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I tried to solve it my answer is $$\frac{-2x^{4/3}(x^{3}+1)^{2/3}+1}{3x(x^3+1)^2}$$ I just want to make sure if I derived it correctly thanks

Mickey
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    you can always check using wolfram alpha http://www.wolframalpha.com/input/?i=D%5Bx%5E%281%2F3%29%2F%28%28x%5E3%2B1%29%5E%281%2F3%29%29%2Cx%5D – Ragnar Jul 28 '14 at 19:34

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Sometimes taking $\ln$ of both sides of the $y=f(x)$, makes the function easier to differentiate. Let $$y=\frac{x^{1/3}} {({x^3+1})^{1/3}}$$ Then $$\ln y=\frac{1}{3}\ln x-\frac{1}{3} \ln ({x^3+1})$$ Therefore $$y'=\frac{y}{3}(\frac{1}{x}-\frac{3x^2}{x^3+1})$$ or $$y'=\frac{1}{3}\cdot \frac{x^{1/3}} {({x^3+1})^{1/3}}(\frac{1}{x}-\frac{3x^2}{x^3+1})$$ simplify this to get a more pretty expression!

Fermat
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A neat way to do this would be to rewrite as $$ \left(\frac{x}{x^3+1}\right)^{1/3} = u^{1/3} $$ then $$ \frac{d}{dx}\left(\frac{x}{x^3+1}\right)^{1/3} = \frac{d}{dx}u^{1/3} = \frac{1}{3}u^{1/3-1}\frac{du}{dx} $$ now $$ u' = \frac{d}{dx}\frac{x}{x^3+1} = \frac{1}{x^3+1} -\frac{3x^3}{\left(x^3+1\right)^2} = \frac{1-2x^3}{\left(x^3+1\right)^2} $$ putting it all together $$ \frac{d}{dx}\left(\left(\frac{x}{x^3+1}\right)^{1/3}\right) = \frac{1}{3}\left(\frac{x}{x^3+1}\right)^{-2/3}\frac{1-2x^3}{\left(x^3+1\right)^2} $$ This derivation is just to show you the result, to compare with your answer and see for yourself. :)

Chinny84
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  • I think its slightly nicer to do $u^{-1/3}$ instead, since then the derivative of $u$ is even more trivial. – Semiclassical Jul 28 '14 at 19:47
  • I think you are right. You fancy updating the post :)? or you can do that as an answer and I will remove this one. – Chinny84 Jul 28 '14 at 19:49
  • Heh, I'm too lazy to do that. Let's just say: "We leave this easy exercise to the reader." :) – Semiclassical Jul 28 '14 at 19:53
  • @chinny84 :/ I didn't understand how you get 1-2x^3/ (x^3+1)^2 :( – Mickey Jul 28 '14 at 20:13
  • In the form used in this answer, it comes from putting both terms over a common denominator. You will get the same result using the standard form of the quotient rule. – colormegone Jul 28 '14 at 23:05
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This is a variation of Chinny84's derivation, using implicit differentiation. If we write $ \ u^3 \ = \ \frac{x}{x^3 \ + \ 1} \ $ , then differentiating both sides with respect to $ \ x \ $ produces

$$ \ \frac{d}{dx} [u^3] \ = \ \frac{d}{dx} \ [ \ \frac{x}{x^3 \ + \ 1} \ ] \ \ \Rightarrow \ \ 3u^2 \ \frac{du}{dx} \ = \ \frac{(x^3 + 1) \cdot x ' \ - \ x \ (x^3 + 1)'}{(x^3 + 1)^2}$$

$$ \Rightarrow \ \ 3u^2 \ \frac{du}{dx} \ = \ \frac{(x^3 + 1) \cdot 1 \ - \ x \cdot 3x^2 }{(x^3 + 1)^2} \ = \ \frac{1 \ - \ 2x^3 }{(x^3 + 1)^2}$$

$$ \Rightarrow \ \frac{du}{dx} \ = \frac{1}{3 \ u^2} \ \cdot \ \frac{1 \ - \ 2x^3 }{(x^3 + 1)^2} \ = \ \frac{(1-2x^3) \ (x^3 + 1)^{2/3} }{3 \ x^{2/3} \ (x^3 + 1)^2} \ \ . $$

I suspect you have a factoring or distribution error somewhere when you applied the Quotient Rule.

colormegone
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  • $x^{1/3}$ also the denominator is raised to 1/3 anyways thank for the effort I appreciate you guys! :) – Mickey Jul 29 '14 at 11:11
  • I have cubed the entire function at the start to eliminate the fractional exponents. They return at the end because the factor in the denominator of the derivative is $ \ u^2 \ $ . The result I have for the derivative of your function is then identical to what Chinny84 shows and what WolframAlpha displays (as suggested by Ragnar). – colormegone Jul 29 '14 at 16:16