高等数学复旦大学出版第三版下册课后答案习题全(陈策提mai供huan)

?z?y?(1?xy)y??yln(1?xy)??y?x?y?(1?xy)??ln(1?xy)?y1?xy??

?(1?xy)y???ln(1?xy)?xy?1?xy??.(6)

?u?lnz?zxy?y ?u?uy?lnz?zxy?x ?z?xy?zxy?1?x? (7)?u1z(x??x1?[(x?y)z]2?z(x?y)?y)z?1?z?11?(x?y)2z. ?uz(x?y)z?1?(?1)z(x?y)z?1?y?1?[(x?y)z]2??1?(x?y)2z.?uzz

?z?(x?y)ln(x?y)(x?y)ln(x?y)1?[(x?y)z]2?1?(x?y)2z.?uyy(8)

?1?x?zxz. ?uy?y?xzlnx?1z?1yzxzlnx.?uyy

?xzlnx????y?yz?z?z2????z2xlnx.9.已知u?x2y2?ux?y，求证：x?x?y?u?y?3u. ?u2xy2(x?y)?x2y2x2y2证明： ?x?(x?y)2??2xy3(x?y)2. ?ux2y2?2yx3由对称性知 ?y?(x?y)2. 于是 x?u?u3x2y2(x?y)?x?y?y?(x?y)2?3u. ???11?10.设z?e?x?y??，求证：x2?z?x?y2?z?y?2z. ?11证明： ?z?x?e???x??y??????????1??1???1?x?1?y??x2?????x2e, 由z关于x,y的对称性得

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???z1???xy? ?e?11??yy2?11??11??11?故 x2?z?x?y2??z21???x?y??21???x?y???????y?x?x2e?y?y2e?2e?xy??2z.

11.设f (x,y) = x+(y-1)arcsinxy，求fx(x,1) . 解：f1x(x,y)?1?(y?1)2?1?1 1???x?2xy?y??y则fx(x,1)?1?0?1.

?x2?y212.求曲线??z??4在点（2，4，5）处的切线与正向x轴所成的倾角.

?y?4解：

?z?x?1?z2x, ?x?1, (2,4,5)设切线与正向x轴的倾角为α, 则tanα=1. 故α=

π4. 13.求下列函数的二阶偏导数： (1)z = x4+ y4-4x2y2; (2)z = arctanyx; (3)z = yx;

(4)z = ex2?y.

解：(1)?z?x?4x3?8xy2， ?z22?2z?x2?12x?8y， ?x?y??16xy 由x,y的对称性知

?2z?y2?12y2?8x2. ?2z?y?x??16xy. (2)

?z12???y???y?x?， 1???y???x2??x2?y2?x?? 198

?2z(x2?y2)?0?y?2?x2??x(x2?y2)2?2xy(x2?y2)2,?z?y?12?1?x1???y?xx2?y2,?x???2z?y2??2xy(x2?y2)2, ?2z(x2?y2)?y?x?y???2yy2?x2(x2?y2)2?(x2?y2)2,?2zx2?y2?x?2xy2?x2?y?x??(x2?y2)2?(x2?y2)2.(3)?z?2?x?yxlny, z?x2?yxln2y, ?z?xyx?1, ?2z?2?y?y2?x(x?1)yx,?2z?yx?1?xyx?1?x?yylny?yx?1(1?xlny), ?2zy?x?yx?1?x?lny?yx?1?yx?1?(1?xlny).(4)

?z?x?ex2?y?2x, ?z2?y?ex?y, ?z?x2?ex2?y?2x?2x?ex2?y?2?2ex2?y(2x2?1),?z

x2?y?zx2?y?zx2?y?y2?e, ?x?y?2xe, ?y?x?2xe.14.设f (x, y, z) = xy2+yz2+zx2,求fxx(0,0,1),fyz(0,?1,0),fzzx(2,0,1).解：f(x,y,z)?y2x?2zx

fxx(x,y,z)?2z, fxx(0,0,1)?2,fy(x,y,z)?2xy?z2fyz(x,y,z)?2z, fyz(0,?1,0)?0,fz(x,y,z)?2yz?x2

fzz(x,y,z)?2yfzzx(x,y,z)?0, fzzx(2,0,1)?0.

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15.设z = x ln ( x y)，求?3z?3z?x2?y及?x?y2.

解：

?z?x?x?yxy?ln(xy)?1?ln(xy), ?2zy1?3?x2?xy?x, z?x2?y?0,?2z3

?x?y?xxy?1y, ?z1?x?y2??y2.16.求下列函数的全微分： (1)z?ex2?y2;

(2)z?yx2?y2;

y(3)u?xyz;

(4)u?xz.

解：(1)∵?z?x?ex2?y2?2x, ?z?y?ex2?y2?2y ∴dz?2xex2?y2dx?2yex2?y2dy?2ex2?y2(xdx?ydy)

(2)∵?z?1?2xxy?x?y????x2?y2???2x2?y2??(x2?y2)3/2 y?zx2?y2?y?x2?y2?y?x2?y2?x2(x2?y2)3/2 ∴ dz??x(x2?y2)3/(2ydx?xyd ).(3)∵

?u?yzxyz?1,?u?xyz?lnx?zyz?1?x?y ?u?lnx?xy2?lny?yz?z ∴du?y2xyz?1dx?xyzlnx?zyz?1dy?lnx?xyz?lny?yzdz.

?uyy(4)∵

?1?x?zxz ?uy1?y?lnx?xz?z 153