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I'm currently taking Calculus. I'm pretty good with derivatives apart from when it comes to logarithmic differentiation etc.

Here is one I'm having problems with, if anyone could help that would be appreciated.

$$ f(x)=\ln (\ln (x) ) .$$

Can someone please explain the derivative of this? Thanks!

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    Note: Whatever "CAL 1" means to you, it's most probably specific to a single school or school system. When you're writing for a worldwide audience (such as here) you cannot expect that putting "CAL 1" in parentheses will tell the reader _anything_ useful.2011-11-26

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The basic formula for the derivative of $\ln x$ is $${d\over dx}\ln x={1\over x}.$$

Recall the chain rule: ${d\over dx}f\bigl(g(x)\bigr) =f'\bigl(g(x)\bigr)\cdot g'(x)$.

So, by the chain rule, with $f(x)=\ln x$ and $g(x)=\ln x$, $${d\over dx}\ln (\ln x)={1\over \ln x}\cdot (\ln x )'={1\over \ln x}\cdot {1\over x}= {1\over x\ln x}.$$

Don't tell anyone I told you this, but you can remember: "the derivative of $\ln$ of something is (1 over the something) times the derivative of the something".

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    Is that easier to remember than the chain rule itself?2011-11-26
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First note that this is going to require an application of the chain rule, where $u=\ln(x)$. So, to find $f'(x)$, one must first find $f'(u)$ and then find $u'(x)$. Rewriting $f(x)$ in terms of $u$ yields $f(u)=\ln(u)$ and $u(x)=\ln(x)$.

Thus, $f'(u)=1/u$ and $u'(x)=1/x$. Therefore, $f'(x)=f'(u)u'(x) = (1/u)(1/x)$. Then substitute $u=\ln(x)$ into the equation and we get $f'(x) = (1/(x\ln(x)))$.

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    A couple of $\LaTeX$ tips: (i) use `\ln` for natural log; that automatically puts it in roman typeface, as it should. (ii) Put the *entire* formula inside the math delimeters, `$...$`, *including* the equal sign; this prevents awkward line breaks, and it also provides appropriate spacing. Don't use `*` for products inside math formulas; use either juxtaposition, or if needed use `\cdot` (as in $f'(u)\cdot u'(x)$), or `\times` (as in $3\times 5$). Finally, don't mix roman typeface and math-italic typeface: x and $x$ look different enough that they may refer to different things!2011-11-26
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    @ArturoMagidin Thanks for the advice!2011-11-26