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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!

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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.

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  • $\begingroup$ 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. $\endgroup$ – user3002473 Aug 15 '14 at 16:25
  • $\begingroup$ Using the equation in your posted comment, then yes, you'd have gotten what you posted as an answer. $\endgroup$ – Namaste Aug 15 '14 at 16:27
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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*}

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

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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.

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  • $\begingroup$ Oops, that was just a copying error on my part. Can you elaborate on why my solution is incorrect? $\endgroup$ – user3002473 Aug 15 '14 at 16:22
  • $\begingroup$ 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. $\endgroup$ – dylan7 Aug 15 '14 at 16:28
  • $\begingroup$ 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. $\endgroup$ – user3002473 Aug 15 '14 at 16:31
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    $\begingroup$ Oh wow I didn't even catch frogeyedpeas did that...my bad. Yeah frogeyedpeas did make a mistake even in the original post. $\endgroup$ – dylan7 Aug 15 '14 at 16:33

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