# Implicit Differentiation???

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:

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

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Your answer is the same as what you're "supposed to get". Multiply top and bottom eacfh by (-1). –  coffeemath Nov 28 '12 at 3:31
So even though it's opposite, it's still correct? How would I know to multiply each side by (-1) if I didn't have the correct answer? –  Courtney Vanoy Nov 28 '12 at 3:33
Also, why are you multiplying out the squared term? You should be able to differentiate without multiplying it out. For example, what if there were thirty terms being squared? Try it without squaring out the quantity and see if you can get the same answer. –  Amzoti Nov 28 '12 at 3:36
Since they are the same, both are correct. There is really no reason to prefer one over the other. Some texts might decide on one version over the other in the answer key because it looks simpler, or has fewer characters to print. –  coffeemath Nov 28 '12 at 3:36
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). –  anorton Nov 28 '12 at 3:56
Your answer is the same as what you're "supposed to get". Multiply top and bottom each by $(-1)$.