## Calculus: Early Transcendentals 8th Edition

${y' = \frac{2y−x}{y−2x}}$
Original equation: ${x^2-4xy+y^2=4}$ Take the derivative of both sides: ${\frac{d}{dx}(x^2-4xy+y^2) = \frac{d}{dx} (4)}$ Take the derivative of each term: ${\frac{d}{dx}(x^2) -\frac{d}{dx}(4xy) + \frac{d}{dx}( y^2) =\frac{d}{dx} 4}$ Simplify using power and product rule: ${2x - 4[x(\frac{dy}{dx}) + 1(y)] + 2y(\frac{dy}{dx}) = 0}$ Simplify: ${2x - 4x(\frac{dy}{dx}) - 4y + 2y(\frac{dy}{dx}) = 0}$ Isolate all terms with ${\frac{dy}{dx}}$ on one side: ${- 4x(\frac{dy}{dx})+ 2y(\frac{dy}{dx}) = -2x + 4y}$ Factor out the ${\frac{dy}{dx}}$: ${\frac{dy}{dx}(-4x+2y) = -2x + 4y}$ Isolate ${\frac{dy}{dx}}$ to find the derivative: ${\frac{dy}{dx} = y' = \frac{-2x+4y}{-4x+2y}}$ Factor out the 2 from the numerator and denominator: ${y' = \frac{2(-x+2y)}{2(-2x+y)}}$ Simplify: ${y' = \frac{2y−x}{y−2x}}$