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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

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

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

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$-xy^3$ 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
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2 Answers 2

up vote 2 down vote accepted

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

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

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but dont we use quotient rule there with chain? why isnt x in that (3y2dydx). –  soniccool Oct 25 '11 at 0:28
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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
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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.

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