find derivative at x=ln2

**genlovesmusic09** · October 4th 2009, 09:27 AM

$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

**Danny** · October 4th 2009, 09:35 AM

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.

**e^(i*pi)** · October 4th 2009, 09:36 AM

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

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

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

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

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

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

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