- #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:)