z=xy xF(u)而u=y x,F(u)为-搜答案-拍题作业网

dz=d[xyP(z)]=yP(z)dx+xP(z)dy+xyP'(z)dz所以dz=[yP(z)dx+xP(z)dy]/[1-xyP'(z)]du=df(x,z)=f'x(x,z)dx+f'z(x,

由z=u²v²,其中u=x-y,v=x+y,题型：求复合函数的偏导数：z=（x-y）²（x+y）²,dz/dx=（x-y）²×2（x+y）+2（x-y

再问：请问怎么变形到4里面这样啊。。

用z'表示z的共轭复数.|(z-u)/(1-z'u)|(分子分母同时乘以z)=|(z-u)z/[z(1-z'u)]|=|(z-u)z/(z-zz'u)|(注意到|z|=1,zz'=|z|^2=1)=|

∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/&#

dz/dx=dz/du*du/dx+dz/dv*dv/dx=vu^(v-1)+u^vlnu=(x-y)(x+2y)^(x-y-1)+(x+2y)^(x-y)ln(x+2y)dz/dy=dz/du*du

试试这样行不行;clear all;clc;u=0:pi/40:3*pi;x=(1+cos(u)).*cos(u);y=(1+cos(u)).*sin(u);z=sin(u);plot3(x

(z对x的偏导)=y+F(u)+x[F'(u)(-y/x^2)](z对y的偏导)=x+F'(u)/x代入,左边=[xy+xF(u)-yF'(u)]+[xy+yF'(u)]=xy+xF(u)+xy=z+

最容易理解的办法,代进去有z=x+y+xy那么对x偏导数有那个偏导数=1+y

第一题是用的拉格朗日数乘法计算条件极值.即在条件a=x+y+z下的乘积xyz的极值.设参数为u,构造拉格朗日函数F（x,y,z,u)=xyz+u(x+y+z-a)分别对四元函数求偏导,使其为零,联立方

symsuv;d=[-5:0.5:5];[uv]=meshgrid(d);x=u.*sin(v),y=u.*cos(v),z=u;surf(x,y,z)

和实数的解法一样,只要实/虚分开就行z+u=1+i1式z-u=lg(5/2)-(lg(5/2))i2式1+2式2z=1+lg2.5+(1-lg2.5)i=lg25+(lg4)i得到z=lg5+(lg2

1.z'x=3x²y²z'y=2x³y2.z'x=4x³z'y=3y³3.z'x=ye^(xy)+2xyz'y=xe^(xy)+x²4.u'

u=ln(xy+z)求du=

u=ln(xy+z)du=d[ln(xy+z)]/dx*dx+d[ln(xy+z)]/dy*dy+d[ln(xy+z)]/dz*dz=y/(xy+z)*dx+x/(xy+z)*dy+1/(xy+z)*

u=f(x-y,y-z,t-z)

分别把x,y,z,t当做为之数,其余都是常数,求就行了再问：具体怎么做呢？麻烦写清楚些

dz/dx是z对x的偏导,这样把u,v都带入的话直接球偏导就好了dz/dx=y*e^(xy)*sin(x+y)+e^(xy)*cos(x+y)同理也可得到dz/dy=x*e^(xy)*sin(x+y)

①偏z/偏x=偏z/偏u偏u/偏x+偏z/偏v偏v/偏x=(2uv-v^2)siny+(2uv-v^2)cosy=(2x^2sinycosy-x^2(cosy)^2)siny+(2x^2sinycos

grad(u)=(∂u/∂x,∂u/∂y,∂u/∂z)=(y^2,2xy,3z^2),所以div(grad(u))=div(y^

设z=cosA+isinAu=1+(cosA+isinA)²=1+cos²A-sin²A+i*2sinAcosA=(1+cos2A)+isin2A|u|²=(1