1. Q about using the product rule with implicit differentiation

 ignore the bit about the product rule in the title 

Hi everyone, this is my first time posting here I hope you guys can help me.
I am trying to get the derivative of x^2 - xy + y^2 = 3

After the first step I got:
2x - y - x(dy/dx) + 2y(dy/dx) = 0

When I get to the step where I would move all like terms to one side, I could either move all non(dy/dx) terms to the right hand side and have the equation
-x(dy/dx) + 2y(dy/dx) = y - 2x

OR if I were to move the (dy/dx) terms to the right hand side I would get:
2x - y = x(dy/dx) - 2y(dy/dx)

The first route will give me the derivative (y - 2x)/(2y - x)
The second route will give me derivative (2x - y)/(x - 2y)

It seems like I attacked both possible routes correctly, but I'm pretty sure one of the final derivatives is incorrect because (y - 2x)/(2y - x) does not equal (2x - y)/(x - 2y).... or do they?

feedback would be much appreciated!!

Originally Posted by robo_robb
...but I'm pretty sure one of the final derivatives is incorrect because (y - 2x)/(2y - x) does not equal (2x - y)/(x - 2y).... or do they?
They are equal. Factor a -1 out of the numerator and denominator:

$\frac{y-2x}{2y-x}$

$=\frac{-(-y+2x)}{-(-2y+x)}$

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

$=\frac{2x-y}{x-2y}.$

It is helpful to remember that, in general, $-(a-b)=(b-a).$

3. Ohh, I see now. Thanks a bunch!

4. Originally Posted by Reckoner
It is helpful to remember that, in general, $-(a-b)=(b-a).$
And it is also helpful to remember that, if this fact was necessary in the homework, it will almost certainly be necessary on the next test!