Question

OK, here is a differentiation problem that is baffling me -- y = {ln[ln(ln x)]} -- What is y prime? I am guessing the chain rule is the key here, but I cannot figure out the solution.

Answer 1

The solution requires repeatedly applying the chain rule:

\[\frac{ d }{ dx } \ln \left( \ln \left( \ln x \right) \right) = \]
\[= \frac{ 1 }{ \ln \left( \ln x \right) } \times \frac{ d }{ dx } \ln \left( \ln x \right) \]
\[= \frac{ 1 }{ \ln \left( \ln x \right) } \times \frac{ 1 }{ \ln x} \times \frac{ d }{ dx } \ln x \]
\[= \frac{ 1 }{ \ln \left( \ln x \right) } \times \frac{ 1 }{ \ln x} \times \frac{ 1 }{ x }\]
\[= \frac{ 1 }{ \ln \left( \ln x \right) \times \ln x \times x } \]

Answer 2

\[\frac{ 1 }{ x \ln(lnx) }\]

Answer 3

First application of the chain rule:
\[\frac{d}{d \ln \left( \ln x \right)} \ln \left[ \ln \left( \ln x \right) \right] \times \frac{d}{dx} \ln \left( \ln x \right)\]
The first part of the expression is the derivative of ln[ln(ln x)] with respect to the inner function, ln(ln x); the second part is the derivative of the inner function, ln(ln x), with respect to x.
Evaluating the first derivative yields:
\[\frac{1}{ \ln \left( \ln x \right)} \times \frac{d}{dx} \ln \left( \ln x \right)\]
since d/da ln(a) = 1/a.
Now comes the second application of the chain rule:
\[\frac{1}{ \ln \left( \ln x \right)} \times \frac{d}{d \ln x} \ln \left( \ln x \right) \times \frac{d}{d x} \ln x \]
This evaluates to:
\[\frac{1}{ \ln \left( \ln x \right)} \times \frac{1}{\ln x} \times \frac{1}{x}\]