How would one diffrentiate ln(lnx)?

Answer 1

Chain rule:
d/dx ln (lnx) = 1 / (ln x) . d/dx (ln x)
= (1 / ln x) (1/x)
= 1 / (x ln x).

Answer 2

Using the chain rule and the fact the (ln(x))1 = 1/x, x >0.

Then, (ln(ln(x))' = 1/ln(x) * (ln(x)' = 1/ln(x) * 1/x = 1/(x*ln(x)), x >1

Answer 3

chain rule

let ln x = t

y = ln(ln x)
= ln t
dy/dx = dy/dt * dt/dx = 1/t dt/dx = 1/t d(ln x)/dx = 1/t 1/x = 1/(x ln x)

Answer 4

Let u = ln x
du/dx = 1 / x
y = ln u
dy/du = 1 / u
dy/dx = (dy/du).(du/dx)
dy/dx = (1 / u).(1 / x)
dy/dx = (1 / ln x).(1 / x)
dy/dx = 1 / ( x ln x )

Answer 5

you should use the chain rule

it states that
(d/dy)x = (d/du)u X (d/dx)x

in your question, let {ln x} be u
hence
(d/dx){ln(ln x)} = (d/du) (ln u) X (d/dx)(u)
........................= (d/du)(ln u) X (d/dx)(ln x)
.................... = 1/u X 1/x
................... = 1/(ln x) X 1/x
.................... = 1/{x(ln x)}