Find the first partial derivatives of f(x,y)=(4x−2y)/(4x+2y) at the point (x,y) = (1, 2).
∂f∂x(1,2)=
∂f∂y(1,2)=
∂f∂x(1,2)=
∂f∂y(1,2)=
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It is easier if altered first.
(4x - 2y)/(4x + 2y) = (2x - y)/(2x + y) = (2x + y - 2y)/(2x + y)
= 1 - 2y/(2x + y) = 1 - 2y*(2x + y)^-1
df/dx = -2y*-1*(2x + y)^-2 * 2 = 4y/(2x + y)^2. At (1, 2) df/dx = 8/(4^2) = 1/2
df/dy = -2*(2x + y)^-1 - 2y*-1*((2x + y)^-2 At (1, 2) df/dy = -2/4 + 4/(4^2) = -1/4
Rs90 has penultimate line correct but -16x/(4x + 2y)^2 = -16/(8^2) = -1/4 at (1, 2)
(4x - 2y)/(4x + 2y) = (2x - y)/(2x + y) = (2x + y - 2y)/(2x + y)
= 1 - 2y/(2x + y) = 1 - 2y*(2x + y)^-1
df/dx = -2y*-1*(2x + y)^-2 * 2 = 4y/(2x + y)^2. At (1, 2) df/dx = 8/(4^2) = 1/2
df/dy = -2*(2x + y)^-1 - 2y*-1*((2x + y)^-2 At (1, 2) df/dy = -2/4 + 4/(4^2) = -1/4
Rs90 has penultimate line correct but -16x/(4x + 2y)^2 = -16/(8^2) = -1/4 at (1, 2)
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∂f/∂x means u differentiate it w.r.t x treating all other variables (y in this case) as constant.Similarly for
∂f/∂y
so ∂f/∂x is [4(4x+2y)-4(4x-2y)] / (4x+2y)^2
= 16y/(4x+2y)^2
= 1/2 at 1,2
∂f/∂y is [-2(4x+2y)-2(4x-2y)] / (4x+2y)^2
= -16x/(4x+2y)^2
= -1/2 at 1,2
∂f/∂y
so ∂f/∂x is [4(4x+2y)-4(4x-2y)] / (4x+2y)^2
= 16y/(4x+2y)^2
= 1/2 at 1,2
∂f/∂y is [-2(4x+2y)-2(4x-2y)] / (4x+2y)^2
= -16x/(4x+2y)^2
= -1/2 at 1,2