Answer to Question #252576 in Calculus for jose

Question

f(x,y)=arctan(2x/(x^2+y^2))

Answer 1

This question is incomplete but we can find $f_x$ here

Given: $f(x,y)=arctan(2x/(x^2+y^2))$

$f_x=\frac{\partial }{\partial x}[arctan(2x/(x^2+y^2))] \\=\frac{1}{1+(2x/(x^2+y^2))^2}. \frac{\partial }{\partial x}[(2x/(x^2+y^2))] \\=\frac{(x^2+y^2)^2}{(x^2+y^2)^2+4x^2}.\frac{(x^2+y^2).\frac{\partial }{\partial x}(2x)-2x.\frac{\partial }{\partial x}(x^2+y^2)}{(x^2+y^2)^2} \\=\frac{2x^2+2y^2-4x^2}{(x^2+y^2)^2+4x^2} \\=\frac{2y^2-2x^2}{(x^2+y^2)^2+4x^2}$