设函数Z＝f(x+y,xy),且f具有二阶连续偏导数,则Z对X求二次偏导为什么

y+y∂z/∂x+z+x∂z/∂x=0∂z/∂x=-(y+z)/(x+y)∂2z/∂x2=【∂

df/dx=f'(xy,yz,x-z)(y+y*dz/dx+1-dz/dx)=0(1-y)dz/dx=f'(xy,yz,x-z)*(y+1)dz/dx=f'(xy,yz,x-z)*(y+1)/(1-y

设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：dz=(∂f/

设u=xy^2;v=x^2y;二阶偏导数：(f'u)*y^2+2(f'u)xy+2(f'v)xy+(f'v)x^2不好好学习啊同志再问：哥们你能上一下步骤求图求真相再答：这就是步骤，这就是答案啊，再问

对方程两边求全微分得：(e^z-1)dz+y^3dx+3xy^2dy=0（方法和求导类似）移项,有dz=-(y^3dx+3xy^2dy)/(e^z-1)

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

x^2+y^2+z^2-3xyz=0两边对x求偏导,2x+2z*dz/dx-3yz-3xydz/dx=0从中解得：dz/dx=(3yz-2x)/(2z-3xy)(1)同理：dz/dy=(3xz-2y)

令G(X,Y,Z)=F(xy,z-2x)GZ'=F'2GX'=yF'1-2F'2∂z/∂x=-GX'/GZ'=(2F'2-yF'1)/F'2Gy'=xF'1∂z/&

设u=sinx,v=xydz/dx=dz/du*du/dx+dz/dv*dv/dx=cosxf1'+yf2'd^2z/dxdy=d(dz/dx)/dy=(-sinx)f1'+cosx*df1'/dx+

两边对x求导1-a*δz/δx=f'(y-bz)*(-bδz/δx)整理得：[a-bf'(y-bz)]δz/δx=-1两边对y求导-a*δz/δy=f'(y-bz)*(1-bδz/δy)整理得：[-a

因为x、y都为自变量,不是宗量,故此题没有全微分,应只有偏微分.详解如下：对方程两边微分：左边：de^z=e^z*dz右边d[xyz+cos(xy)]=xydz+yzdx+xzdy-(sinxy)*(

设u=xy,v=y/x,则z=f(u,v),所以ðz/ðx=f'1*ðu/ðx+f'2*ðv/ðx=yf'1-yf'2/x^2,注意到f'1

你想说这个问题?z=e^(x^2+2xy)应该是y=e^(x^2+2xy)(2x+2y）i+e^(x^2+2xy)2xj

当点(x,y)沿x轴和y轴趋于(0,0)时,f(z)的极限都是0.但它沿直线y=mx趋于(0,0)时,limf(x,y)=lim(mx*x/(x*x+m*m*x*x))=m/(1+m*m),对于不同的

y+y∂z/∂x+z+x∂z/∂x=0∂z/∂x=-(y+z)/(x+y)y∂2z/∂x2+2ͦ

f对第1个变量的偏导函数记作f1,第2个变量的偏导函数记作f2,dz=f1*d(xz)+f2*d(z/y)...[注：写完整的话是f1(xz,z/y),f2也如此]=f1*(xdz+zdx)+f2*(

(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+

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)