一道很简单的大一高数题目,但是我不会,题目如图所示,虽然有点丑,将就着看吧因为我的悟性比较差

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方程两边分别取对数,化为
1/2 ln(x^2+y^2)=arctan(y/x)
两边对x求导,(1/2)(2x+2yy')/(x^2+y^2)=(xy'-y)/x^2/(1+y^2/x^2)
整理得y'=(x+y)/(x-y)
故dy=(x+y)/(x-y)dx
======以下答案可供参考======
供参考答案1:
方程两侧同时对x求导：
(1/2)*(x^+y^)^(-1/2) * d(x^+y^)/dx =e^[arctan(y/x)]*{1/[1+(y/x)^)]}
*d(y/x)/dx
1/[2√(x^+y^)] * (2x+2y*dy/dx)=e^[arctan(y/x)]* [x^/(x^+y^)]*
[(x*dy/dx - y)/x^]
最后化简可得：
dy/dx = {y*[e^arctan(y/x)] + x√(x^+y^)} / {x*[e^(arctan(y/x)] - y√(x^+y^)}
只要把dx移到等号右边，就可以得出dy了~~~