How do you differentiate $log_xe$ ?

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Answer 1 · 2016-09-11T08:46:02

$= - 1/ (x ln^2 x)$

you can use a base switch (see below for how it works), switching everything into base $e$
we have

$y = log_x e$

$= (ln e)/(ln x) = 1/(ln x)$

so $y' = - 1/ (ln^2 x) * 1/x$

$= - 1/ (x ln^2 x)$

Base Switching

let $y = log_a b$

$implies a^y = b$

We'll switch to base $Q$

So $log_Q a^y = log_Q b$

$y log_Q a = log_Q b$

$y = (log_Q b)/(log_Q a )$

How do you differentiate log_xelog_xe?