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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$\begingroup$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".
$\endgroup$ 1 $\begingroup$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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