# Implicit differentiation (2 different answers!)

• Feb 2nd 2008, 12:00 PM
pingleby
Implicit differentiation (2 different answers!)
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?
• Feb 2nd 2008, 12:35 PM
Opalg
Quote:

Originally Posted by pingleby
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?

These answers are algebraically equivalent! Use the initial equation from Method 1 (x^2 = 5x - 5y^2) to make the substitution y^2 = x(1–(x/5)) in the second answer, and see it reduce to the first answer.
• Feb 2nd 2008, 12:37 PM
TrevorP
^^ Like he said ^^

Here's how I would do it:

$\frac{x^2}{x-y^2} = 5$

$x^2 = 5x - 5y^2$

$2x = 5 - 10y\tfrac{dy}{dx}$

$\tfrac{dy}{dx} = \frac{2x - 5}{-10y} = \boxed{\frac{5-2x}{10y}}$

Which is your method 1.