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Thread: Question x^x

  1. #1
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    Question x^x

    If $\displaystyle
    \frac{{df}}{{dx}}=x^x
    $ for x>0, and $\displaystyle
    g(x)=f(x^2)$ then $\displaystyle
    \frac{{dg}}{{dx}} =
    $ ?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chaddy View Post
    If $\displaystyle
    \frac{{df}}{{dx}}=x^x
    $ for x>0, and $\displaystyle
    g(x)=f(x^2)$ then $\displaystyle
    \frac{{dg}}{{dx}} =
    $ ?
    by the chain rule: $\displaystyle \frac {dg}{dx} = f'(x^2) \cdot 2x$

    now continue
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  3. #3
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    Okay, so $\displaystyle
    \frac {dg}{dx} = f'(x^2) \cdot 2x
    $ which means that $\displaystyle
    \frac {dg}{dx} = x^{2x^2} \cdot 2x
    $
    Thus $\displaystyle
    \frac {dg}{dx} = 2x^{2x^2+1}
    $
    Correct?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chaddy View Post
    Okay, so $\displaystyle
    \frac {dg}{dx} = f'(x^2) \cdot 2x
    $ which means that $\displaystyle
    \frac {dg}{dx} = x^{2x^2} \cdot 2x
    $
    Thus $\displaystyle
    \frac {dg}{dx} = 2x^{2x^2+1}
    $
    Correct?
    yup
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