- #1
tompenny
- 15
- 3
- 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:)
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:)