Find the first partial derivatives of f(x,y) = [(2x − 2y)/(2x +2y)] at the point (x,y) = (3, 1).

[(∂f)/(∂x)](3, 1) = ???

[(∂f)/(∂y)](3, 1) = ???

Question

Find the first partial derivatives of f(x,y) = [(2x − 2y)/(2x +2y)] at the point (x,y) = (3, 1).

[(∂f)/(∂x)](3, 1) = ???

[(∂f)/(∂y)](3, 1) = ???

turingtest · Answer

w/respect to y:
[-2(2x+2y)-2(2x-2y)]/(2x+2y)^2
=-4x/(2x+2y)^2=-x/(x+y)^2
evaluated at (3,1)
-3/16

turingtest · Answer

partial w/respect to x: 
[2(2x+2y)-2(2x-2y )]/(2x+2y)^2
=4y/(2x+2y)^2=y/(x+y)^2
evaluated at (3,1)=1/16

turingtest · Answer

sorry i forgot to square the bottom the first time. my bad...