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Find dy/dx by implicit differentiation

x^2-4xy+y^2=4

I know to take the derivatives of both sides, which would be:

d/dx[x^2-4xy+y^2]=0

I'm not sure if I did it right, but I then got:

2x-4*(xdy/dx)+y+2y(dy/dx)=0

I don't know where to go from here, or even if the previous step is correct. Please help!

Edit: I have followed the advice given and I ended up with:

(x-2y)/(2x-1)

However this was incorrect. Someone please tell me what I am missing here.

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  • $\begingroup$ If you simply expand the parentheses following $-4$ to include $+y$, so that you have $$2x-4\left(x\frac{dy}{dx} +y\right)+ 2y\frac{dy}{dx}=2x-4x\frac{dy}{dx} -4y+ 2y\frac{dy}{dx} $$ then all is good! $\endgroup$ – amWhy Sep 28 '16 at 20:43
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You are close! You forgot a coefficient on the $y$ term. You should have $$ 2x-4y-4x\frac{dy}{dx}+2y\frac{dy}{dx}=0 $$ Now you can solve for $\frac{dy}{dx}$ like any other variable.

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  • $\begingroup$ I did that and got (2x-4y)/(4x-2) is this correct? I don't think you can reduce that. $\endgroup$ – Maggie Sep 28 '16 at 20:52
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    $\begingroup$ Well, "2" can be factored out of both numerator and denominator so it certainly can be reduced. $\endgroup$ – user247327 Sep 28 '16 at 21:37
  • $\begingroup$ (x-2y)/(2x-1) was incorrect. What am I missing here? $\endgroup$ – Maggie Sep 28 '16 at 21:47
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    $\begingroup$ @Maggie To repeat this part of the answer: You forgot a coefficient on the y term. $\endgroup$ – dxiv Sep 28 '16 at 22:08
  • $\begingroup$ The $2x-1$ in the denomintor should be a $2x-y$. It looks like you just lost track of the $y$ $\endgroup$ – ASKASK Sep 28 '16 at 22:08

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