为dz1y2.221yx

17、122dxxdx2dzxy2xdx2xy2Tdyy,2xydx-2dy1y2xy；2dy1y2,y2xy3dx0.02,12dxdy.3x2y2dy0.02,y2xy3x3x20.01时全微分为0.080.120.2.0.01时的全微分.2,x0.01,y0.03时的全微分和全增量,xy22xy22、d(xy)xyd(xy)22xy(ydx+xdy)xy(2xdx2ydy)233一2.xyydx+x+xydy所以当x2,y1dz3y2x而当x2,全增量与全微分之差为1.设U0.01y0.03时全微分的值为32x+x+xyy(x,y)(2,1)x0.01,y0.030.01,0.250.0277

18、77,0.03时的全增量为zdzsint,0.028252xy-2x0.028252,V(x,y)(2,1)0.010.030.0277770.000475.解duudxxdtudyydt2.设zarccos(u解dzzduudx3.设zzdvvdx4x222x(4x2uv3)24.设z习题74,3tdut，求证.2y.cost(uxcosy,v一2uvvx2ex2y3t23x,求.dxv)2xsin2.，、3xsinycosy(cosysiny)2uv12x2cosy33x(sin2sin2ycosy3cosu3xsint2t3exsinyy2cos2,_2、(cost6t).(Iv)232

19、uvsiny2u2uvxcosyysiny).5.6(3x2y)ln6.设uf(u,x,y)ln(u222ulnv3v-(3xx2y)2ulnv2ysinx),12uysinxy14(3x2y)ln-(3xxy-ycosxuysinx2y)2.7.设z2x2e-2Te2e2xy./2sin(x(rs(rs(rs,xarctan,yycosxysinxsinxysinx),stststtr)(str)(r12uysinx1-sinxysinxt)t)tr)(rs)uv,y2xrsstutr,zrst,求2y(st)c/22zstcos(xz2)2.2rstcos2x2y(r22rstcos2x2

20、y(s22rstcos(rt)(rr)(rt)2(rs22zrtcos(xst)2(rs22zrscos(xst)2(rsstststtr)2z2)tr)2z2)tr)2(rst)2,(rst)2,(rst)28.设zf(x,y,t)(u2(uv)2v)(uv)2xsint,cost,解dzzdxxdtdyydt2xcost2y(sint)9.求下列函数的一阶偏导数（其中f具有一阶连续偏导数）(1)z-22f(xy)；(2)u(3)uf(x,xy,xyz)(4)解（1）z2xfx(x2(2)2yf(x2、y)；(3)(4)10.设zf1f2f2f1yzf3xf2xzf32xf1xyyef22y

21、f1xyxF(u),而x2ydzdt2sinf(x2v2.v2t1.xyx,e,lnx).f1xyf3；*yrxef2.F（u）为可导函数，证明:yzxy.yxF (u) yxyxyF(u)xF(u)xyxxyF(u)-F(u)xyxF(u)xyxF(u)xyzxy.11.设zycos(xy),试证：.xyy、丁z证一xysin(xy)cos(xy)sin(xy)12.设ucos(xy)-y且函数F-,具有一阶连续偏导数,xx试证:u y-yku .uxx、丁u证xkxk1FFiz2xF2k-xF21F1k-xF1uxxuzzkkkxFx1zF11yF11yF11zF1ku.13.设zsiny

22、f(sinsiny),试证:zsecxxzdsecy-1.yfcosx，ycosy(cosy)f,zzsecxsecyxysecxcosxfsecycosysecy(cosy)f1.14.求下列函数的二阶偏导数222-z,-2（其中f具有二阶连续偏导数）xxyy(1)zf(xy,y)；,22zf(xy)；第七章多元函数微分法及其应用习题详解i6(3)zf(x2y,xy2)；(4)zf(sinx,cosy,exy).解(1)令sxy,ty,则zf(xy,y),s和t是中间变量fizsdtyf1，f1f2xf1yydyf2.因为f(s,t)是s和t的函数，所以f1和f2也是s和t的函数,从而f1和

23、f2是以s和t为中间变量的x和y的函数.故2 z2 x一yfixyfii一 yfi yfifiifi2dtdyfixyfiiyfi22z2 yxfiyf2xfiifi2dtdyf2if2*x2fii 2xfi2f22(2)令 s x2 y2,则 z一一22.f(xy)是以s为中间变量的x和y的函数.z's-'zsf2xf,f-y fi 2xyf2 xyfxxyy因为f(s)是s的函数，所以f也是s的函数，从而f是以s中间变量的x和y的函数.故22f 2xf 2x 2f 4x2 fzz2xfxxxx2zz2xf2xf2y4xyf,xyyxy2-zz2yf2f2yf2y2f4y2f

24、yyyy(3)令sxy2tx2y,贝Ust2zfi-f2yfi2xyf2,xxyst2fif2-2xyfixf2.yy2zz""2"xxx第七章多元函数微分法及其应用习题详解I7(4)2z2x、,2fsftyfiifi2xx2yf22yfi2yfi2yfi2xf12xf12xf1dufiTdx2yf22xyf21f222c，2r22rfii2xyfi22yf22xyy£?143,yfii4xyfi2,224xyf22,y2fi2xyf22xyf22fii-y2xy3f123-2xf22xyfii一2xyf1ys2xyfii一y2xy2xyfiix22-4

25、xyfiisinxf1sinxf1exy2xf2yfi22xf225xy2fI2f22xyf2i2xy2xyf2132xyf22,f122f124x3yfi2cosy,cosxf1cosxficosxcosxsinxfif212xyf2iXf22.exyf3，exyx2f22dvdysf22乂旺22f3xysinyf2ef3.dufiiVdxfi3xdu3i展cosxf11cos2xfiiyfl3exyexycosxf31exyf33cosxfiex2exycosxfi3f33，第七章 多元函数微分法及其应用习题详解y f dvf32 dy_dv_wcosxf12f13dyy77cosxXyx

26、ysinyf12ef13ef3xyxyesinyf32ef33exyf3cosxsinyf12eycosxf13exysinyf322z2ysinyf2exycosyf2sinyf22dyf23exyexycosyf2sinysinyf22xyef23exyf3cosyf2sin2yf222exysinyf23f32dyf33sinyf32xyref33习题751.设xcos y ex2y 0,求曳 dx解设F(x, y)x 2cosy e x ydydxFxFex 2xysin y2.设 xylnIndydx解设F(x, y)xyln y当x 1时，由xyex 2xysin ydydxFyl

27、n y ln x 1 知 y1,所以dy dx2xy3.设lnjxy2arctan"，求dy.xdx解设F(x,y)ln7x2y2arctan,则xdydxFxFyx4.设cos21xx2cos2cos解设F(x,y,z)2cosx2cos2ycos5.设方程F(xx2yy2yy2yx2y6.设由方程y(x,z),z证因为所以y2cosxsin2coszsiny乙xyFxFzFiFiyzF(x,y,z)0z(x,y)证明:FyFxsin2xsin2zFyFz2cosysiny2coszsinzsin2ysin2zzx)0确定了函数z(yz)F2(yx)F2FyFzz(x,y),其中F

28、存在偏导函数,F1(xz)F2F1(yx)F2分别可确定具有连续偏导数的函数xx(y,z),FzFy7.设(U,V)具有连续偏导数f(x,y)满足abc.xyFzFyFxFz证明由方程(cxaz,cybz)0所确定的函数证令ucxaz,vcybz,8.设cv，xyzc.解设F(x,y,z)Fx9.设2z-2.xxyz,则Fxyz,Fzxy.yzyzz盘、y(exy)xyzzexy2zexyyzt2yexy22z32yze2xyzxyz(x,y)是由方程解设F(x,y,z)zexzxz0所确定的隐函数,(0,1)Fxx,Fy2y.日_zZExFxFzFFz2yezzze2yzex2yezx2yz

29、ezzex由ezxzy20,知z(0,1)0,得一z-l2.xy(0,1)10.求由方程xyz西z2J2所确定的函数zz(x,y)在点(1,0,1)处的全微分dz.解设F(x,y,z)xyzJx2yz./FxxFzxy小xzzFy2:y2z2J2,则x222.'222xyzyz.xyzxzxy；x2y2z2z222xyzy222222xyzxzxyzyyFzxy-=zxy.222xyz222zzyzJxyzxdzdxdy.xyxy,x2y2z2:dz(1,0,1)dx2dy.11.求由卜列方程组所确定的函数的导数或偏导数：，222xyzz222(xzxyzy-dx-Y=-dy,'

30、;222乙xyxyzz22(1)设zxy,才222x2y3z20,xuyv0,u(2)设求，-yuxv1,xu、厅xeusinv,u(3)设口求一uyeuucosv,x解(1)分别在两个方程两端对dzdx2x称项，得%dxdxuvv一，一，yxy，,?yxyx求导，得2x2y%dxdydz4y6z0.dxdxdydz2y2x，dxdxcdycdz2y3zxdxdx在解方程组得2y 12y 3z6yz 2y0的条件下,2x1dyx3z6xzxx(6z1)dxD6yz2y2y(3z1)2y2xdz2yx2xyxdxD6yz2y3z1(2)此方程组确定两个二元隐函数uu(x,y),vv(x,y),将

31、所给方程的两边对x求导并移项，得uvxyu,xxuvyxv.xxxy22c_、在Jxy0的条件下，yxuyuvxxuyv22，xxyxyyxxuvyvyuxv1-22,xxyxyyx将所给方程的两边对y求导，用同样方法在Jx2y20的条件下可得uxvyuvxuyv22,22yxyyxy(3)此方程组确定两个二元隐函数uu(x,y),vv(x,y)是已知函数的反函数，令u1(F,G)11ucosvsinvxJ(x,v)J0usinvue(sinvcosv)1u1(F,G)10ucosvcosvyJ(y,v)J1usinveu(sinvcosv)1v1(F,G)1u.一esinv1ucosvexJ

32、(u,x)Juecosv0ueu(sinvcosv)1v1(F,G)1u.一esinv0.一usinvexJ(u,x)Juecosv1ueu(sinvcosv)1习题76组得FxGxF(x,y,u,v) x1, Fy0, Gy(F,G)(u,v)u sin v, G(x,y,u,v) yue u cosv.0,u eu eFuGusin vcosvsin v , Fvu cosv,cosvucosvusin v,Gv u sin v.ueu (sin v cosv)u 0的条件下，解方程1.求下列曲线在指定点处的切线方程和法平面方程:(1)(2)一3.t在(1,0,1)处;2.zt在t1的对应

33、点处;(3)sint ,cost , z 4sin -在点21,1,2 衣处;2(4)2 x2 y2 y2 z10100, 在点(1,1,3)处.0,(1)因为xt2t, yt1 ， Zt(1,0,1)所对应的参数t 1,所以(2,1,3).于是，切线方程为法平面方程为2(x1)y3(z1)0,2xy3z50.(2)因为xt12)(1t)ytt(it)t2zt2t,t1对应着点11,2,1,所以24,1,2.于是，切线方程为1x21法平面方程为2x8y16z10.(3)因为x所以2切线方程为法平面方程为(4)将2x2y由此得2y2zcost10100,0,ytsint,zt2cos-2占八、1

34、,1,2.2对应在的参数1,1,2.的两边对x求导并移项，得2x,dydxx一,ydz0,dzdx0.2y2y2x2y02y2z4xy4yzdydx(1,1,3)dzdx(1,1,3)从而故所求切线方程为法平面方程为1T1,1-.3x1y1z333-V3x3yz30.2.在曲线xt,yt22斛(1) F (x, y, z) 3x y z 27 ,ztn (Fx, Fy,Fz) (6x,2y, 2z),n(3,i,i)(18,2, 2).所以在点(3,1,1)处的切平面方程为9(x 3) (y 1) (z 1) 0, 即9x y z 27 0.法线方程为(2) F(x,y, z) ln(1 x2

35、 2y2) z,上求一点，使此点的切线平行于平面x2yz4.解因为xt1,yt2t,zt3t2,设所求点对应的参数为t0,于是曲线在该点处的切向量可取为T(1,2to,3t2).已知平面的法向量为n(1,2,1),由切线与平面平行，得2-，1Tn0,即14t03to0,解得to1和.于是所求点为(1,1,1)或3111.3'9'273.求下列曲面在指定点处的切平面和法线方程：(1) 3x2y2z227在点(3,1,1)处；22.(2) zln(1x2y)在点(1,1,ln4)处；(3) zarctany在点1,1-处.x4n(Fx,Fy,Fz)2x22c21x2y-4y-,1,

36、1x22y2所以在点(1,1,ln4)处的切平面方程为法线方程为y(3)F(x,y,z)arctan上x所以在点1,1-4法线方程为4.求曲面11(1,1,ln4)2,i,2y2z3z,n(Fx，Fy,Fz)处的切平面方程为2y23z21,1l44ln20.F(x,y,z)n(Fx,Fy,Fz)行，得代入曲面方程得解得所以切点为所求切平面方程为2ln22zyx2,2yx0.-,1，y21上平行于平面2y23z221,x4y6z0的切平面方程.则曲面在点(x,y,z)处的一个法向量(2x,4y,6z).已知平面的法向量为(1,4,6),由已知平面与所求切平面平2x11-z,2yz.2,2z23z

37、221.1,2,2.4y6z21xy5 .证明：曲面F(xaz,ybz)0上任息点处的切平面与直线z平行(a,bab为常数，函数F(u,v)可微).证曲面F(xaz,ybz)0的法向量为n(RE,aFibF?),而直线的方向向量s(a,b,1),由ns0知ns,即曲面F0上任意点的切平面与已知直线Jxyz平行.ab6 .求旋转椭球面3x2y2z216上点(1,2,3)处的切平面与xOy面的夹角的余弦.解令F(x,y,z)3x2y2z216,曲面的法向量为n(Fx,Fy,Fz)(6x,2y,2z),曲面在点(1,2,3)处的法向量为n1n,一q、(6,4,6),xOy面的法向量(1,2,3)n2

38、(0,0,1),记n1与山的夹角为，则所求的余弦值为n1n263COSr;jccc.n11ln2|4621V22按题意，方向l2.求函数z解依题意,3.求函数z向的方向导数.(1,2)(1,73),ez2y,y2,(1,2)ln(x2ei(1,2)2244,（1,1）解先求切线斜率:法线斜率为内法线方向1,32,-2123.2y2）在点（1,1）处沿与x轴正向夹角为60“的方向的方向导数.12,22x2y(1,2)112yb2dydx(b,（1,1）在点2yb22y2y处沿曲线1两端分别对x求导,2x2ab2x2ay2ydyb2dxdydxab=,-=22ab'a),ei,a2b2，.a22-y21在这点的内法线方ba_一b2ab2/24.求函数解因为所以5.求函数.2ab2,2ab了中逆b1,2(a2b2).abxyz2在点(1,0,1)处沿该点到(3,1,3)方向的方向导数.2x2,(1,0,1)2,(1,0,1)u2z,z1,(1,0,1)2.1,2(1,0,1)ei2z在曲线yt2,.3zt上点(1,1,1)处，沿曲线在该点的切线正方向(对应于t增大的方向)的方向导数解先求曲线在