?z?x2?y2,27. 分别求曲线?在xOy面及yOz面的投影.
z?1?解：消去变量z，得x?y?1,
22?x2?y2?1,故曲线在xOy面内的投影曲线为?
?z?1,消去变量x，得z=1,y?1.故曲线在yOz面内的投影为?28. 求z?y绕z轴旋转所得旋转曲面的方程？ 解：方程为z?x?y.
2222?z?1, (?1?y?1).
?x?0?z2?5x,29. 曲线?绕x轴旋转所得旋转曲面方程及名称为何？
?y?0答：旋转曲面方程为y?z?5x，它称为旋转抛物面.
222230. 画出曲面z?1?x?y与z?x?y所围空间图形.
221 z O x y
四、 多元函数的微分学
1.表达式limf?x,y??lim?limf?x,y??成立吗？
x?x0x?x0y?y0??y?y0??答：不一定. 例如：limx?0y?0?xyxy?limlim?0. 不存在，而
x?0?y?0x2?y2?x2?y2??2. 已知f?x,y??3x?2y，求f[xy,f(x,y)].
第36页
解：f[xy,f(x,y)]=3(xy)?2f?x,y?=3(xy)?2(3x?2y)?3xy?6x?4y. 3. 求limsinxy.
x?0xy?2解：limsinxysinxysinu?y=lim=lim?limy?2.
x?0x?0u?0y?2xyxuy?2y?24?x2?y2ln(x2?y2?1)的定义域, 并画出定义域的图形.
4. 求函数z?22??4?x?y?0,22221?x?y?4D?(x,y)|1?x?y?4. 解：由?2得，故定义域为2??x?y?1?0,??如下图：
y O 1 2 x
385.f(x,y)?xy，求fx(1,0)，fy(1,1). 28解：fx(x,y)?3xy, fy(x,y)?8xy,
28故 fx(1,0)?3?1?0?0, fy(1,1)?8?1?1?8.
37276.u?esinxy, 求
x?u?x,(0,1)?u?y.
(1,0)解：因
?u?exsinxy?excosxy?y?ex(sinxy?ycosxy), ?x?u?excosxy?x, ?y??u?x?u?yy?e0(sin0?cos0)?1,
(0,1)?e(cos0?1)?e.
(1,0)7.z?x，求
?z?z，. ?x?y第37页
解：
?z?zy?1=yx, =xylnx.
?y?x8.z?lnxy，求
?z?z，. ?x?y解：
11?z1=?(xy)'x??y?,
xyx?xxy?z111=?(xy)'y??x?.
xyy?yxy?z?2z?z9.z?xe，求，，.
?x?x2?y8y?z?2z?z7y6y7y8y(8xe)'?56xe解：=8xe, =, =xe. x2?y?x?x10.z= sin(2x?3y)，求zx，zy，zxx，zyy，zxy. 解：zx=cos(2x?3y)?(2x?3y)'x?2cos(2x?3y),
zy=cos(2x?3y)?(2x?3y)'y?3cos(2x?3y), ?zxx=?2cos(2x?3y)?x??4sin(2x?3y), ?zyy=?3cos(2x?3y)?y??9sin(2x?3y),
?zxy=?2cos(2x?3y)?y??6sin(2x?3y).
11.若z?(1?x)，求
xy?z?z，. ?x?y解：取对数得lnz?xyln(1?x), 两边对x求导，得
1?z1, ??yln(1?x)?xy?z?x1?x??zxy???(1?x)xy?yln(1?x)??, ?x1?x???z?(1?x)xy(xy)'yln(1?x)?x(1?x)xyln(1?x). ?y12.若f(x,y)?x?(y?1)lnsinx，求fx(x,1). y第38页
解：fx(x,1)=[f(x,1)]?x =(x)?x =1. 13.z?exycosxy，求
?z?z,. ?x?y解:
??z??exyxcosxy?exy?cosxy?x?yexy?cosxy?sinxy?, ?x????z??exyycosxy?exy?cosxy?y?xexy?cosxy?sinxy?. ?y??14. u??x?2y?3z?，求
2?u?u?u,,. ?x?y?z解：
?u??2?x?2y?3z???x?2y?3z?x?2?x?2y?3z?, ?x
?u??2?x?2y?3z???x?2y?3z?y?4?x?2y?3z?, ?y
?u??2?x?2y?3z???x?2y?3z?z?6?x?2y?3z?. ?z
15.设z?xylny，试用两种方法求dz.
解法一： ??z?z1?ylny,?xlny?xy??x?lny?1?, ?x?yy?z?zdx?dy?ylny?dx?x?lny?1?dy. ?x?y?dz?解法二：dz?d?xylny?
?ylnydx?xd?ylny?
?ylny?dx?x?lny?dy?y?dlny? ?ylnydx?x?lny?1?dy.
16.设z?y,当x?2,y?1,?x?0.1,?y??0.2, 求?z及dz. x?y??yy?????y?1?x??xx??x?2y?1?x?0.1?y??0.2解：?zx?21?0?215?????0.119.
2?0.1242?x?0.1?y??0.2第39页
y1dx?dy,
xx211?dzx?2,y?1??2?0.1????0.2???0.125.
?x?0.1,?y??0.222?dz??17.z?xyexy?x3y4，求dz.
解：dz?dxye?xy??d?xy?
34?exyd?xy??xydexy?y4dx3?x3dy4 ?exy?1?xy?dxy?3y4x2dx?4x3y3dy ?exy?1?xy??xdy?ydx??3x2y4dx?4x3y3dy ?3x2y4?y?xy2exydx?4x3y3?x?x2yexydy.
18.设z?x2y2ln(2x?y)，求
?????????z?z ，.
?y?x解一 令u?x ，v?2x?y，原式可写成z?u2lnv， y由复合函数求导法则，得
?z?z?u?z?v，即 ?????x?u?x?v?x?z1u22x2x2 ， ?2ulnv???2=2ln(2x?y)?2?xyvyy(2x?y)2x2x2?z?z?u?z?vxu2. ln(2x?y)?2????=2ulnv?(?2)??(?1)=?v?y?u?y?v?yy3y(2x?y)y解二 利用一元函数求导法则求偏导，可直接求出两个偏导数
?z?z ，.即
?y?x2x2x2?z2x2x2?z= 2ln(2x?y)?2，=?. ln(2x?y)?2?yy3?xy(2x?y)yy(2x?y)y?z?z19.设z?x2f(,sinxy)，求 ，.
?yx?x解：此题为抽象函数，所以只能用多元函数求导法则. 令 u?y , v?sinxy , 则z?x2f(u,v) ，于是 x?f?u?f?v?z=2xf(u,v)+x2fx(u,v)=2xf(u,v)+x2[???] ?x?u?x?v?x =2xf(u,v)+x2[
yx?fy?f1?(?2)??cosxy??y] ?u?vx2xy =2xf(,sinxy)+x2（?y?f1?2?u2xy?f）， cosxyx?v第40页