# How can I find the derivative of $y = \ln [\ln(\ln(x^2 +1))]$?

My question is, how can I solve the following derivative question?

$y = \ln [\ln(\ln(x^2 +1))]$

-
Are you familiar with the chain rule? –  Andrey Rekalo Jan 1 '11 at 11:28

Assuming you're familiar with the chain rule, recall that for a function $f(x)$, if $y=\ln(f(x))$, then $y'=\frac{f'(x)}{f(x)}$.
So in this particular case, let $f(x)=\ln(\ln(x^2+1))$, so $$y'=\frac{f'(x)}{\ln(\ln(x^2+1))}.$$ Now you must calculate $f'(x)$. Try writing $f(x)$ as $f(x)=\ln(g(x))$ with $g(x)=\ln(x^2+1)$ and apply the same principle as before.
Use the chain rule: $$h(x)=f(g(x))\implies h'(x)=f'(g(x))\cdot g'(x)$$
For example, the derivative of $\ln x$ is $\frac{1}{x}$, so the derivative of $\ln(\text{something})$ is $\frac{1}{\text{something}}\cdot(\text{derivative of that something})$. Since you've got several layers of nesting (functions inside functions inside functions...), you should expect to use the chain rule multiple times.