$\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
$\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
That change of base rule doesn't like right
$\displaystyle log_2e^x = \frac{ln(e^x)}{ln(2)}$
$\displaystyle \frac{d}{dx}\left(\frac{x}{ln(2)}\right)$
$\displaystyle u = x\: \rightarrow \: u' = 1$
$\displaystyle v = ln(2) \: \rightarrow \: v' = 0$
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$\displaystyle y' = \frac{ln(2) - 0}{(ln(2))^2} = \frac{1}{ln(2)}$
$\displaystyle f'(ln(2)) = \frac{1}{ln(2)}$