Confusion about implicit differentiation.

Question

I want to implicitly differentiate $Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$. This is not an exceedingly difficult task, and when I solved it I got

$$ y' = -\frac{2Ax + Cy + D}{2By + Cx + E} $$ But my confusion comes from the fact that in this answer by frogeyedpeas, he says it is equal to $$ -\frac{2Ax + D}{2By + Cx + E}. $$ The confusion comes from the $Cxy$ term. The product rule says that $$\frac{d}{dx}Cxy = C(\frac{d}{dx}x\cdot y + x \cdot \frac{d}{dx}y) = C(y + xy'),$$ and wolfram alpha can verify this (just input $xy = 1$). Did frogeyedpeas accidentally make a mistake, or is there something I'm missing that makes this scenario different?

EDIT: Finally it's all correct, I copied correctly and the coefficients are fixed. Thanks to everyone who pointed out the errors!

Answer 1

You are correct that there is an error in the post you link to, but there's also an error in your result.

You wrote: $$y' = -\frac{2Ax + By + D}{2Bx + Cy + E}.$$

In fact,

$$y' = -\frac{2Ax +Cy +D}{2By+Cx+E}$$

Somehow you mixed up which coefficient is located where.

Answer 2

That's not the same answer frogeyedpeas gave from the link you posted. There wasn't a $2Cy $. There was just a $ Cy$. And yes frogeyedpeas is correct. In addition therr are other wromg terms you posted. Go back to ur link and look at the answer again.

Answer 3

frogeyedpeas did indeed forget to apply product rule. Implicitly differentiating, we get: \begin{align*} 2Ax + 2Byy' + (Cxy' + Cy) + D + Ey' &= 0 \\ (2By + Cx + E)y' &= -2Ax - Cy - D \\ y' &= \frac{-2Ax - Cy - D}{2By + Cx + E} \end{align*}

Answer 4

$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$ $$2Ax + 2Byy' + Cy+Cxy' + D + Ey'= 0$$ $$y'(2By + Cx+E)=-2Ax-Cy-D$$ $$y'=\frac{-2Ax-Cy-D}{2By + Cx+E}$$