z=x.ln(xy),求

{（x,y）|y^2-4x+4＞0}再问：这是答案？再答：是呀再问：真不好意思，能解释下吗，没明白，加分再答：对数函数中，真数必须大于0，所以y^2-4x+4应该大于0

方法一（微分法）d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即（xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²

两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)

z'x=(-y/x^2)/(y/x)=-1/xz'y=(1/x)/(y/x)=1/ydz=z'xdx+z'ydyu=ln(x^2+y^2+z^2)u'x=2x/(x^2+y^2+z^2)u'y=2y/

x=z(lny-lnz)对x求导1=∂z/∂x*(lny-lnz)+z*(0-1/z*∂z/∂x)1=∂z/∂x(lny-lnz

x=z(lnz-lny)=zlnz-zlny令F(x,y,z)=zlnz-zlny-xaF/ax=-1aF/ay=-z/yaF/az=lnz+1-lny所以az/ax=-Fx/Fz=1/(lnz+1-

δz/δx=1/(xy+x/y)*(y+1/y)=(y²+1)/(xy²+x)=1/xδ^2z/δxδy=δ(δz/δx)/δy=0

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

设Y=y'降阶：Y'=(Y/x)ln(Y/x)这就是一个一阶齐次方程.设Y/x=u,所以Y=ux,Y'=u+x(du/dx),代回原方程,解得：lnu=C1x+1Y=xe^(C1x+1)所以y=[(C

x/z=ln(z/y),求微分：（zdx-xdz)/z^2=y/z*(ydz-zdy)/y^2=(ydz-zdy)/(yz),∴yzdx-xydz=yzdz-z^2dy,∴z'=yz/(xy+yz)=

∂z/∂x=(1/(x²+y))(2x)=2x/(x²+y)∂²f/∂x∂y=∂[∂z

嗯,是对的

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)*

z=(x^2)*ln(2xy),Zx=(2x)ln(2xy)+(x^2)/2xy*(2xy)'=(2x)ln(2xy)+xZxx=2ln(2xy)+(2x)/2xy*(2xy)'+1=2ln(2xy)

z=lnx^z+lny^x=zlnx+xlnyz=xlny/(1-lnx)先关于x求偏导,把y看做常数,再对y求偏导,把x看做常数dz=0dx+x/y(1-lnx)dy（此处省略了一些计算过程,）dz

dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(

应该是∂z/∂x吧!令u=x+y^2+z=>du/dx=1+dz/dxu=lnu^(1/2)=1/2*lnudu/dx=1/2*1/u*du/dx=>du/dx=u/(1/2+

两边求导得y'·e^y+(y+xy')/(xy)+e^(-x)=0

[x,y]=meshgrid(0:0.01:2);z=1000.*[log(x)./log(y)];mesh(x,y,z)

x/z=ln(y/z),x=zlny-zlnz两端对x求偏导得1=z'lny-z'lnz-z'两端对y求偏导得0=z'lny+z/y-z'lnz-z'