z(x,y)的微分-搜答案-灵鹊做题网

我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y

dz=y*x^(y-1)/cosz*dx+x^y*lnx/cosz*dy

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

z'x=2e^(2x+y)z'y=e^(2x+y)所以dz=2e^(2x+y)dx+e^(2x+y)dy

e^(-xy)-2z+e^z=0-ye^(-xy)-2z'(x)+e^zz'(x)=0z'(x)=ye^(-xy)/(e^z-2)-xe^(-xy)-2z'(y)+e^zz'(y)=0z'(y)=xe

symsxydiff(z,x,1)

z=f(x,y∧2,z)两边取全微分,dz=f'xdx+(f'y)*2ydy+f'zdz所以dz=[(f'x)/(1-f'z)]dx+[2y(f'y)/(1-f'z)]dy

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

z=3x²y+x/yzx=6xy+1/yzy=3x²-x/y²所以dz=zxdx+zydy=(6xy+1/y)dx+(3x²-x/y²)dy

x‘=11²=x∴x=1

代入：2z-2z+lnz=0--->z=1,所以z'(y)=-z/y从而dz=z'(x)dx+z'(y)dy=(e^x-yz)/(xy)

两边即对数得：lnz=xy*ln(lnu),不妨记u=x^2+y^2z'x/z=yln(lnu)+2x^2y/lnu,z'x=z[yln(lnu)+2x^2y/lnu]z'y/z=xln(lnu)+2

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+(&#

看到dy,deltay,∂y,初学的话就别管区别,都是一个事：y的变化量还有你的公式有问题dz不是等于∂z/∂x+∂z/∂y,是等于(

dz=1/y/(1+x^2/y^2)*dx-x/y^2/(1+x^2/y^2)*dy

dz=x的偏导数乘德尔塔x+y的偏导数乘德尔塔y这是最基础的题呀,直接套公式啊

dz=2xydx+x^2dy再问：有全过程吗再答：en我想知道这里的X^2Y是指的X得平方乘以Y吗？如果是过程如下：dz/dx=2xydz/dy=x^2dz=2xydx+x^2dy再问：是X的2Y次方

dz=(y+1/y)dx+(x-x/y^2)dy