# Confusion about implicit differentiation.

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!

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.

• Yeah I mixed up a lot of the coefficients. I'm used to $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ as opposed to $Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$, so when I tried to convert I must've messed up somewhere. – user3002473 Aug 15 '14 at 16:25
• Using the equation in your posted comment, then yes, you'd have gotten what you posted as an answer. – Namaste Aug 15 '14 at 16:27

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*}

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

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.

• Oops, that was just a copying error on my part. Can you elaborate on why my solution is incorrect? – user3002473 Aug 15 '14 at 16:22
• Actually most of the answer you copied was incorrect, so I would look back at the link. I actually do not know how you arrived at your answer. Some of the terms in your answer look like them came out of nowhere. For example: there is no $Bx$ term at all. Just a $2By$ term on the bottom. I would look back at the link you posted for clarification. If that doesn't help, I'll explain more. – dylan7 Aug 15 '14 at 16:28
• I got confused about the difference between the equations me and frogeyed peas used (which switched the $B$ and $C$ coefficients), and when converting the coefficients I messed up somewhere. The relevant part of this question is the confusion about the use of the product rule. – user3002473 Aug 15 '14 at 16:31
• Oh wow I didn't even catch frogeyedpeas did that...my bad. Yeah frogeyedpeas did make a mistake even in the original post. – dylan7 Aug 15 '14 at 16:33