Question

设u=(x+z)/(y+z),其中z=z(x,y)由方程ze^z=xe^x+ye^y所确定,求∂u

设u=(x+z)/(y+z),其中z=z(x,y)由方程ze^z=xe^x+ye^y所确定,求∂u/∂x,∂u/∂y

Answer 1

ze^x=xe^x+ye^y
ze^x + e^x∂z/∂x = (1+x)e^x
∂z/∂x =(1+x)/(z+1)

ze^x=xe^x+ye^y
e^x ∂z/∂y = (1+y)e^y
∂z/∂y = (1+y)e^(y-x)

u=（x+z）/（y+z ）
∂u/∂x
=[(y+z)∂/∂x(x+z) -(x+z)∂/∂x(y+z) ]/(y+z)^2
=[(y+z)(1+∂z/∂x) -(x+z)∂z/∂x) ]/(y+z)^2
=[(y+z)(1+(1+x)/(z+1)) -(x+z)(1+x)/(z+1)) ]/(y+z)^2

∂u/∂y
=[(y+z)∂/∂y(x+z) -(x+z)∂/∂y(y+z) ]/(y+z)^2
=[(y+z)∂z/∂y -(x+z)(1+∂z/∂y) ]/(y+z)^2
=[(y+z)(1+y)e^(y-x) -(x+z)(1+(1+y)e^(y-x)) ]/(y+z)^2

追问

是ze^z,不是x，不过还是很感谢