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I'm given $x^3+y^3=6xy$. It's stated that $y$ is a function of $x$ and I'm tasked to differentiate with respect to $x$.

The implicit differentiation is: $$3x^2+3y^2y'=6xy'+6y$$

Now simplify and express in terms of $y'$. I'm going to number these steps.

  1. $x^2+y^2y'=2xy'+2y$
  2. $y'=\frac{2xy'+2y-x^2}{y^2}$
  3. $y'=\frac{2xy'}{y^2}+\frac{2y-x^2}{y^2}$
  4. $y'=y'*\frac{2x}{y^2}+\frac{2y-x^2}{y^2}$
  5. $\frac{y'}{y'}=\frac{2x}{y^2}+\frac{2y-x^2}{y^2}.$ We know that $\frac{y'}{y'}=1$ and therefore I've made a mistake in my algebra. But I'm not sure what is wrong.

Here is the correct simplification, taking it from step 1 again:

  1. $x^2+y^2y'=2xy'+2y$
  2. $y^2y'-2xy'=2y-x^2$
  3. $y'(y^2-2x)=2y-x^2$
  4. $y'=\frac{2y-x^2}{y^2-2x}$

This makes complete sense. What was wrong with my simplification? If it's not wrong, how can I arrive at the correct final expression of $y'$?

Source of problem: Stewart, James. Calculus Early Transcendentals. 7th Ed. 2012. Page 211.

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  • $\begingroup$ What error? I don't see anything wrong, there's no $y'/y'$ in you steps, so I don't understand step "5" $\endgroup$ – Simply Beautiful Art Dec 22 '16 at 17:16
  • $\begingroup$ Put $y'{2x\over y^2}$ in the LHS and you'll get the same result $\endgroup$ – MattG88 Dec 22 '16 at 17:19
  • $\begingroup$ There's no algebraic mistake. Unless you were thinking about dividing both sides by y', and only dividing one term, which would have been a huge no no. $\endgroup$ – Kaynex Dec 22 '16 at 17:23
  • $\begingroup$ @MattG88, When I put $y'\frac{2x}{y^2}$ in the LHS, and then divide both sides by $\frac{2x}{y^2}$ to isolate the $y'$, I'm left with $y'-y'$ in the LHS. So I'm not sure how that path is correct? $\endgroup$ – baverso Dec 22 '16 at 17:38
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    $\begingroup$ @baverso if you put $y'{2x\over y^2}$ in the LHS, you must collect $y'$ so: $y'(1-{2x\over y^2})={2y-x^2\over y^2}$. $\endgroup$ – MattG88 Dec 22 '16 at 17:45
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Gather the $y'$ terms together to get $y'(6x-3y^2) = 3x^2-6y$.

Then divide: $y' = {3x^2-6y \over 6x-3y^2}$.

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  • $\begingroup$ Agreed, however this is essentially the textbook answer which I provided. It definitely makes sense. However, I'm trying to learn from my mistake, or how to complete the algebra I had made. $\endgroup$ – baverso Dec 22 '16 at 17:40
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The correct answer to my algebraic simplification is this comment made here by MattG88. It starts from step 4 in my simplification.: Implicit differentiation for $y'$

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