习题1—1解答 1． 设f(x,y)?xy?x11x1，求f(?x,?y),f(,),f(xy,), yxyyf(x,y)解f(?x,?y)?xy?x111yx1y；f(,)? ?;f(xy,)?x2?y2;?2yxyxyxyf(x,y)xy?x2． 设f(x,y)?lnxlny，证明：f(xy,uv)?f(x,u)?f(x,v)?f(y,u)?f(y,v)
f(xy,uv)?ln(xy)?ln(uv)?(lnx?lny)(lnu?lnv) ?lnx?lnu?lnx?lnv?lny?lnu?lny?lnv?f(x,u)?f(x,v)?f(y,u)?f(y,v)3． 求下列函数的定义域，并画出定义域的图形：
2（1）f(x,y)?1?x?y2?1;
（2）f(x,y)?4x?y2ln(1?x?y)22;
x2y2z2（3）f(x,y)?1?2?2?2;
abc（4）f(x,y,z)?x?y?z1?x?y?z222.
解（1）D?{(x,y)x?1,y?1?
y 1 -1 O -1 1 x 222 （2）D?(x,y)0?x?y?1,y?4x?? y 1
-1 O -1 1 x 1
??x2y2z2（3）D??(x,y)2?2?2?1? abc??
z c -a -b a x O b y 222 （4）D?(x,y,z)x?0,y?0,z?0,x?y?z?1
??
4．求下列各极限： （1）limx?0y?1z １ O x
１ y １ 1?xy1?0?1 =22x?y0?1ln(x?ey)x?y22（2）limx?1y?0?ln(1?e0)1?0?ln2
（3）lim2?xy?4(2?xy?4)(2?xy?4)1?lim??
x?0x?0xy4xy(2?xy?4)y?0y?0（4）limsin(xy)sin(xy)?lim?x?2
x?2x?2yxyy?0y?05．证明下列极限不存在：
x2y2x?y（1）lim ; （2）lim22x?0xy?(x?y)2x?0x?yy?0y?0（1）证明 如果动点P(x,y)沿y?2x趋向则lim x?yx?2x?lim??3；
x?0x?0x?2xx?yy?2x?0，则lim如果动点P(x,y)沿x?2y趋向
x?y3y?lim?3
y?0y?0x?yyx?2y?0 2
所以极限不存在。
（2）证明： 如果动点P(x,y)沿y?x趋向
x2y2x4?lim4?1； 则lim222x?0x?0xxy?(x?y)y?x?0如果动点P(x,y)沿y?2x趋向所以极限不存在。
6．指出下列函数的间断点：
x2y24x4?lim4?0 ，则lim222x?0x?04x?x2xy?(x?y)y?2x?0y2?2x（1）f(x,y)?； （2）z?lnx?y。
y?2x解 （1）为使函数表达式有意义，需y?2x?0，所以在y?2x?0处，函数间断。 （2）为使函数表达式有意义，需x?y，所以在x?y处，函数间断。 习题1—2 1．（1）z?xy? yx?z1x?z1y??2；??2.
?yxy?xyx (2)
?z?ycos(xy)?2ycos(xy)sin(xy)?y[cos(xy)?sin(2xy)] ?x?z?xcos(xy)?2xcos(xy)sin(xy)?x[cos(xy)?sin(2xy)] ?y(3)
?z?y(1?xy)y?1y?y2(1?xy)y?1, ?x，两边同时对y求偏导得
lnz=yln
1?zx?ln(1?xy)?y, z?y1?xyxyxy?z?z[ln1(?xy)?]?(1?xy)y[ln1(?xy)?]; ?y1?xy1?xy2y3x3?2y?zx??(4), y?xx(x3?y)x?2x1? 3
?z??y1x2x?yx2?1;x3?y
y?uyz?1?u1z?u?x,?xlnx,??2xzlnx(5)?x; z?yz?zzyyy?uz(x?y)z?1(6), ?2z?x1?(x?y)z(x?y)z?1?u , ??2z?y1?(x?y)?u(x?y)zln(x?y); ?2z?z1?(x?y)2.(1)
zx?y,zy?x,zxx?0,zxy?1,zyy?0;
(2) zx?asin2(ax?by),zy?bsin2(ax?by),
zxx?2a2cos2(ax?by),zxy?2abcos2(ax?by),zyy?2b2cos2(ax?by).
3 fx?y?2xz,fy?2xy?z,fz?2yz?x,fxx?2z,fxz?2x,fyz?2z,
222fxx(0,0,1)?2,fxz(1,0,2)?2,fyz(0,?1,0)?0.
4
ttttzx??2sin2(x?),zt?sin2(x?),zxt?2cos2(x?),ztt??cos2(x?)
2222tt2ztt?zxt??2cos2(x?)?2cos2(x?)?0.
22yyyy115.(1) zx??2ex, zy?ex,dz??2exdx? exdy;
xxxx (2) z?yyyy1xxln(x2?y2),zx?2dz? dx?dy; ,,z?y22222222x?yx?yx?yx?yy12?ydx?xdyyxxx (3)zx? , ,; dz?z????2yy2x2?y2y2x2?y2x?y21?()1?()xx? 4
(4) ux?yzxyz?1,uy?zxyzlnx,uz?yxyzlnx,
du?yzxyz?1dx?zxyzlnxdy?yxyzlnxdz.
6. 设对角线为z,则z?x2?y2,zx?xx?y22,zy?yx?y22, dz?xdx?ydyx?y22
当x?6,y?8,?x?0.05,?y??0.1时,?z?dz?6?0.05?8?(?0.1)6?822 =-0.05(m).
7. 设两腰分别为x、y,斜边为z,则z?x2?y2,
, dz?zx?xx?y22，zy?yx?y22xdx?ydyx?y22,
设x、y、z的绝对误差分别为?x、?y、?z,
当x?7,y?24,?x??x?0.1,?y??y?0.1时, z?72?242?25
?z?dz?7?0.1?24?0.17?2422 =0.124,z的绝对误差?z?0.124
z的相对误差
0.124?z?0.496%. ?25z8. 设内半径为r，内高为h，容积为V，则
V??r2h,Vr?2?rh,Vh??r2,dV?2?rhdr??r2dh,
当r?4,h?20,?r?0.1,?h?0.1时,
?V?dV?2?3.14?4?20?0.1?3.14?42?0.1?55.264(cm3).
习题1—3
xyyx?2du?fdx?fdy?fdzzzz?2a(ax?1) 1.????? ?aeax?xyxyxydx?xdx?ydx?zdx1?()21?()21?()2zzzy[z?axz?2axy(ax?1)](ax?1)eax(1?a2x2)==. 222422axz?xy(ax?1)?xe??z?f???f????2.=?x???x???x1??24x3?arcsin??4=422x?y1?x?y?x 5