# Derivative using the product rule

• Sep 30th 2009, 10:28 PM
cynlix
Derivative using the product rule
I am familiar with the equation, but I am having trouble working out the entire answer.

"Differentiate using the Product Rule":

h(t)= $\sqrt[3]{t}(t^2+4)$

Thanks!
• Sep 30th 2009, 10:48 PM
cynlix
This is what I have so far...

= $(\frac{1}{3t^\frac{-2}{3}})(t^2+4) + (t^\frac{1}{3}) (1)$

= $\frac{t^2+4}{3t^\frac{-2}{3}}+ (t^\frac{1}{3})$

I'm not too sure where to go after that.

I try to get the LCD, but I keep on coming up with the wrong answer.
• Sep 30th 2009, 11:00 PM
mr fantastic
Quote:

Originally Posted by cynlix
This is what I have so far...

= $(\frac{1}{3t^\frac{-2}{3}})(t^2+4) + (t^\frac{1}{3}) (1)$ Mr F says: This should either be ${\color{red}(\frac{1}{3t^\frac{2}{3}})(t^2+4) + (t^\frac{1}{3}) (2t)}$ or ${\color{red}(\frac{1}{3} t^\frac{-2}{3})(t^2+4) + (t^\frac{1}{3}) (2t)}$. Everything that followed will be wrong.

[snip]

Putting 1 for the derivative of $t^2 + 4$ was a particularly bad mistake.
• Oct 1st 2009, 12:52 AM
creatively12