x^2+y^2=e^(arctan（y/x）),求dy/dx

题目

x^2+y^2=e^(arctan（y/x）),求dy/dx
令F(x,y)=x^2+y^2-e^arctan(y/x)=0
对x求偏导 Fx = 2x-e^arctan(y/x) * 1/[1+(y/x)^2] * [-y/(x^2)]=2x + y/(x^2+y^2) * e^arctan(y/x)
然后e^arctan(y/x)用题目中的x^2+y^2代替 就可以把
即Fx=2x+y
对y求偏导 Fy=2y-e^arctan(y/x)*1/[1+(y/x)^2]*(1/x)=2y-x/(x^2+y^2) * e^arctan(y/x)
同上把e^arctan(y/x)用题目中的x^2+y^2代替 得 Fy=2y-x
∴ dy/dx= - Fx/Fy= - (2x+y) / (2y-x) = (2x+y)/(x-2y)

答案