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Math Help - Derivative using the product rule

  1. #1
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    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!
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by cynlix View Post
    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.
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  4. #4
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    if your answer is (7/2)(t^5/2)+6t^1/2

    then I have the answer
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  5. #5
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    h(t)=

    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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