Trying to find dy/dx for x^2/(x-y^2)=5
Method 1: x^2 = 5x - 5y^2
2x = 5 - 10y(dy/dx)
dy/dx = (5-2x)/10y
Method 2: Quotient rule
2x(x-y^2) - x^2(1-2y(dy/dx)) all over (x-y^2)^2 =0
2x(x-y^2) = x^2(1-2y(dy/dx))
2(x-y^2)/x = 1 - 2y (dy/dx)
2y(dy/dx) = 1 - 2(x-y^2)/x = 2(y^2-x)/x + 1 = (2y^2 - x)/x
dy/dx = (2y^2 - x)/2xy
Two different answers, not algebraically equivalent. Where have I gone wrong? Which one is right?