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Math Help - logarithmic differentiation

  1. #1
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    logarithmic differentiation

    What does ln2 equal? I'm stuck at that. Is there a formula for the derivative of ln(2^x)?

    The problem is:

    Use logarithmic differentiation to differentiate the following functions:
    f(x)=2^x.

    I got: ln(f(x))=ln(2^x)
    d/dx [lnf(x)]=[f'(x)]/[f(x)=[x(2^(x-1)]/(2^x)
    f'(x)=f(x)[x(2^(x-1)]/(2^x)=[x2^(x-1)]

    But the answer says it's (2^x)ln2
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  2. #2
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    Quote Originally Posted by fabxx View Post
    What does ln2 equal? I'm stuck at that. Is there a formula for the derivative of ln(2^x)?

    The problem is:

    Use logarithmic differentiation to differentiate the following functions:
    f(x)=2^x.

    I got: ln(f(x))=ln(2^x)
    d/dx [lnf(x)]=[f'(x)]/[f(x)=[x(2^(x-1)]/(2^x)
    f'(x)=f(x)[x(2^(x-1)]/(2^x)=[x2^(x-1)]

    But the answer says it's (2^x)ln2

    \log_ax^n=n\log_ax\Longrightarrow \ln(f(x))=\ln 2^x=x\ln 2\Longrightarrow  \frac{f'(x)}{f(x)}=(\ln f(x))'=(x\ln 2)'=\ln 2 ...and etc.

    Tonio
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  3. #3
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    To be clear: the whole point of using logarithmic differentiation is to move terms out of the exponent. So, in this case:

    f(x)=2^x

    \ln f(x)=\ln (2^x)

    \ln f(x)=x \ln 2

    The last step is the important step, because now the right-hand side is a constant (ln 2) times x, which is easy to differentiate.
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  4. #4
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    I still don't get where I'm doing it wrong:


    The problem is: f(x) = 2^x

    So I got:

    ln f(x)=ln2^x = xln2

    d/dx [lnf(x)] = f'(x)/f(x) = (ln2 x 1)/ (xln2)=1/x

    f'(x)=f(x) [ ln2/ln(2^x) = 2^x [ln2/(ln2x)

    Any help would be appreciated.
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  5. #5
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    Quote Originally Posted by tonio View Post
    \log_ax^n=n\log_ax\Longrightarrow \ln(f(x))=\ln 2^x=x\ln 2\Longrightarrow  \frac{f'(x)}{f(x)}=(\ln f(x))'=(x\ln 2)'=\ln 2 ...and etc.

    Tonio

    Shouldn't it be f'(x)/f(x)=ln2/xln2 ?
    Last edited by fabxx; April 13th 2010 at 09:07 PM.
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  6. #6
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    Quote Originally Posted by fabxx View Post
    Shouldn't it be f'(x)/f(x)=ln2/xln2 ?

    No since \frac{\ln 2}{x\ln 2}=\frac{1}{x} ...

    Tonio
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