全微分精通者帮忙！ 设z=z(x,y)由方程e的z次方-xy的2次方+sin(y+z)=0确定，求dz

Question

如题所述

Answer 1

方程e的z次方-xy的2次方+sin(y+z)=0
e^z-xy^2+sin(y+z)=0
设F＝e^z-xy^2+sin(y+z)
F分别对x、y、z求导
Fx^'=-y^2
Fy^'=-2xy+cos(y+z)
Fz^'=e^z+cos(y+z)
Z对x的偏导数为　　-Fx^'/Fz^'＝y^2/[e^z+cos(y+z)]
Z对y的偏导数为　　-Fy^'/Fz^'＝-[-2xy+cos(y+z)]/[e^z+cos(y+z)]＝[2xy-cos(y+z)]/[e^z+cos(y+z)]
dz=-Fx^'/Fz^'dx+-Fy^'/Fz^'dy
=y^2/[e^z+cos(y+z)]dx+[2xy-cos(y+z)]/[e^z+cos(y+z)]dy

Answer 2

(y^2+2xy-cos(y+z))/(e^z+cos(y+z))追问

没有过程吗？

追答

求导：
e^z * dz -y^2-2xy+cos(y+z)(1+dz)=0
把含有dz的项移到一起：
(e^z+cos(y+z))dz=y^2+2xy-cos(y+z)
把dz的系数除到等式右边：
dz=(y^2+2xy-cos(y+z))/(e^z+cos(y+z))

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