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Assume that $y$ is a function of $x$. Find $y' = \dfrac{dy}{dx}$ for $(x-y)^2 = x + y - 1$.

I've worked out the problem multiple times, but I continue to get a different answer than the correct answer. First I multiply out the $(x-y)^2$ to $(x-y)(x-y)$ and work it out from there; after that I take the derivative of both sides, etc.

This is supposed to be the correct answer:

$y' = \frac{2y-2x+1}{2y-2x-1}$

But I keep getting:

$y' = \frac{2x-2y-1}{2x-2y+1}$

Help please? Thank you!

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    The multiplication that @coffeemath made isn't to multiply each **side** by (-1), rather to multiply the **numerator** and the **denominator** of the fraction by (-1).2012-11-28

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This question was answered in the comments; this CW answer intends to remove the question from the Unanswered queue.


As coffeemath remarked:

Your answer is the same as what you're "supposed to get". Multiply top and bottom each by $(-1)$.

Cheers!

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    (Apologies to the one that upvoted for being mislead by the italicized remark, while I forgot to mark the answer CW.)2013-05-24