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I've an equation that I like to use for implicit differentiation. The equation is:$x^2 = \frac{(x+2y)}{(x-2y)}$

I used two different methods but got two different answers for same equation.

Can anyone kindly tell me where I am wrong? Why am I getting two different answers?

The first method I used is:

$$ \begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{ [after simplification] } \end{align*} $$

Then I used the same equation but multiplied both sides by $(x-2y)$. So here's the second procedure:

$$ \begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ x^2 (x-2y) =& x + 2y \\ x^3 - 2x^2y - x - 2y =& 0 \\ 3x^2 - 2x^2 \frac{dy}{dx} - 4xy - 1 - 2 \frac{dy}{dx} =& 0 \\ -2x^2 \frac{dy}{dx} - 2 \frac{dy}{dx} =& -3x^2 + 4xy + 1 \\ \frac{dy}{dx}(2x^2 + 2) =& 3x^2 - 4xy - 1 \\ \frac{dy}{dx} =& \frac{3x^2 - 4xy - 1}{2x^2 + 2}\end{align*} $$

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1 Answer 1

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They may not be different. Remember, there's an equation relating $x$ and $y$. If you use that equation, you might be able to get one of the answers to look like the other one.

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    $\begingroup$ That is, take the difference between your two expressions for $dy/dx$, put it over a common denominator, and the numerator will be some polynomial in $x$ and $y$ times $x^2(x-2y) - (x+2y)$. $\endgroup$
    – Robert Israel
    Sep 5, 2012 at 1:53

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