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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Gather the $y'$ terms together to get $y'(6x-3y^2) = 3x^2-6y$.

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

  • $\begingroup$ 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

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 $y'$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.