第五节复合函数微分法与隐

第五节复合函数微分法与隐函数微的情形.下面分几种情况来讨论.分布图★★★★★★★★例★例★例★例★★★★例★例★★习题-★内容要一、z

f(uvuu(tvv(tz

f[u(t),dz

zduu

zdvv

(5.1)dz称为全导数z

f(uv),uu(x,y),vv(x,yzf[u(x

v(x, z zxuxvxzzuzv

u v定理3如果函数uu(xy)在点(x,yxy的偏导数,函数vvyyz

f(uv在对应点(uv具有连续偏导数,zf[u(x

v(在对应点(x,y的两个偏导数存在, zxux

z zyuyvdy

zz

f[u(x,yx,yy

zf(uxy中的uyx的偏导数

与

f(u,v)

f2

f(u,v)

2f 1表示对第一个变量u2表示对第二个变量v

等等二、.zf(u,v)

uu(x,

vv(x, z zv z zvdzxdxydyuxvxdxuyvy z z uxdxydyvxdxy zduz u、v是中间变量，但全微分dzxy是自变量时的表达式在形式上完全一致.这个性质称为全微分形式不变性.适当应用这个性质，会收到很好的效果.F(x,y)

来求它所确定的隐函数的导数的方法.这里将进一步从理论上阐明隐函数的存在性，并通过定理4设函数F(xy)在点P(x0y0)的某一邻域内具有连续的偏导数,Fy(x0,y0)0,F(x0,y0)0,则方程F(x,y)0在点P(x0,y0)的某一邻域能唯一定续且具有连续导数的函数y

f

y0

f(x0),dyFx

5F(xyzP(x0y0z0的某一邻域内有连续的偏导数,F(x0,y0,z0) Fz(x0,y0,z0)则方程F(x,y,z)0在点P(x0,y0,z0)的某一邻域能唯一确定续且具有连续z

f(xy,z0

f(x0,y0,并zFx

zFy

例题选1E01)zuvsint而uetv

求导 dzzduzdv

vetusint u v etcostetsintcostet(costsint)2E02)zeusinv而uxyvx

zz zzuzveusinvyeucosv u veu(ysinvcosv)exy[ysin(xy)cos(xzzuz

eusinvxeucosv u veu(xsinvcosv)exy[xsin(xy)cos(x3z3x2y24x2y的偏导数 设u3x2y2,v4x2y,则zuv可 zvuv1,zuv u6x,u2y,v

v zzuzvvuv16xuvlnu u v6x(4x2y)(3x2y2)4x2y14(3x2y2)4x2yln(3x2y2zzuz

vuv12yuvlnu u v2y(4x2y)(3x2y2)4x2y12(3x2y2)4x2yln(3x2y24设u

f(x,y,z)ex2y2z2,zx2sin

uu u

z2xex2y2z22zex2y2z22xsin z2x(12x2sin2y)ex2y2x4sin2yu

z2yex2y2z22zex2y2z2x2cos z2(yx4sinycosy)ex2y2x4sin2y5E03)zxy

u(x,

z

2zx2

2 xyxyx(x,2z

z

xyxx

xx(x,2

z

yx

yyx1xy1xy(x, zf(exy,x2y2

z26

其 有连续的二阶偏导数,求y,y2 设exy,x2y2,zf

yexyf2x

2 xyf

fxyy

y2xxy22xy22222f 2 f2xy2 4xy e

2y

2x 7E04)设w

f(xyz,

fx和xz 令uxyz,vxyz,

f(u,v)

2f(u,

wfuf

fyzf u

v 2w

z(f1yzf2)zyf2yzzf1f1uf1vfxyf u

v f2f2uf2

fxyf

v 2w

fxyf

yfyz(f

xyff)fy(xz)f

xy2zf

yf28zeusin

而u

vx

zxzy dzd(eusinv)eusinvdueudud(xy)ydxxdy,dvd(xy)dxdy,代入后归并含dxdy的项,dz(eusinvyeucosv)dx(eusinvxeu zdxzdyexy[ysin(xy)cos(xy)]dxexy[xsin(xy)cos(x 比较上式两边的dxdy的系数,

zxexy[ysin(xy)cos(xzyexy[xsin(xy)cos(x2的结果一样x9E05)利用一阶全微分形式的不变性求函数ux2y2z2的偏导数 (x2y2z2)dxx(2xdx2ydy2zdz)(x2y2z2)2

(x2y2z2)dxxd(x2y2z2)(x2y2z2)2(y2z2x2)dx2xydy.(x2y2z2所以

y2z2,(x2y2z2

u

,(x2y2z2

(x2y2z2)210zarctanxy的全微分1 设uxy,v1xy,则zarctanuv于是dzzduzdv

1du

udv

(vdu

1(u)2v

1v

)2

v2

u2由uxy,v1xy,dudxdy,dvydxxdy代入上式,dz

(xy)2(1

[(1xy)(dxdy)(xy)(ydxxdy) 1

dy.1y211(E06)已知exy2zez

解d(exy2zezexyd(xy)2dzezdz(ez2)dzexy(xdyye xe (ez2)dx(ez2)

,ez2

xexyez12E07)x2y210在点(0,1)x0y1yf(x,x0的值 令F(x,y)x2y21,Fx2x,Fy2y,Fx(0,1)0,Fy(0,1)2x2y210在点(0,1)的某领域内能唯一确定一个有连续导数,xy1yf(xdy

xy

d2

yxyy

yx(xyy

1

d2d2

例13求由方程xyexey0所确定的隐函数y的导数dy, dx

令Fxyexey,则Fxyex,Fyxey

ex

xeyexxeyxexxey

14E08)xlnzzf(xy的偏导数z,z x解F(xyz)

xz

zF(xyz)0y

1,1,y 1,x1zz yyz y ，

zzFx

x

z

zy(xz)z

例15z33xyza3a是常数)z

f(x,y)z 令F(x,y,z)z33xyza3,则Fx3yz,Fy3xz,Fz3z23xy.显然都是续.Fz3z23xy0z

3z2

z2z

3z2

z216E09)

4z0,

2zx2 令F(x,y,z)x2y2z24z,Fx2x,Fz2z

Fx

2x,2

(2z)x

(2z)x

(2

.

2

(2

(2

(217z

f(xyz,

zxyxy z看成x,y的函数对x求偏导数z

z

yzxyz u

v

z

fuyzfv,1fuxyfvxzyy 0fuy1fvxzyzy x

fuxzfvfuyxzz1

y

xyxzyu v

zy1fuxyfv fu

zz(x,

2z2z218

x2xyy2