Answer
${y' = \frac{2y−x}{y−2x}}$
Work Step by Step
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:
${y' = \frac{2y−x}{y−2x}}$
