Determine for which x the derivative exists of: ##f(x)=\ln|\sin(x)|##

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The function f(x) = ln|sin(x)| has a derivative f'(x) = cos(x)/sin(x), which is defined in the intervals (πn, π + πn) for all integers n. The original function f(x) is well-defined in the intervals (2πn, π + 2πn) and (-π + 2πn, 2πn). The derivative exists in the domain of f'(x), which is limited to the intervals where sin(x) is non-zero. Therefore, the derivative exists for all x in the intervals (πn, π + πn), excluding points where sin(x) = 0. The conclusion is that the derivative exists for all real numbers except at points where sin(x) equals zero.
tompenny
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Homework Statement
Determine for which x the derivative exists of: ##f(x)=\ln|\sin(x)|##
Relevant Equations
$$f(x)=\ln|\sin(x)|$$
Hi there.

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:)
 
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for all real numbers except for 0 right?
 
tompenny said:
My solution is that the derivative exists in the domain of the derivative $$(\pi n, \pi+\pi n )$$
Union thereof for all n, yes.
tompenny said:
for all real numbers except for 0 right?
No, for all real numbers except...?
 
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