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Math Help - Derivative Of A Natural Logarithm

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    Derivative Of A Natural Logarithm

    I am given the function f(x) = ln(lnx^2) to differentiate; but I am not entirely certain how to do this. Could someone possibly help me?

    Thank you
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    Re: Derivative Of A Natural Logarithm

    Quote Originally Posted by Bashyboy View Post
    I am given the function f(x) = ln(lnx^2) to differentiate; but I am not entirely certain how to do this. Could someone possibly help me?
    Note that f(x)=\ln(2)+\ln(\ln(x)) so f'(x)=\frac{1}{x\ln(x)}~.
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    Re: Derivative Of A Natural Logarithm

    You need to know this: the derivative of f(x)= ln(u(x)) is [tex]\frac{1}{u}\frac{du}{dx}[/itex].

    As Plato said, ln(ln(x^2)= ln(2ln(x))= ln(ln(x))+ ln(2). "ln(2)" is a constant and its derivative is 0.

    so you have ln(u(x)) with u(x)= ln(x).
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