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Math Help - Trouble taking the derivative. Hung up on the algebra

  1. #1
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    Trouble taking the derivative. Hung up on the algebra

    I don't know if I should post this here or on the algebra forum.
    Trying to find the derivative of f (X) = √t (t + 4)
    Let √t be a
    Let (t + 4) be b
    f(X)' = a'b +b'a
    a' =⅓t٨-⅔
    b' = 2t
    thus f(x)' =(⅓t٨-⅔)(t + 4) + 2t √t
    or (t+4/3t٨⅔) +2t*t٨⅓
    I do not understand how to get from here to the answer (7t+4)/(3t٨⅔)

    Thanks for any help or advice.
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  2. #2
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    Re: Trouble taking the derivative. Hung up on the algebra

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

    distribute ...

    t^{\frac{7}{3}} + 4t^{\frac{1}{3}}

    take the derivative w/r to t using the power rule ...

    \frac{7}{3} t^{\frac{4}{3}} + \frac{4}{3} t^{-\frac{2}{3}}

    \frac{7t^{\frac{4}{3}}}{3} + \frac{4}{3t^{\frac{2}{3}}}

    common denominator is 3t^{\frac{2}{3}} ...

    \frac{7t^{\frac{4}{3}} \cdot t^{\frac{2}{3}}}{3t^{\frac{2}{3}}} + \frac{4}{3t^{\frac{2}{3}}}

    \frac{7t^2+4}{3t^{\frac{2}{3}}}
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  3. #3
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    Re: Trouble taking the derivative. Hung up on the algebra

    Thank you for the help. Evidently, the fundamental problem is that I do not understand dealing with these fractional powers. I have searched it on the web and gained some knowledge, but why t⅓ X t is t٨7/3 alludes me. Can you explain or direct me to an explanation? Also, where do you find the fonts to write out this math so elegantly? (At 71, I am finding it a struggle to re-tool in math, so I can go on an learn some calculus!)
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  4. #4
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    Re: Trouble taking the derivative. Hung up on the algebra

    you should already know the fundamental rule that t^a \cdot t^b = t^{a+b}

    t^{\frac{1}{3}} \cdot t^2 = t^{\frac{1}{3}} \cdot t^{\frac{6}{3}} = t^{\frac{1}{3} + \frac{6}{3}} = t^{\frac{7}{3}}
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