xdy?ydx.

x2?y22y?2u? . 4. 设u?yf()?xg(x),其中f,g具有二阶连续偏导数,则x?u?yxy?x?y?x2y?y?解：?u?yf'()??2??g(x)?xg'(x)1,

?xx?x?yyyy?y2?y2y?2ux1x1x1()xg' '(), 2?yf''(?)4??yf'()3?g'(?)g'?x?x?xxyyyyyy2?x ?y?y2?y2y?2uxxxx2xx 2??f''()?3??f'()2?g'()2?g''()3?g'()2,

x?x?xxyyyyyy?x22?u?所以x2?yu?0.

?x?y?x5. 若函数z=f(x,y)在点(x0,y0)处的偏导数存在，则在该点处函数z?f(x,y) ( D )

A 有极限 B连续

C 可微 D以上三项都不成立

解：因为偏导数存在，不能推出极限存在，所以ABC三项不一定正确. 6. 偏导数fx(x0,y0)，fy(x0,y0)存在是函数z=f(x,y)在点(x0,y0)连续的( D ) A 充分条件 B 必要条件

C 充要条件 D 即非充分也非必要条件 解：同5.

22

7. 设函数f(x,y)=1－x+y,则下列结论正确的是( D )

A 点(0,0)是f(x,y)的极小值点 B 点(0,0)是f(x,y)的极大值点 C 点(0,0)不是f(x,y)的驻点 D f(0,0)不是f(x,y)的极值 8. 求下列极限： (1)

lim(x2?y2)sin1； (2) (x,y)?(0,0)xy(x,y)?(0,0)(x,y)?(0, 0)limxy?1?1.

x?y解：(1) 因为 (2)

(x,y)?(0, 0)lim(x2?y2)?0，而sin11有界.所以lim(x2?y2)sin?0.

(x,y)?(0,0)xyxylimxy?1?1(xy?1?1)(xy?1?1)xy?lim?lim (x,y)?(0, 0)(x,y)?(0, 0)x?y(x?y)(xy?1?1)(x?y)(xy?1?1) =0

9. 设u=e3xy，而x2+y=t2,x－y=t+2,求du.

dtt?0－

解：由x2+y=t2,x－y=t+2,可得 dxdydxdy2x??2t,??1,所以 dtdtdtdtdx2t?1dy2t?2x?,?. dt2x?1dt2x?1dy2t?13x?y2t?2x. 因此，du?dudx?du?3e3x?y?edtdxdtdydt2x?12x?1令t?0,得x??2,y??4或x?1,y??1.

?2du5e4故?e或. dtt?03322?z?z?10. 设z=f(x,y)由方程xy+yz+xz=1所确定,求,2,z. ?x?x?x?y解：两边同时对x求偏导，得

y?z?z?z?z?zx?z.

y?y?z?x?0,因此??,由对称性可得???x?x?xx?y?yx?yy?z?z(x?y)?(y?z)?(x?y)?y?zx?y2y?2z?2z?x?????. 2222?x(x?y)(x?y)(x?y)(1??z)(x?y)?(y?z)(1?x?z)(x?y)?y?z2?yx?y?z2z?????. 22?x?y(x?y)(x?y)(x?y)2?2f?2f111. 设f(u,v)具有二阶连续偏导数,且满足2?2?1，又g(x,y)?f[xy,(x2?y2)],

2?u?v试证

?2g?2g?2?x2?y2. 2?x?y1证：设u?xy,v?(x2?y2),则g(x,y)?f(u,v).则

2?g?f?u?f?v?f?f?g?f?u?f?v?f?f???y?x,???x?y, ?x?u?x?v?x?u?v?y?u?y?v?y?u?v?2g?2f?u?2f?y?2?x2?u2?x?v?2g?2f?u?2f?x?2?y2?u2?y?v?f?2f2?2f2?f?vx??y?2x?, ?x?v?u2?v?v?f?2f2?2f2?f?vy??x?2y?, ?y?v?u2?v?v?2g?2g所以2?2?x2?y2.

?x?y12. 求函数f(x,y)=x2(2+y2)+ylny的极值.

2??fx(x,y)?2x(2?y)?0解：先解方程组? 2f(x,y)?2xy?lny?1?0??y得驻点为(0,1).

1fxx?2(2?y2),fxy?x,y??4xy,fyy?x,y??2x2?,

y在点(0，1)处,Δ=AC-B2=6×1-0>0,又A>0,所以函数在(0，1)处有极小值f(0，1)=0.

（B）

－?z?1. 设z=ex+f(x－2y)，且已知y=0时，z=x2,则 .

?x解：令y?0得f(x)?x2?ex,因此，z?ex?(x?2y)2?ex?2y,

?z所以?ex?2(x?2y)?ex?2y.

?x2. 设f(x,y,z)=exyz2，其中z=z(x,y)是由x+y+z+xyz=0确定的隐函数，则fx(0,1,?1)? .

?z?z解：由x?y?z?xyz?0可得1??y(z?x)?0.

?x?x1?yz故?z??. ?x1?xy1?yz?zfx(x,y,z)?y(exz2?ex?2z)?y(exz2?2exz)

?x1?xy因此fx(0,1,?1)?1.

3. 设z?ln(x?y),则x?z?y?z? .

?x?y11112y2x?z?z,?解：?，

?xx?y?yx?y1(x?y)?z?z1?. 所以x?y?2?x?y2x?y214. 设z?f(xy)?yg(x?y),,其中f,g具有二阶连续偏导数,则?z? .

x?x?yy解：?z??12f(xy)?f'(xy)?yg'(x?y),

?xxxy?2z11??f'(xy)?f'(xy)?f''(xy)x?g'(x?y)?yg''(x?y) ?x?yxxx ?yf''(xy)?g'(?xy?)yg''?(x. y5. 函数f(x,y)?eA B C D

x2?y4在点(0,0)处的偏导数存在的情况是( C ).

fx(0,0)，fy(0,0)都存在 fx(0,0)存在，fy(0,0)不存在

fx(0,0)不存在，fy(0,0)存在 fx(0,0)，fy(0,0)都不存在

2?x2f(0??x,0)?f(0,0)e?1?x解：fx(0,0)?lim?lim=lim不存在, ?x?0?x?0?x?0?x?x?x?y4f(0,0??y)?f(0,0)e?y?1fy(0,0)?lim?lim=lim=0.

?x?0?x?0?x?0?y?y?y46. 设f(x,y),g(x,y)均为可微函数，且gy(x,y)≠0，已知(x0,y0)是f(x,y)在约束条件

g(x,y)=0下的一个极值点，下列结论正确的是( D ) A 若fx(x0,y0)=0,则fy(x0,y0)=0 B 若fx(x0,y0)=0,则fy(x0,y0)≠0 C 若fx(x0,y0)≠0,则fy(x0,y0)=0 D 若fx(x0,y0)≠0,则fy(x0,y0)≠0

解：作拉格朗日函数L(x,y,?)?f(x,y)??g(x,y)，则有 Lx(x,y,?)?fx(x0,y0)??gx(x0,y0)?0， Ly(x,y,?)?fy(x0,y0)??gy(x0,y0)?0.

由于gy(x,y)≠0，所以当fx(x0,y0)≠0，??0,因此?gy(x0,y0)?0，从而fy(x0,y0)≠0. 7. 设函数u=f(x,y,z)有连续偏导数，且z=z(x,y)是由xex－yey=zez所确定的隐函数，求du.

yy?zz?zex?xex?z?e?yexyzxxz?z解：由xe－ye=ze可得e?xe?e?ze. .故?,同理?z?x?x?xez?zez?ye?zez因此du?fxdx?fydy?fzdz

xxey?yeye?xe ?fxdx?fydy?fz(zdx?zdy)

e?zeze?zezey?yeyex?xex

?(fx?fzz)dx?(fy?fzz)dy.

e?zeze?zez8. 设函数u=f(x,y,z)有连续偏导数，且y=y(x),z=z(x)分别由下列两式确定：

x?zexy?xy?2,ex??sintdt，

0t求

du. dxxyxydydydyexyy?yy解：由e?xy?2,可得e(y?x)?(y?x)?0,因此，???. xydxdxdxx?exxx?zsintsin(x?z)ex(x?z)dzdzxx由e??. dt,可得e?(1?)，因此?1?0tx?zdxdxsin(x?z)dyyex(x?z)dudz故?fx?fy?fz?fx?fy?fz[1?]. dxdxdxxsin(x?z)9. 设z=z(x,y)由方程x2+y2－z=g(x+y+z)所确定，其中g具有二阶连续偏导数且g′≠－1. (1) 求dz;

?u.1(?z??z)(2) u(x,y)?,求

x?y?x?y?x解：(1)由x2?y2－z?g?x?y?z?,两边分别同时对x、y求偏导得

2x?因此dz??z?z?z?z?g'?x?y?z?(1?),2y??g'?x?y?z?(1?). ?x?x?y?y?z2x?g'?x?y?z??z2y?g'?x?y?z??,?. ?xg'?x?y?z??1?yg'?x?y?z??12x?g'?x?y?z?2y?g'?x?y?z?dx?dy.

g'?x?y?z??1g'?x?y?z??1(2) u(x,y)?2x?2y1?z?z12, (?)??x?y?x?yx?yg'?x?y?z??1g'?x?y?z??12x?g'?x?y?z??z?2g''(x?y?z)[1?]?2g''(x?y?z)(1?)g'x?y?z?1???u?x??. 2?x[g'(x?y?z)?1][g'(x?y?z)?1]210. 求函数u=x2+y2+z2在约束条件z=x2+y2和x+y+z=4下的最大值和最小值.解：由z=x2?y2,x?y?z?4可得x2?y2?4?x?y.因此，问题转化为求 (4?x?2y)在约束条件 u?4?x?y?22x?y?4

?x下的极值问题?y.

令L(x,y,?)?4?x?y?(4?x?y)2??(x2?y2?4?x?y), Lx(x,y,?)??1?2(4?x?y)?2?x???0， Ly(x,y,?)??1?2(4?x?y)?2?y???0.

x2?y2?4?x?y?0,

解得: x??2,y??2或x?1,y?1.因此, z=8或z=2.

又f(?2,?2,8)?72,f(1,1,2)?6.