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Thread: Tricky derivative

  1. #1
    Senior Member DivideBy0's Avatar
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    Tricky derivative

    I need help to do:

    $\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}}$

    The answer says
    $\displaystyle \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}$, but I don't think that is correct

    Thanks
    Last edited by DivideBy0; Aug 12th 2007 at 12:36 AM.
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  2. #2
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    Hello, DivideBy0!

    I assume we are to differentiate the fuction . . .


    $\displaystyle y\:=\:\sqrt[x]{\frac{1-x}{1+x}}$
    The answer says: .$\displaystyle \frac{1-x-x^2}{(1+x) (1-x^2)^{\frac{1}{2}}}$ .but I don't think that is correct
    Please check the problem for typos . . . Something is terribly wrong!

    If that is really an $\displaystyle x^{th}$ root, the answer cannot have a square root in it.

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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    I need help to do:

    $\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}}$

    The answer says
    $\displaystyle \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}$, but I don't think that is correct

    Thanks
    You are correct. It isn't right.

    The derivative of $\displaystyle a^x$ is $\displaystyle ln(a) \cdot a^x$, so using the chain rule on
    $\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}} = \left ( \frac{1 - x}{1 + x} \right ) ^{1/x}$
    we obtain:
    $\displaystyle y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left [ \frac{1 - x}{1 + x} \right ]^{\prime}$

    $\displaystyle y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left ( \frac{-2}{(1 + x)^2} \right )$

    which I would simply leave as:
    $\displaystyle y^{\prime} = \left ( \frac{2}{x^2(x + 1)^2} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot ln \left ( \frac{1 - x}{1 + x} \right )$

    -Dan
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