# Math Help - find derivative at x=ln2

1. ## find derivative at x=ln2

$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

2. Originally Posted by genlovesmusic09
$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

$
\log_2 e^x = x \log_2 e
$
and $
\log_2 e
$
is constant.

3. Originally Posted by genlovesmusic09
$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

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

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

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

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

-----------

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

$f'(ln(2)) = \frac{1}{ln(2)}$

4. sorry i didn't put the second e

$f(x) = {\log _2}({e^{{e^x}}})
$

5. Originally Posted by genlovesmusic09
sorry i didn't put the second e

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