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Thread: find derivative at x=ln2

  1. #1
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    find derivative at x=ln2

    $\displaystyle f(x) = {\log _2}({e^{{e^x}}}) = \frac{{\ln {e^{{e^x}}}}}
    {{\ln 3}} = \frac{{{e^x}\ln e}}
    {{\ln 3}} = \frac{{{e^x}}}
    {{\ln 3}}
    $

    Find the derivative at x=ln2

    i'm not sure where to go next
    Last edited by genlovesmusic09; Oct 4th 2009 at 09:42 AM.
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    $\displaystyle f(x) = {\log _2}({e^{^x}}) = \frac{{\ln {e^{{e^x}}}}}
    {{\ln 3}} = \frac{{{e^x}\ln e}}
    {{\ln 3}} = \frac{{{e^x}}}
    {{\ln 3}}
    $

    Find the derivative at x=ln2

    i'm not sure where to go next
    It might be easier to realize that

    $\displaystyle
    \log_2 e^x = x \log_2 e
    $ and $\displaystyle
    \log_2 e
    $ is constant.
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  3. #3
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    Quote Originally Posted by genlovesmusic09 View Post
    $\displaystyle f(x) = {\log _2}({e^{^x}}) = \frac{{\ln {e^{{e^x}}}}}
    {{\ln 3}} = \frac{{{e^x}\ln e}}
    {{\ln 3}} = \frac{{{e^x}}}
    {{\ln 3}}
    $

    Find the derivative at x=ln2

    i'm not sure where to go next
    That change of base rule doesn't like right

    $\displaystyle log_2e^x = \frac{ln(e^x)}{ln(2)}$

    $\displaystyle \frac{d}{dx}\left(\frac{x}{ln(2)}\right)$


    $\displaystyle u = x\: \rightarrow \: u' = 1$

    $\displaystyle v = ln(2) \: \rightarrow \: v' = 0$

    -----------

    $\displaystyle y' = \frac{ln(2) - 0}{(ln(2))^2} = \frac{1}{ln(2)}$

    $\displaystyle f'(ln(2)) = \frac{1}{ln(2)}$
    Last edited by e^(i*pi); Oct 4th 2009 at 09:41 AM. Reason: clarification
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  4. #4
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    sorry i didn't put the second e

    $\displaystyle f(x) = {\log _2}({e^{{e^x}}})
    $
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  5. #5
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    Quote Originally Posted by genlovesmusic09 View Post
    sorry i didn't put the second e

    $\displaystyle f(x) = {\log _2}({e^{{e^x}}})
    $
    So $\displaystyle f(x) = e^x \log_2 e = c e^x$ where $\displaystyle c = \log_2 e$. Now differentiate and substitute.
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