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

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

Thanks in advance

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  • $\begingroup$ Are you familiar with the chain rule? $\endgroup$ Jan 1, 2011 at 11:28

2 Answers 2

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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.

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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.

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