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Thread: 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?

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

    Step 1: Chain rule?

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

    Step 2: Simplify

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

    so ...

    $\displaystyle y'=(x+1)^\frac{-1}{3}$?
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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 $\displaystyle y' = (3x+4)^{-\frac{2}{3}} $

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