http://i43.tinypic.com/261dxqe.jpg

I explained here^...this is just a practice quiz but I'm not sure if I have the right answer and they messed up the choices, OR if mine is just wrong.

Thanks!

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- Apr 22nd 2009, 01:05 PMjanedoeMult. choice implicit differentiation
http://i43.tinypic.com/261dxqe.jpg

I explained here^...this is just a practice quiz but I'm not sure if I have the right answer and they messed up the choices, OR if mine is just wrong.

Thanks! - Apr 22nd 2009, 01:12 PMderfleurer
I'm getting choice d.

Convert it to xy^2 -xy + x - y = 0

Take -(d/dx)/(d/dy)

d/dx = y^2 - y + 1

d/dy = 2xy - x - 1 - Apr 22nd 2009, 01:20 PMjanedoe
I'm not sure what you did but never mind I got it--I forgot to make something negative. D is right.

Thanks anyway :] - Apr 22nd 2009, 01:33 PMderfleurer
Ok, move the y to the other side of the equals sign so that you get this:

$\displaystyle xy^2 -xy + x - y = 0$

Now, implicitly derive (i'll group each piece)

$\displaystyle (y^2 + 2xyy') - (y + xy') + 1 - y' = 0$

Now you move y^2 - y + 1 to the other side of the equals sign

$\displaystyle 2xyy' + xy' - y' = -(y^2 - y + 1)$

Now factor out y'

$\displaystyle y'(2xy + x - 1) = -(y^2 - y + 1)$

And finally divide by (2xy + x - 1)

If you noticed, what you essentially did was move d/dx to the other side of the equals sign and divide d/dy.

Hence -(d/dx)/(d/dy) being the shortcut to implicitely deriving.