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I have the following function:

$f(x)=\ln|\sin(x)|$

I've caculated the derivative to:

$f'(x)=\frac{\cos(x)}{\sin(x)}$

And the domain of $f(x)$ to: $(2\pi n, \pi+2\pi n ) \cup (-\pi + 2\pi n, 2\pi n)$

And the domain of $f'(x)$ to: $(\pi n, \pi+\pi n )$

I want to determine for which x the derivative exists.

My solution is that the derivative exists in the domain of the derivative $(\pi n, \pi+\pi n )$ because the original function f is well defined on that intervall.

Am I thinking correct or am I wrong?

Any help would be greatly appreciated:)

freya
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    You use "$n$", but do not indicate what it is or where it represents. You probably want to say the domain of $f'$ is $\Bbb{R} \smallsetminus {n \pi : n \in \Bbb{Z}}$ -- that is, the domain is all real numbers except for those giving a zero of the sine function. – Eric Towers Sep 12 '20 at 18:07
  • Yes, I forgot to explain n. thank you. – freya Sep 12 '20 at 18:13

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