# Thread: Derivative help

1. ## Derivative help

Pretty sure I got this right, my online math homework program says it's wrong though

Find the derivative of the function
$\displaystyle y=(x^2+2)(\sqrt[3]{x^2+3})$

After working through it using chain rule and product rule, I found

$\displaystyle \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}$

What did I do wrong?
Thanks!

2. without looking at your work, its not possible to tell where you went wrong.

Note: You have to use the product rule here.

$\displaystyle u=(x^2+2) \; and\; v = (x^2+3)^{\frac{1}{3}}$

3. Originally Posted by beanus
Pretty sure I got this right, my online math homework program says it's wrong though

Find the derivative of the function
$\displaystyle y=(x^2+2)(\sqrt[3]{x^2+3})$

After working through it using chain rule and product rule, I found

$\displaystyle \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}$

What did I do wrong?
Thanks!
wolfram alpha agrees with your result (it just has the numerator factored).

derivative of &#40;x&#94;2&#43;2&#41;&#42;&#40;x&#94;2&#43;3&#41 ;&#94;&#40;1&#47;3&#41; - Wolfram|Alpha

4. $\displaystyle (x^2+2)\frac{d}{dx} [x^2+3]^\frac{1}{3} + \frac{d}{dx} [x^2+2]$$\displaystyle (x^2+3)^\frac{1}{3}$

$\displaystyle y= (x^2+3)^\frac{1}{3}$
$\displaystyle u= (x^2+3)$

$\displaystyle \frac{dy}{du} = \frac{1}{3u^\frac{2}{3}}$
$\displaystyle \frac{du}{dx} = 2x$

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

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

$\displaystyle \frac{2x^3+4x}{3(x^2+3)^\frac{2}{3}} + \frac{6x^3+18x}{3(x^2+3)^\frac{2}{3}}$

$\displaystyle \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}$

5. Originally Posted by beanus

$\displaystyle \frac{8x^3+22x}{3(x^2+3)^\frac{2}{3}}$

that actually is the correct answer.