Hello all.

I'm starting calculus and I'm stumped by this exercise. I've arrived at multiple answers only to realise that, according to my text, I'm lost.

Find the derivative of:

$\displaystyle \frac{x^3}{\sqrt[3]{3x^2-1}}$

Here is my working:

Change fractions to exponents.

$\displaystyle x^3(3x^2-1)^\frac{-1}{3}$

Apply the product rule and then the chain rule for the derivative of the bracketed (3x...) term.

$\displaystyle 3x^2(3x^2-1)^\frac{-1}{3} + x^3(-2x(3x^2-1)^\frac{-4}{3})$

Expand.

$\displaystyle \frac{3x^2}{(3x^2-1)^\frac{1}{3}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}$

At that point it just doesn't seem right. I'm way off track according to the book's answer but I'm just not sure where I'm going wrong.

Any help would be much appreciated.