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Math Help - How to take the derivative of ln(lnx)?

  1. #1
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    How to take the derivative of ln(lnx)?

    How do you take the derivative of ln(lnx)?

    This is where I go with it then I get confuse:

    y=ln(lnx)
    e^y=lnx.

    From here I get lost. Everything I try does not work.
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  2. #2
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    Re: How to take the derivative of ln(lnx)?

    \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}

    now, let u = \ln{x}
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  3. #3
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    Re: How to take the derivative of ln(lnx)?

    thanks I got it. It is (x ln(x))^-1. However, I do not understand why 1 over u?
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  4. #4
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    Re: How to take the derivative of ln(lnx)?

    Quote Originally Posted by Barthayn View Post
    How do you take the derivative of ln(lnx)?

    This is where I go with it then I get confuse:

    y=ln(lnx)
    e^y=lnx.

    From here I get lost. Everything I try does not work.
    Alternatively, you can just do implicit differentiation from there.

    e^y\frac{dy}{dx}=\frac{1}{x}

    \frac{dy}{dx}=\frac{1}{e^yx}

    \frac{dy}{dx}=\frac{1}{e^{ln(ln(x))}x}

    And you can see how this simplifies.
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