Question

How do you differentiate `log_xe`?

Answer 1

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

Explanation:

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 )`