1. ## derivative

yea so my tutor screwed this one up and she did it wrong.. can someone show a step by step for this derivative please.

(x^3-2x)/(sqrt(2x+1))

2. Originally Posted by tbenne3
yea so my tutor screwed this one up and she did it wrong.. can someone show a step by step for this derivative please.

(x^3-2x)/(sqrt(2x+1))
$\displaystyle \frac{d}{dx} \frac{x^3-2x}{\sqrt{2x+1}} = \frac{\sqrt{2x+1} \cdot (3x^2-2) - (x^3-2x) \cdot \frac{1}{\sqrt{2x+1}}}{2x+1}$

multiply by $\displaystyle \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$ to clear the inner fraction ...

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

$\displaystyle \frac{6x^3+3x^2-4x-2 - x^3 + 2x}{(2x+1)^{\frac{3}{2}}}$

$\displaystyle \frac{5x^3 + 3x^2 - 2x - 2}{(2x+1)^{\frac{3}{2}}}$

3. thanks but i have a question.. did you chain rule the derivative of sqrt(2x+1)

4. Originally Posted by tbenne3
thanks but i have a question.. did you chain rule the derivative of sqrt(2x+1)
sure did ...

$\displaystyle \frac{d}{dx} (2x+1)^{\frac{1}{2}} = \frac{1}{2}(2x+1)^{-\frac{1}{2}} \cdot 2 = (2x+1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x+1}}$