Find the slope of the curve

Question

Find the slope of the curve (x^2+y^2)^2 = (x-y)^2 at the point (1, -1)

Answer 1

(x²+y²)² = (x-y)²

Need to find dy/dx=y' at (1,-1)

x⁴+2x²y²+y⁴ = x²-2xy+y²

Doing implicit derivation:

4x³+4x²yy'+4xy²+4y³y' = 2x-2xy'-2y+2yy'

y'(4x²y+4y³+2x-2y) = -4x³-4xy²+2x-2y

dy/dx = (-4x³-4xy²+2x-2y)/(4x²y+4y³+2x-2y)

at (1,-1)

dy/dx = (-4-4+2+2)/(-4-4+2+2) = -4/-4 = 1

--or--

Take square root of both sides first

x²+y²=x-y

2x+2yy'=1-y'

y'(2y+1)=1-2x

y'=(1-2x)/(2y+1)

at (1,-1)

dy/dx = (1-2)/(-2+1) = -1/-1 = 1