Existence of Derivatives in a Neighborhood

In summary, the conversation discusses the relationship between the differentiability and continuity of a real-valued function f defined on an open interval U. It is established that if f is twice differentiable at a point u, then f' is continuous at u. However, this does not guarantee that f' will be differentiable in a neighborhood of u. The proof for this involves the epsilon-delta definition of the derivative, which assumes the function is already properly defined. It is also noted that a derivative cannot exist at a point of discontinuity.
  • #1
e(ho0n3
1,357
0
Simple question: Let f be a real-valued function defined an open interval U. If f is twice differentiable at u, then f' is continuous at u right? Does that mean that there exists a neighborhood around u where f' exists for all points in this neighborhood?
 
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  • #2
That is true, however, keep in mind that f' may not be differentiable in a neighborhood of u (ie., think of the infamous function f(x) = x for rational x, and 0 otherwise, which is only continuous at 0).
 
  • #3
What would be the proof of this though? If I use the epsilon-delta definition of the derivative for f' at u, we get the following:

For any e > 0 there is a d > 0 such that |x - u| < d implies |f'(x) - f'(u) - f''(u)(x - u)| < e|x - u|.

This doesn't say that f'(x) exists for x satisfying |x - u| <d because the epsilon-delta definition assumes the function is already defined properly, so as a matter of fact, we can't even employ the epsilon-delta definition.
 
  • #4
e(ho0n3 said:
Simple question: Let f be a real-valued function defined an open interval U. If f is twice differentiable at u, then f' is continuous at u right?(2) Does that mean that there exists a neighborhood around u where f' exists for all points in this neighborhood?(1)

An answer to 1:

Consider the following question: "If [tex]f'(x_0)[/tex] exists, then does [tex]f(x)[/tex] exist for [tex]x_0-\delta < x < x_0+\delta[/tex] with some delta?"

Do you see why that is confusing question? The derivative is defined by the formula

[tex]
f'(x_0) = \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h},
[/tex]

so you cannot define the derivative in the first place, if f doesn't exist in some environment of [tex]x_0[/tex]. In any case, the answer to the question is "yes", because by definition, the f must have existed before the derivative was defined. It is not a kind of thing you prove, instead you read it from the definition.

In your original question you made this one step trickier by assuming that [tex]f''(x_0)[/tex] exists, and then asked about [tex]f'(x)[/tex] in the environment. But it's the same thing.

An answer to 2:

A derivative never exists in a point of discontinuity.
 
  • #5
jostpuur said:
An answer to 1:

Consider the following question: "If [tex]f'(x_0)[/tex] exists, then does [tex]f(x)[/tex] exist for [tex]x_0-\delta < x < x_0+\delta[/tex] with some delta?"

Do you see why that is confusing question? The derivative is defined by the formula

[tex]
f'(x_0) = \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h},
[/tex]

so you cannot define the derivative in the first place, if f doesn't exist in some environment of [tex]x_0[/tex]. In any case, the answer to the question is "yes", because by definition, the f must have existed before the derivative was defined. It is not a kind of thing you prove, instead you read it from the definition.

In your original question you made this one step trickier by assuming that [tex]f''(x_0)[/tex] exists, and then asked about [tex]f'(x)[/tex] in the environment. But it's the same thing.
Thanks for the clarification. I think I can also argue by contradiction: if f' was not defined for any neighborhood around u, then f''(u) would not exists by definition.
 

1. What are derivatives in a neighborhood?

Derivatives in a neighborhood refer to the rate of change of a function at a specific point within a small interval or neighborhood. It represents the slope of the tangent line to the function at that point.

2. Why is the existence of derivatives in a neighborhood important?

The existence of derivatives in a neighborhood is important because it allows us to understand the behavior of a function in a small interval around a given point. It helps us to determine the instantaneous rate of change and the direction of the function at that point.

3. How can we prove the existence of derivatives in a neighborhood?

The existence of derivatives in a neighborhood can be proven by using the limit definition of a derivative. This involves finding the limit of the difference quotient as the interval around the given point approaches zero. If the limit exists, then the derivative exists in that neighborhood.

4. What is the relationship between continuity and derivatives in a neighborhood?

A function is said to be differentiable at a point if it is continuous at that point. This means that if a function has a derivative in a neighborhood, then it must also be continuous in that neighborhood. However, the converse is not always true.

5. Can a function have a derivative at a point but not in its neighborhood?

Yes, it is possible for a function to have a derivative at a specific point but not in its neighborhood. This can happen when the function is not continuous at that point. In such cases, the existence of the derivative is limited to that specific point only.

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