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Math Help - derivative logarithm

  1. #1
    chg
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    derivative logarithm

    Hi guys, I'm stuck

    can someone please help me understand this:

    f(x) = e^(2x) log(x^2)

    we are given the answer which should be : f(x)' = ( e^(2x) / ln(10) ) * (ln(x^2) + 1)


    this is where I'm stuck (I'm not even sure this is right), after the product rule:

    f(x)' = 2e^(2x) * ( ln(x^2)) / ln(10) ) + ( (e^(2x) * 2x) / (x^2 ln(10)) )

    I am on the right track here? Thanks
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  2. #2
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    Re: derivative logarithm

    f(x) = e^{2x} \cdot \log{x^2}

    f(x) = e^{2x} \cdot 2\log{x}

    log is base 10, change base ...

    f(x) = e^{2x} \cdot \frac{2}{\ln(10)} \cdot \ln{x}

    f(x) = \frac{2}{\ln(10)}\left(e^{2x} \cdot \ln{x}\right)

    f'(x) = \frac{2}{\ln(10)}\left(e^{2x} \cdot \frac{1}{x} + 2e^{2x} \cdot \ln{x}\right)

    f'(x) = \frac{2e^{2x}}{\ln(10)}\left(\frac{1}{x} + 2\ln{x}\right)
    Thanks from chg
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  3. #3
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    Re: derivative logarithm

    Hi,
    my first question:
    What is the basis fo logaritm, 10 or e(Euler number) or what?
    Than is the derivate of f(x) with the product- und chain Regula to solve.

    When lg=log with basis 10 and ln=log with basis e, we have:
    lg(x)= lg(e)*ln(x)
    When log = log of basis b, we have
    log(x)= log(10)*lg(x)= log(e)*ln(x)

    For derivating transform log in ln.
    (e^2x)'= 2*e^2x
    ( ln(x))' = 1/x, where x >0
    Don't help it?
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  4. #4
    chg
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    Re: derivative logarithm

    Thank you very much kind sir! This really helps to finally see what I was doing wrong. Much appreciated!
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