Derivative of ln(ln(1/x))? please explain in detail... really confused thank you?

Chain rule: d/dx ln(u) = 1/u * d/dx (u)

You'll need to use chain rule twice:

d/dx [ln(ln(1/x))

= 1/ln(1/x) * d/dx [ln(1/x)]

= 1/ln(1/x) * 1/(1/x) * d/dx (1/x)

= 1/ln(1/x) * x * −1/x²

= −1/(x ln(1/x))

= 1/(x ln x)

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To answer your other question:

L = lim[x→0⁺] (ln(1/x))^x

ln L = lim[x→0⁺] ln((ln(1/x))^x)

ln L = lim[x→0⁺] x ln(ln(1/x))

ln L = lim[x→0⁺] ln(ln(1/x)) / (1/x) = ∞/∞

Use L'Hopital's Rule

ln L = lim[x→0⁺] (1/(x ln x)) / (1/x²)

ln L = lim[x→0⁺] x / ln x

ln L = 0 / −∞

ln L = 0

L = lim[x→0⁺] (ln(1/x))^x = e^0 = 1

y = ln u

dy/du = 1/u

u = 1/x

du/dx = - 1/x²

dy/dx = [ 1/u ] [ - 1/x² ] = - 1 / x

For your case, write

f = ln ( ln ( 1 / x ) )

in a way that makes application of chain rule obvious:

f = ln ( ln ( u ) );............... where u = 1/x is a function of x

f = ln ( w ) ;...................... where w = ln (u)

so, chain rule takes the form

df/dx = df/dw * dw / du * du /dx

where

u = 1/x implies .................du / dx = -1 / x^2

w = ln ( u ) implies ...........dw / du = 1 / u = 1 / (1 / x) = x; recall u = 1/x

f = ln (w) implies.............. df / dw = 1 / w = 1 / ln (u) = 1 / ln ( 1 / x)

multiplying those derivatives,

df/dx =1 / ln (1 / x) * x * (-1 / x^2)

or

df/dx = - 1 / [ x ln (1 / x) ] = 1 / [x ln (x) ] ............. [Ans.]