Derivative using the product rule

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September 30th 2009, 09:28 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!
September 30th 2009, 09:48 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.
September 30th 2009, 10:00 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.
September 30th 2009, 11:52 PM
creatively12

if your answer is (7/2)(t^5/2)+6t^1/2

then I have the answer
October 1st 2009, 12:11 AM
robertocusato

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!

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