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Math Help - Derivatives (checking work)

  1. #1
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    Jan 2008
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    Derivatives (checking work)

    Here is the equation and how I worked it out. Am I correct?

    y=(3x+4)^\frac{1}{3} what is y'

    Step 1: Chain rule?

    y=(3) \frac{1}{3}(3x+4)^\frac{-1}{3}

    Step 2: Simplify

    (x+1)^\frac{-1}{3}. Since I believe the 3 will take care of the 1/3 multiple.

    so ...

    y'=(x+1)^\frac{-1}{3}?
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  2. #2
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    Sep 2007
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    Nearly. Look closely at the power of the function.

    It should be  y' = (3x+4)^{-\frac{2}{3}}

    Remember you're taking 1 off the power, and \frac{1}{3} -1 = -\frac{2}{3}
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