Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the problem $2x^3 + x^2y-xy^3 = 2$ and i am supposed to implicity differentiate the problem but i am getting lost at $-xy^3$ in theproblem and it got me stuck. How do i work this problem out i got the idea down but its just murdering me in that little part, do i use the quotient rule? Or do i combine quotient with chain? And how does that work? maybe im just missing it.

So far i got this

$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$

$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \text{Here i am lost}) = 0$$

share|cite|improve this question
$-xy^3$ is just the product of $-x$ and $y^3$. You applied the product rule correctly in the middle, I don't see why you don't think you can just do it again (because you can and should). – anon Oct 25 '11 at 0:25
up vote 2 down vote accepted

Remember that $y$ is implicitly a function of $x$, so there is a chain rule:

$$ \frac{d}{dx}(xy^3)=y^3+x\left(3y^2\frac{dy}{dx}\right). $$

share|cite|improve this answer
but dont we use quotient rule there with chain? why isnt x in that (3y2dydx). – soniccool Oct 25 '11 at 0:28
Quotient rule? You use the quotient rule when there's a quotient to use it on. Where do you see a quotient? – Gerry Myerson Oct 25 '11 at 0:39

Getting the derivative of that third term is done in the same was as doing the second term. There is a product rule involved. The only one other minor complication: when you find the derivative of $y^3$, there is a chain rule involved (because $y$ is a function of $x$).

Try it out and post your attempt so we can help you further.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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