Question

求解一道数学题，高手帮帮忙

设y为x的函数是由方程ln根号（x平方加y平方）＝arctan y/x 确定的，求y'

Answer 1

令F(x,y)=ln根号(x^2+y^2)-arctan(y/x)
对F(x,y)求x的偏导数Fx=(x + y)/(x^2 + y^2)
对F(x,y)求y的偏导数Fy=-(x - y)/(x^2 + y^2)
根据隐函数求导公式可得：
y'=-Fx/Fy
=(x + y)/(x - y)

Answer 2

(1/2)ln(x^2+y^2)=arcy/x
同时求导数
[2x+2y*y']/[2(x^2+y^2)]=(y'x-y)/[x^2(1+y^2/x^2)]
化简2x^3(1+y^2/x^2)+2yy'(x^2(1+y^2/x^2)=2(x^2+y^2)(y'x-y)
2x^3+2xy^2+2yy'x^2+2y^3y'=2x^2y'x+2y^2y'x-2x^2y-y^3