# Existence of Derivatives in a Neighborhood

1. Nov 24, 2008

### e(ho0n3

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?

2. Nov 24, 2008

### slider142

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. Nov 24, 2008

### e(ho0n3

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. Nov 24, 2008

### jostpuur

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

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

$$f'(x_0) = \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h},$$

so you cannot define the derivative in the first place, if f doesn't exist in some environment of $$x_0$$. 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 $$f''(x_0)$$ exists, and then asked about $$f'(x)$$ in the environment. But it's the same thing.