函数z=x^2y y^2的全微分-搜答案-智慧作业帮

∂z/∂x=2xy∂z/∂u=x²所以dz=2xydx+x²dy

Z=（1/2）ln(1+x²+y²)dz=(1/2)2x/(1+x²+y²)dx+(1/2)2y/(1+x²+y²)dy=x/(1+x&su

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

解；z(x)=2x+2y²z(y)=4xy+12y²dz=(2x+2y²)dx+(4xy+12y²)dy

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

再问：非常感谢，还要问大侠一道题面目。曲线y=x³+3x的拐点坐标为？？？再答：y'=3x²+3y''=3x令y"=0，得x=3当x=3时，y=36所以拐点坐标（3，36）

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

dz=Z'xdx+Z'ydy=2xcos(x^2+y^2)dx+2ycos(x^2+y^2)dy

dz=[2e^(2x+y)]dx+[e^(2x+y)]dy

再问：啊不好意思搞错了。。是z=e^(x^2+y^2)，求dz，谢谢你帮我解答一下吧。。再答：

dz=2e^(2x+y^2)dx+2ye^(2x+y^2)dy把对x和对y的偏导分别求了出来再乘以各自的微分项即可.

u'x=2x/(x^2+y^2+z^2)u'y=2y/(x^2+y^2+z^2)u'z=2z/(x^2+y^2+z^2)du=2xdx/(x^2+y^2+z^2)+2ydy/(x^2+y^2+z^2)

z=1/2*ln(x^2+y^2+4)Z'x=1/2*1/(x^2+y^2+4)*(2x)=x/(x^2+y^2+4)Z'y=1/2*1/(x^2+y^2+4)*(2y)=y/(x^2+y^2+4)所

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

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

2^(x^2+y^2)*In2*(2xdx+2ydy)

z=x^3y^2dz=3x^2*y^2dx+2x^3*ydy,在（1,1）处的全微分dz=3dx+2dy.

对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】

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