Question

设函数u=f(x,y,z)有连续偏导数，且z=z(x,y)由方程xe^x-ye^y=ze^z所确定

设函数u=f(x,y,z)有连续偏导数，且z=z(x,y)由方程xe^x-ye^y=ze^z所确定，求du.

Answer 1

du = ∂f/∂x dx + ∂f/∂y dy + ∂f/∂z dz.............................................................(1)
且 z=z(x,y) 由方程 xe^x-ye^y=ze^z 所确定
ze^z = xe^x-ye^y...................................................................................................(2)
z'xe^z + zz'xe^z = e^x+xe^x ............z'x = (1+x)e^x/(1+z)e^z.......(3)
类似地：............................................................z'y = (1+y)e^y/(1+z)e^z.......(4)
dz = z'x dx + z'y dy.................................................................................................(5)
将(3、4、5)代入(1):
du = ∂f/∂xdx+∂f/∂ydy+∂f/∂z{}(1)
= [∂f/∂x + (1+x)/(1+z)e^(x-z)]dx +
+ [∂f/∂y + (1+y)/(1+z)e^(y-z)]dy............................(6)
请再检查一遍.

Answer 2

简单计算一下即可，答案如图所示

Answer 3

引用yxue的回答：
du = ∂f/∂x dx + ∂f/∂y dy + ∂f/∂z dz.............................................................(1)
且 z=z(x,y) 由方程 xe^x-ye^y=ze^z 所确定
ze^z = xe^x-ye^y...................................................................................................(2)
z'xe^z + zz'xe^z = e^x+xe^x ............z'x = (1+x)e^x/(1+z)e^z.......(3)
类似地：............................................................z'y = (1+y)e^y/(1+z)e^z.......(4)
dz = z'x dx + z'y dy.................................................................................................(5)
将(3、4、5)代入(1):
du = ∂f/∂xdx+∂f/∂ydy+∂f/∂z{}(1)
= [∂f/∂x + (1+x)/(1+z)e^(x-z)]dx +
+ [∂f/∂y + (1+y)/(1+z)e^(y-z)]dy............................(6)
请再检查一遍.

du = ∂f/∂xdx+∂f/∂ydy+∂f/∂z{}(1)
= [∂f/∂x +∂f/∂z (1+x)/(1+z)e^(x-z)]dx +
+ [∂f/∂y －∂f/∂z (1+y)/(1+z)e^(y-z)]dy.