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Thread: f(x)=x^2/3 (x^2-4)...simplifiy then get f'(x)

  1. March 26th 2009, 12:25 PM #1
    calcconfused
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    f(x)=x^2/3 (x^2-4)...simplifiy then get f'(x)

    i know it simplifies to x^8/3 - 4x^2/3 but how does one get to that (esp. the x^8/3)


    also, how do you get to f'(x) from here?



    Thank you in advance to those smarter than I...
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  2. March 26th 2009, 01:39 PM #2
    sinewave85
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    Quote Originally Posted by calcconfused View Post
    i know it simplifies to x^8/3 - 4x^2/3 but how does one get to that (esp. the x^8/3)


    also, how do you get to f'(x) from here?



    Thank you in advance to those smarter than I...
    Hi, calcconfused. To simplify the basic equation, remember that x^{a}x^{b} = x^{a + b} and the rules for adding fractions:

    f(x) = x^{\frac{2}{3}}(x^{2} - 4)

    f(x) = (x^{\frac{2}{3}})(x^{2}) - (x^{\frac{2}{3}})(4)

    f(x) = (x^{\frac{2}{3}})(x^{\frac{6}{3}}) - (x^{\frac{2}{3}})(4)

    f(x) = x^{\frac{8}{3}} - 4x^{\frac{2}{3}}

    And now the derivative:

    f^{\prime}(x) = \left(\frac{8}{3}\right)x^{(\frac{8}{3} - 1)} - \left(\frac{2}{3}\right)4x^{(\frac{2}{3} - 1)}

    f^{\prime}(x) = \frac{8}{3}x^{\frac{5}{3}} - \frac{8}{3}x^{-\frac{1}{3}}

    f^{\prime}(x) = \frac{8}{3}(x^{\frac{5}{3}} - x^{-\frac{1}{3}})

    Hope that helps!
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