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

[tompenny]

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:)

 

[PeroK]

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?

 

[tompenny]

But what I have hard to understand is that the function is defined at [0,1] but does the derivative exists at the endpoints?

 

[tompenny]

the derivative at 0 is undefined right?

 

[PeroK]

It looks that way.

 

[tompenny]

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?

 

[PeroK]

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.

 

[Mark44]

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 $f(x) =\text{arcsin}(x)$.

 

[vela]

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.

 

[PeroK]

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 $f'(0) = 0$.

 

[Mark44]

I wasn't thinking in terms of piecewise-defined functions.

 

[PeroK]

I wasn't thinking in terms of piecewise-defined functions.

Is my argument in post #7 valid otherwise? I'm not entirely sure.

 

[vela]

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?

 

[vela]

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 $f'(1)=3$.

 

[PeroK]

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.

 

[PeroK]

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 $x = 0$: $f(0) = 0$

 

[haruspex]

The function is defined at $x = 0$: $f(0) = 0$

Sorry, missed that.