Originally Posted by
skeeter $\displaystyle \frac{d}{dx} \frac{x^3-2x}{\sqrt{2x+1}} = \frac{\sqrt{2x+1} \cdot (3x^2-2) - (x^3-2x) \cdot \frac{1}{\sqrt{2x+1}}}{2x+1}$
multiply by $\displaystyle \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$ to clear the inner fraction ...
$\displaystyle \frac{(2x+1)(3x^2-2) - (x^3-2x)}{(2x+1)^{\frac{3}{2}}}
$
$\displaystyle \frac{6x^3+3x^2-4x-2 - x^3 + 2x}{(2x+1)^{\frac{3}{2}}}$
$\displaystyle \frac{5x^3 + 3x^2 - 2x - 2}{(2x+1)^{\frac{3}{2}}}$