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Existence of Derivatives in a Neighborhood

  1. Nov 24, 2008 #1
    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. jcsd
  3. Nov 24, 2008 #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).
  4. Nov 24, 2008 #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.
  5. Nov 24, 2008 #4
    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

    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 [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.
  6. Nov 25, 2008 #5
    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.
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