To differentiate `f(x) = ln(ln(ln(x)))` use the Chain Rule for differentiation of nested function.

The rule is that ` `if `f(x) = u(v(w(x)))` where `u, v` and `w` are functions

`f'(x) = u'(v(w(x)))v'(w(x))w'(x)`

Using the fact that `d/dx lnx = 1/x` we have that

**` ``f'(x) = 1/ln(ln(x))1/ln(x)1/(x)` **** answer to...**

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To differentiate `f(x) = ln(ln(ln(x)))` use the Chain Rule for differentiation of nested function.

The rule is that ` `if `f(x) = u(v(w(x)))` where `u, v` and `w` are functions

`f'(x) = u'(v(w(x)))v'(w(x))w'(x)`

Using the fact that `d/dx lnx = 1/x` we have that

**` ``f'(x) = 1/ln(ln(x))1/ln(x)1/(x)` ****answer to first question**

To find the domain of `f(x)` we proceed as follows

The domain of `x in (0,oo)`

So we need `ln(ln(ln(x)))` such that `ln(ln(x)) in (0,oo)`

`implies ln(x) in (1, oo)`

`implies x in (e^1,oo)` **answer to second question**