Answer to Question #339592 in Calculus for Partial Derivative

Question

Find the first partial derivative of

f(x,y) = $(4x-y)/(4x+y)$ at the point (4,2) for both x and y.

Answer 1

$f(x,y)=(4x-y)/(4x+y)$

$f_x=\frac{(4x-y)_x'(4x+y)-(4x+y)_x'(4x-y)}{(4x+y)^2}=\frac{4(4x+y)-4(4x-y)}{(4x+y)^2}=\frac{8y}{(4x+y)^2}$

At the point (4,2)

$f_x=\frac{8(2)}{(4(4)+2)^2}=16/324=4/81$

$f_y=\frac{(4x-y)_y'(4x+y)-(4x+y)_y'(4x-y)}{(4x+y)^2}=\frac{-1(4x+y)-1(4x-y)}{(4x+y)^2}=\frac{-8x}{(4x+y)^2}$

At the point (4,2)

$f_y=\frac{-8(4)}{(4(4)+2)^2}=-32/324=-8/81$