一個微分的題目

ln(x^2+y^2)=x+y

d ln(x^2+y^2) / dx = d(x+y)/dx

[d ln(x^2+y^2) / d (x^2+y^2)] [ d (x^2+y^2) /dx ] = 1+ dy/dx

[1/(x^2+y^2)] [2x +2y (dy/dx) ] = 1+ dy/dx

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

(y^2 - 2y) dy/dx =2x -x^2

dy/dx = -(x^2-2x)/(y^2 - 2y)

2014-11-12 09:51:53 補充：
更正...倒數第二行
(x^2+y^2 - 2y) dy/dx =2x -x^2 -y^2
dy/dx = - (x^2+y^2-2x)/(x^2+y^2 - 2y)