二元函数z=1 ln(x y)的定义域为-搜答案-百度作业网

(1)在(x,y)=(0,0)(2)在x=0或y=0在上面没定义

[x,y]=meshgrid(-9:0.5:9);z = 2*sin(x).*sin(y)./(x.*y); % z的表达式sinyy是什么?mesh(x,y,

先对x求导y*dz/dx+z+x*dz/dx+y=0所以dz/dx=-(z+y)/(x+y)同理得dz/dy=-(z+x)/(x+y)所以dz=-(z+y)/(x+y)dx-(z+x)/(x+y)dy

两边同时微分zdx+xdz+zdy+ydz+xdy+ydx=0(x+y)dz+(y+z)dx+(z+x)dy=0dz=-[(y+z)dx+(z+x)dy]/(x+y)

Z=e^xy在x处的导函数为ye^(xy)在y处的导函数为xe^(xy)dz=ye^(xy)dx+xe^(xy)dy=2e^2dx+e^2dy

已知二元函数z=f[x²-y²,e^(xy)]求∂²z/∂x∂y设z=f(u,v),u=x²-y²,v=e^(xy

dz=d(xyln(xy))=xyd(ln(xy))+ln(xy)d(xy)=xyd(xy)/(xy)+ln(xy)d(xy)=d(xy)+ln(xy)d(xy)=(1+ln(xy))d(xy)=(1

在几何画板中不太好画.可以借助GEOGEBRA或者MATHSLAB等软件.我用GEOGEBRA,通过借助滑杆提供的参数,可以画出上述图象.

x＞0,y＞z

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

全微分,分别对x和y求微分,再综合起来

1.函数f(x,y)=ln(y-x)+1/√(y+x)的定义域?y>x,且y>-x2.如果函数z=sinvlnu,u=x+y,v=xyz=sin(xy)ln(x+y)3.设z=arctan

ln(x+y+1)≠0【它充当分式的分母,当然不能为0】也就是ln(x+y+1)≠0=ln1x+y+1≠1且x+y+1＞0【对数的真数必须大于0】联合得到：x+y∈（-1,0）∪（0,+∞）

dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy

z偏x=-sin3xy*3y+1/(x+y+1)z偏y=-sin3xy*3x+1/(x+y+1)dz=[-sin3xy*3y+1/(x+y+1)]dx+[sin3xy*3x+1/(x+y+1)]dy

(x+1)y>0(1)x+1>0且y>0,得到x>-1且y>0;(2)x+1

两边求微分就行了2cosx*sinx*dx+2cosy*siny*dy+2cosz*sinz*dz=0dz=-(2cosx*sinx*dx+2cosy*siny*dy)/2cosz*sinz

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

z=xy+lnxy=xy+lnx+lny所以zy=x+1/y对的.