微积分（二）课后题答案，复旦大学出版社

?z??z??u??z??v?veuv1?2y?ueuv?1?1?y?u?y?v?yx2?y22x2?y21?(yx

x)2x2?y2?arctanyx ?euvln22(yv?xu)?ex?yx2arctany?xlnx2?y2)?y2(xx

３．求下列函数的一阶偏导数，其中f可微： (1) u?f (

xy,yz)； (2) z?f(x2?y2)； (3) u?f(x, xy, xyz)．

解：（1）

?u11?x?f1??y?f2??0?yf1?

?u?y?f1???x?y2?f2??1z?1zf2?xy2f1?

?u?f??0?f?y?z12???y?z2z2f2

（2）令u?x2?y2,则z?f(u)

?z?df??u?f?(u)?2x?2xf?(x2?y2)?xdu?x

?z?df??u?ydu?y?f?(u)?2y?2yf?(x2?y2)

（3）令t?x,v?xy,w?xyz,则u?f(t,v,w).

?u??f?dt??f??v??f??w?ff?x?tdx?v?x?w?x1??1?f2??y?3??yz?f1??yf2??yzf3??u??f?dt?y?tdy??f??v??f??w?v?y?w?y?f1??0?f2??x?f3??xz?xf2??xzf3?

?u??f?dt??f??v??f??w?f?0?f?xy?xyf?z?tdz?v?z?w?z1?2??0?f3?3?

４．设z?xy?x2Ｆ(u),u?

y，Ｆ（u）可导．证明：

xx?z?y?z?x?y?2z．

证：??z?y?2xF(u)?x2F?(u)??y?y?2xF(u)?yF?(u)?xx2

?z21?y?x?xF?(u)?x?x?xF?(u)

9

?x?z?x?y?z?y?xy?2xF(u)?xyF?(u)?xy?xyF?(u)

2 ?2[xy?xF(u)]?? z５．利用全微分形式不变性求全微分：

(1) z?(x2?y2)sin(2x?y)； (2) u?

yf(x?y?z)2222，f可微．

v解：（1）令u?x?y,v?sin(2x?y)，则z?u

22dz??z?udu??z?vdv?vuv?1d(x?y)?u?lnudsin(2x?y)

22v

?vuvv?1(2xdx?2ydy)?ulnu?cos(2x?y)d(2x?y)?2(xdx?ydy)?lnu?cos(2x?y)(2dx?dy)]?2sin(2x?y)?22(xdx?ydy)?cos(2x?y)ln(x?y)(2dx?dy)??22x?y???1f22v?u[vu?(x?y)22sin(2x?y)（2）du?1fdy?y?df?1f2dy?yf2f?(x?y?z)d(x?y?z)

222222?1fdy?yf?(x?y?z)f122(2xdx?2ydy?2zdz)222

?f(x?y?z)222dy?2yf?(x?y?z)f(x?y?z)2222

(xdx?ydy?zdz)６．求下列隐函数的导数：

(1) 设ex?y?xyz?ex，求z?x,z?y； (2)设

xzx?ln

zy，求

?z,?z．

?x?y解：（1）设F(x,y,z)?e Fx??e 故zx???x?yx?y?xyz?ex?0，则

?yz?e,Fy??ex?y?xz,Fz??xy

Fx?Fz?e?exx?y?yzxyxzyzzy,zy???FyFz??ex?y?xzxy

（2）设F(x,y,z)??ln?0,则

Fx??

1z,Fy?????zy2?1y,Fz???xz2?yz?1y??xz2?1z

10

1故

?z?x??Fx?Fz????zxz2?1z?zx?z

1

?z?y??Fy?Fz????yxz2?1z?z2y(x?z)

７．设x?z?yf(x2?z2)，其中f可微，证明：z?zz?x?y??y?x．

证：设F(x,y,z)?x?z?yf(x2?z2)

则Fxyf?(x2?z2x??1?2)

22

Fy???f(x?z)Fz??1?2yzf?(x2?z2

)22故

?z??Fx??2xyf?(x?z)?1?xFz?1?2yzf?(x2?z2)

?zFy?f(x2?y2

???)?y

Fz?1?2yzf?(x2?z2)2222从而z?zyzf?(x?z)?zyf(x?y)?x?y?z?x?y?1?2yzf?(x2?z2)?1?2yzf?(x2?z2)2222?2xyzf?(x?z)?z?yf(x?y)1?2yzf?(x2?z2)2xyzf?(x2?z2 ?)?z?x?z1?2yzf?(x2?z2)

x[2yzf?(x2?z2?)?1]?x1?2yzf?(x2?z2)８．设x?eucosv, y?eusinv, z?uv,求

?z及

?z．

?x?y解法一：由x?eucosv,y?eusinv得

u?1ln(x2?y2),v?arctany,z?uv2x 故

?z??z?u?x?u?x??z?vxv?yu?v?x?x2?y2?(vcosv?usinv)e?u 11

?z?y??z?u?u?y??z?v?v?yu?yv?xux?y22?(vsinv?ucosv)e?u

??x?ecosv解法二：设方程组?确定了函数u?u(x,y),v?v(x,y),对方程组的两个方程关u??y?esinv于x求偏导得

?u?v?uu1?ecosv?esinv???x?x ??0?eusinv?u?eucosv?v??x?x???u?u?ecosv???x解方程组得?

??v??e?usinv??x?又方程组的两个方程关于y求偏导得

?u?v?uu0?ecosv?esinv??y?y? ??u?vu?1?eusinv?ecosv??y?y?解方程组得：

??u?u?esinv???y ??v?u??ecosv???y从而

?z?x??z?u?z?u??u?x???z?v?v?x??z?v??e?v?y?u(vcosv?usinv)

?z?y??u?ye??u(vsivn?ucovs

)９．设u?f(x,y,z)有连续偏导数，y?y(x)和z?z(x)分别由方程e解：方程exyxy?y?0和e?xz?0确定，求

z

dudx．

?y?0两边对x求导得

exy(y?xdydx)?dydx?0，解得

dydx?yexyxy1?xe?y21?xy

方程e?xz?0两边对x求导得

z 12