# Finding dy/dx by implicit differentiation

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.

• 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! – amWhy Sep 28 '16 at 20:43

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.
• @Maggie To repeat this part of the answer: You forgot a coefficient on the y term. – dxiv Sep 28 '16 at 22:08
• The $2x-1$ in the denomintor should be a $2x-y$. It looks like you just lost track of the $y$ – ASKASK Sep 28 '16 at 22:08