Thread: find derivative at x=ln2

1. 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

2. Originally Posted by genlovesmusic09
$\displaystyle f(x) = {\log _2}({e^{^x}}) = \frac{{\ln {e^{{e^x}}}}} {{\ln 3}} = \frac{{{e^x}\ln e}} {{\ln 3}} = \frac{{{e^x}}} {{\ln 3}}$

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.

3. Originally Posted by genlovesmusic09
$\displaystyle f(x) = {\log _2}({e^{^x}}) = \frac{{\ln {e^{{e^x}}}}} {{\ln 3}} = \frac{{{e^x}\ln e}} {{\ln 3}} = \frac{{{e^x}}} {{\ln 3}}$

Find the derivative at x=ln2

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

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

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

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

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

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

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

4. sorry i didn't put the second e

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

5. Originally Posted by genlovesmusic09
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.