I am familiar with the equation, but I am having trouble working out the entire answer.

"Differentiate using the Product Rule":

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

Thanks!

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- Sep 30th 2009, 09:28 PMcynlixDerivative 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)=$\displaystyle \sqrt[3]{t}(t^2+4)$

Thanks! - Sep 30th 2009, 09:48 PMcynlix
This is what I have so far...

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

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

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, 10:00 PMmr fantastic
- Sep 30th 2009, 11:52 PMcreatively12
if your answer is (7/2)(t^5/2)+6t^1/2

then I have the answer - Oct 1st 2009, 12:11 AMrobertocusato
h(t)=http://www.mathhelpforum.com/math-he...307aa591-1.gif

so theproduct rule = (t^1/3)(2t) + (t^2+4)(1/3)t^(-2/3)

= 2t^(4/3) + 1/3t^(4/3) + 4/3t^(-2/3)

=7/3t^(4/3) + 4/3t^(-2/3)

hope this helps!