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

• tompenny
In summary, the conversation discusses the function f(x) = ln|sin(x)| and its derivative, f'(x) = cos(x)/sin(x). The domain of f(x) is determined to be (2πn, π+2πn) ∪ (-π+2πn, 2πn), while the domain of f'(x) is (πn, π+πn). The question is asked for which values of x does the derivative exist, and the solution is that it exists in the domain of the derivative. The speaker also asks for confirmation that their thinking is correct, to which the response is that the derivative exists for all real numbers except for 0.
tompenny
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:)

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...?

## 1. What is the function f(x)?

The function f(x) is ln|sin(x)|, which represents the natural logarithm of the absolute value of the sine of x.

## 2. How do you determine the derivative of f(x)?

To determine the derivative of f(x), we use the chain rule and the fact that the derivative of ln(x) is 1/x. The derivative of f(x) is given by f'(x) = 1/(sin(x)) * cos(x).

## 3. Is the derivative of f(x) defined for all values of x?

No, the derivative of f(x) is not defined for all values of x. It is only defined for values of x where sin(x) is non-zero, which means that x cannot be equal to 0, π, 2π, etc.

## 4. How do you determine the values of x for which the derivative of f(x) exists?

We can determine the values of x for which the derivative of f(x) exists by finding the values of x where sin(x) is non-zero. These values are any real numbers except for 0, π, 2π, etc.

## 5. Can you provide an example of a value of x for which the derivative of f(x) exists?

Yes, for example, if we choose x = π/4, then the derivative of f(x) exists and is equal to f'(π/4) = 1/(sin(π/4)) * cos(π/4) = 1/(√2) * (√2/2) = 1.

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