# What is the derivative of ln(ln(ln(x)))?

May 13, 2015

We need to rewrite the expression via chain rule.

First: $\ln \left(x\right) = u$.
Now, we have $\ln \left(\ln \left(u\right)\right)$ as our original function.

Second: $\ln \left(u\right) = z$.
Now, we have $\ln \left(z\right)$ as our original function.

We now have to derive $\ln \left(z\right)$.

$\frac{\mathrm{dl} n \left(z\right)}{\mathrm{dx}} = \frac{z '}{z}$, where $z '$ is the derivative of $z$.

However, we know $z$: it's $\ln \left(u\right)$.

So, $\frac{\mathrm{dl} n \left(z\right)}{\mathrm{dx}} = \frac{\left[\ln \left(u\right)\right] '}{\ln \left(u\right)}$, where $\left[\ln \left(u\right)\right] '$ stands for the derivative of $\ln \left(u\right)$.

But we know that the derivative of a $\ln f \left(x\right)$ is $\frac{f ' \left(x\right)}{f} \left(x\right)$, so we can rewrite:

$\frac{\mathrm{dl} n \left(z\right)}{\mathrm{dx}} = \frac{\frac{u '}{u}}{\ln} \left(u\right)$.

Again, we know $u$. It's $\ln \left(x\right)$, isn't it?

Substituting...

$\frac{\mathrm{dl} n \left(z\right)}{\mathrm{dx}} = \frac{\left[\ln \left(x\right)\right] '}{\ln x} / \left(\ln \left(\ln \left(x\right)\right)\right)$

Going part by part, now.

$\left[\ln \left(x\right)\right] ' = \frac{1}{x}$, thus $\frac{\left[\ln \left(x\right)\right] '}{\ln x} = \frac{1}{x . \ln \left(x\right)}$.

Going back to our original derivation again,

$\frac{\mathrm{dl} n \left(\ln \left(\ln \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{1}{x . \ln \left(x\right)} / \left(\ln \left(\ln \left(x\right)\right)\right) = \frac{1}{x . \ln \left(x\right) . \left(\ln \left(\ln \left(x\right)\right)\right)}$

:)