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Math Help - find derivative at x=ln2

  1. #1
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    find derivative at x=ln2

    f(x) = {\log _2}({e^{{e^x}}}) = \frac{{\ln {e^{{e^x}}}}}<br />
{{\ln 3}} = \frac{{{e^x}\ln e}}<br />
{{\ln 3}} = \frac{{{e^x}}}<br />
{{\ln 3}}<br />

    Find the derivative at x=ln2

    i'm not sure where to go next
    Last edited by genlovesmusic09; October 4th 2009 at 10:42 AM.
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    f(x) = {\log _2}({e^{^x}}) = \frac{{\ln {e^{{e^x}}}}}<br />
{{\ln 3}} = \frac{{{e^x}\ln e}}<br />
{{\ln 3}} = \frac{{{e^x}}}<br />
{{\ln 3}}<br />

    Find the derivative at x=ln2

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

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

    Find the derivative at x=ln2

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

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

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


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

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

    -----------

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

    f'(ln(2)) = \frac{1}{ln(2)}
    Last edited by e^(i*pi); October 4th 2009 at 10:41 AM. Reason: clarification
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  4. #4
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    sorry i didn't put the second e

    f(x) = {\log _2}({e^{{e^x}}})<br />
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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

    f(x) = {\log _2}({e^{{e^x}}})<br />
    So f(x) = e^x \log_2 e = c e^x where c = \log_2 e. Now differentiate and substitute.
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