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Math Help - Derivative help

  1. #1
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    Derivative help

    differentiate e^x^ln(x)

    u = x^ln(x)
    derivative e^u= e^u

    Now, how do i take derivative of x^ln(x) where a function is the power of another function?
    Last edited by mr fantastic; August 13th 2009 at 03:28 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by justin016 View Post
    sorry wrong title, i meant derivative

    differentiate e^x^ln(x)

    u = x^ln(x)
    derivative e^u= e^u

    Lets assume you mean: differentiate f(x)=e^{x^{\ln(x)}} with respect to x.

    OK so you put u=x^{\ln(x)}, then:

     <br />
\frac{df}{dx}=\left(\frac{d}{du}e^u\right) \frac{du}{dx}=e^u \frac{d}{dx}(x^{\ln(x)})<br />

    So now we see what you are asking, so try

    x^{\ln(x)}=(e^{\ln(x)})^{\ln(x)}=e^{(\ln(x))^2}

    CB
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  3. #3
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    Quote Originally Posted by justin016 View Post
    sorry wrong title, i meant derivative

    differentiate e^x^ln(x)

    u = x^ln(x)
    derivative e^u= e^u

    Now, how do i take derivative of x^ln(x) where a function is the power of another function?
    \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}

    if u = x^{\ln{x}}

    \ln{u} = \ln\left(x^{\ln{x}}\right)

    \ln{u} = (\ln{x})^2

    \frac{1}{u} \cdot \frac{du}{dx} = 2\ln{x} \cdot \frac{1}{x}

    \frac{du}{dx} = \frac{2u \cdot \ln{x}}{x} = \frac{2x^{\ln{x}} \cdot \ln{x}}{x}<br />

    so ...

    \frac{d}{dx} e^{x^{\ln{x}}} = e^{x^{\ln{x}}} \cdot \frac{2x^{\ln{x}} \cdot \ln{x}}{x}
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