Determine for which x the derivative exists for $$f(x)=arcsin(\sqrt x)$$

In summary, the conversation discusses the concept of finding the domain of a derivative and the existence of the derivative at the endpoints of a closed interval. It is stated that the derivative will always have a domain that is a subset of the base function's domain. However, it is also mentioned that this may not always be the case, as shown by a counterexample. The conversation also touches on the rules of differentiation and the possibility of them leading to incorrect results if not applied carefully.
  • #1
15
3
Homework Statement
Determine for which x the derivative exists
Relevant Equations
$$f(x)=arcsin(\sqrt x)$$
Hi there.

I have the following function:

$$f(x)=arcsin(\sqrt x)$$

I've caculated the derivative to:

$$f'(x)=\frac{1}{2\sqrt x\sqrt{ (1-x}}$$

And the domain of f(x) to: $$[0, 1]$$

And the domain of f'(x) to: $$(0, 1)$$

I want to determine for which x the derivative exists but I'm not really sure if I should use the domain of the original function or the derivative of the function?

Any help would be greatly appreciated:)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You can only differentiate a function at a point if it is defined at that point.

If ##f## is differentiable at ##0## what is the derivative?
 
  • Like
Likes tompenny
  • #3
But what I have hard to understand is that the function is defined at [0,1] but does the derivative exists at the endpoints?
 
  • #4
the derivative at 0 is undefined right?
 
  • #5
It looks that way.
 
  • #6
I understand :) Thank you! Is it enough "proof" to show the derivative exists by calculating the domain of the derivative? or is there other ways?
 
  • #7
tompenny said:
I understand :) Thank you! Is it enough "proof" to show the derivative exists by calculating the domain of the derivative? or is there other ways?
You could argue that if the derivative of ##f## at ##x## exists then it must satisfy the equation you calculated. Therefore, the derivative at ##0## and ##1## cannot exist.
 
  • Love
Likes tompenny
  • #8
tompenny said:
I understand :) Thank you! Is it enough "proof" to show the derivative exists by calculating the domain of the derivative? or is there other ways?
You're not asked to prove anything. You know the domain of the function, and you have calculated the derivative of the function. The domain of the derivative of a function will always be a subset (possibly a proper subset) of the domain of the base function.
By examining the formula of the derivative, you can see that the points 0 and 1 will make the derivative undefined.
It's also useful to look at the graph of ##f(x) =\text{arcsin}(x)##.
 
  • #9
tompenny said:
But what I have hard to understand is that the function is defined at [0,1] but does the derivative exists at the endpoints?
If you're using the common definition of the derivative from Calc 1, then derivatives don't exist at the endpoints of a closed interval. When you take a limit, you have to be able to calculate the limit coming from both the left and the right (and they have to be equal). At an endpoint, you can only approach the point from one side. The limit of the difference quotient doesn't exist at an endpoint for this reason, so the derivative doesn't exist. To get around this technicality, you may define the derivative at an endpoint to be the one-sided limit.
 
  • #10
Mark44 said:
You're not asked to prove anything. You know the domain of the function, and you have calculated the derivative of the function. The domain of the derivative of a function will always be a subset (possibly a proper subset) of the domain of the base function.
By examining the formula of the derivative, you can see that the points 0 and 1 will make the derivative undefined.
It's also useful to look at the graph of ##f(x) =\text{arcsin}(x)##.
A counterexample to this idea is:
$$f(x) = x^2\sin(\frac 1 {x^3}) \ \ (x \ne 0) \ \ \ \text{and} \ f(0) = 0$$
The derivative is:
$$f'(x) = 2x\sin(\frac 1 {x^3}) - \frac 3 {x^2}\cos(\frac 1 {x^3}) \ \ (x \ne 0)$$
It looks like the derivative at ##0## doesn't exist. Yet ##f'(0) = 0##.
 
  • #11
I wasn't thinking in terms of piecewise-defined functions.
 
  • #12
Mark44 said:
I wasn't thinking in terms of piecewise-defined functions.
Is my argument in post #7 valid otherwise? I'm not entirely sure.
 
  • #13
If applying the rules of differentiation leads to a result that isn't actually the derivative, wouldn't that imply that one or more of the theorems establishing the product rule, quotient rule, and chain rule is wrong?
 
Last edited:
  • #14
tompenny said:
Is it enough "proof" to show the derivative exists by calculating the domain of the derivative?
I'm always wary about trying to come up with a generalized rule in mathematics. Unless I can prove an idea, I worry there's some sort of pathological counterexample I didn't think of.

One way your suggestion could lead to an error is with this function, for example:
$$f(x) = \frac{x^3-1}{x-1}.$$ It's not defined at ##x=1## so ##f'## doesn't exist at ##x=1## either. But a student will typically find ##f'## by first simplifying ##f## to ##x^2+x+1## and then differentiating. The expression ##2x+1## is clearly defined at ##x=1##, but it's not true that ##f'(1)=3##.
 
  • Like
Likes PeroK
  • #15
vela said:
If applying the rules of differentiation leads to a result that isn't actually the derivative, wouldn't that imply that one or more of the theorems establishing the product rule, quotient rule, and chain rule is wrong?
We'd have to find a way (other than by defining ##f## by a different formula at ##x = 0##) for the differentiation process to have to exclude ##x = 0##. Somehow to produce a formula for the derivative that explicitly has had to exclude ##x = 0##. But, when we look directly at the function, the derivative exists at ##x= 0##.

I must admit I can't see a way to do that.
 
  • #16
haruspex said:
No, you can only say ##\lim _{x\to 0}f'(x)=0##, no matter which side you approach from. But the definition of a derivative at x requires that the function is defined at x.
The function is defined at ##x = 0##: ##f(0) = 0##
 
  • #17
PeroK said:
The function is defined at ##x = 0##: ##f(0) = 0##
Sorry, missed that.
 

1. What is the derivative of f(x)?

The derivative of f(x) is the rate of change of the function with respect to x. In this case, the derivative of f(x) is 1 / (2*sqrt(x*(1-x))).

2. What is the domain of the function f(x)?

The domain of the function f(x) is [0,1], as the argument of the arcsin function must be between -1 and 1.

3. How do you determine if the derivative exists for a given x value?

The derivative exists for a given x value if the function is continuous and differentiable at that point. In this case, the function is continuous on its entire domain and differentiable on the open interval (0,1).

4. Is the derivative of f(x) defined for all values of x in its domain?

No, the derivative of f(x) is not defined for x=0 and x=1, as the function is not differentiable at these points.

5. How can the derivative of f(x) be used to find the maximum and minimum values of the function?

The maximum and minimum values of the function can be found by setting the derivative equal to 0 and solving for x. The resulting x values correspond to the points where the function has a maximum or minimum value. Additionally, the sign of the derivative can indicate whether the function is increasing or decreasing at a given point, which can help determine the location of extrema.

Suggested for: Determine for which x the derivative exists for $$f(x)=arcsin(\sqrt x)$$

Back
Top