how do u find log [ln(x)]?

Question

so far i have [1/ (ln 10)] * [1/ (ln (ln x))] * e^x, but iono if this is right, but the answer is [1/(x (lnx) ln10)]. can someone please explain step by step how to do the problem?

i'm trying to find y' when given the expression log[ln(x)]

Answer 1

I assume you're trying to differentiate it.

Recall the chain rule. We know d/dx ln x = 1/x, and d/dx log x = 1/(x ln 10). The chain rule says

d/dx log (f(x)) = 1/(f(x) ln 10) . f'(x)
so d/dx log (ln x) = (1/ (ln x)(ln 10)) . (d/dx ln x)
= 1 / [x ln x ln 10]

Answer 2

this is not an equation, therefore it is not solvable.
are you trying to put it in a different form? if so what form do you need it in?