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I'm not understanding what you're saying. The standard definition of the derivative of f at c is f'(c) = lim_{h → 0} (f(c + h) - f(c))/h The definition would

I'm not understanding what you're saying. The standard definition of the derivative of f at c is f'(c) = lim_{h → 0} (f(c + h) - f(c))/h The definition would not make sense if f wasn't defined at c (note the "f(c)" in the numerator). For instance, it can't be applied to your f(x) = (x² - 1)/(x - 1) at x = 1, because f(1) is not defined. And it's a standard result (even stated in Calc 1 classes) that if a function is differentiable at a point, then it's continuous there. For example: 5.2 Theorem. Let f be defined on [a, b]. If f is differentiable at a point x ∈ [a, b], then f is continuous at x. (Walter Rudin, "Principles of Mathematical Analysis", 3rd edition, p. 104) Or: Theorem 2.1 If f is differentiable at x = a, then f is continuous at x = a. (Robert Smith and Roland Minton, "Calculus -Early Transcendentals", 4th edition, p. 140) It's true that your f(x) = (x² - 1)/(x - 1) has a removable discontinuity at x = 1, since if we define g(x) = f(x) for x ≠ 1 and g(1) = 2, then g is continuous. Was this what you meant?

terminalbraid 2y ago
This is correct. You cannot have a discontinuity with any accepted definition of a derivative (and your definition is explicit about this: the value f(c) must exist). Just allowing the limits on both sides to be equal already has a mathematical definition which is that of a functional limit, the function in this case being (f(x) - flim(c))/ (x-c) where flim(c) is the value of a (different) functional limit of f(x): x->c (as f(c) doesn't exist). and yes, by defining a new function with that hole explicitly filled in with a defined value to make it continuous is the typical prescription. It does not imply the derivative exists for the other function as the other post posits.
dwattttt 2y ago
https://en.m.wikipedia.org/wiki/Classification_of_discontinuities https://en.m.wikipedia.org/wiki/Classification_of_discontinu... is responsive and quite accessible. It notes that there doesn't have to be an undefined point for a function to be discontinuous (and that terminology often conflates the two), and matches what I recall of determining that if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists.
- bikenaga 2y ago
  > ... there doesn't have to be an undefined point for a function to be discontinuous. That's right. In the example f(x) = (x² - 1)/(x - 1) for x ≠ 1, if we further define f(1) = 0, the function is now defined at x = 1, but discontinuous there. > ... if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists. (You probably mean "both sides of the point", since if there's a discontinuity there the derivative can't exist.) Your point that, if the left and right-hand limits both exist and are equal, then the derivative exists (and equals their common value) is true for all limits. Also, there's a difference between the use of the word "continuous" in calc courses and in topology. In calc courses where functions tend to take real numbers to real numbers, a function may be said to be "not continuous" at a point where it isn't defined. So f(x) = 1/(x - 2) is "not continuous at 2". But in topology, you only consider continuity for points in the domain of the function. So since the (natural) domain of f(x) = 1/(x - 2) is x ≠ 2, the function is continuous everywhere (that it's defined).
  - dwattttt 2y ago
    I was actually aiming for the situation where a function is defined on all reals but still discontinuous (e.g. the piecewise function in the wiki article for the removable discontinuity). So there's a discontinuity (x=1), however the function is defined everywhere.
smokedetector1 2y ago
The standard definition of a derivative c involves the assumption that f is defined at c. However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard.
- deleted 2y ago
  [deleted]
- JadeNB 2y ago
  > However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard. Although this gives the right answer whenever f is differentiable at c, it can wrongly think that a function is differentiable when it isn't, as for the absolute-value function at c = 0.
  - smokedetector1 2y ago
    Good point. So this is probably one of the reasons why the version I stated isn't used.
    - JadeNB 2y ago
      It is used, just with the caveat in mind that it may exist when the derivative doesn't. It is usually called the symmetric derivative (https://en.wikipedia.org/wiki/Symmetric_derivative https://en.wikipedia.org/wiki/Symmetric_derivative).
      - deleted 2y ago
        [deleted]

terminalbraid 2y ago

This is correct. You cannot have a discontinuity with any accepted definition of a derivative (and your definition is explicit about this: the value f(c) must exist). Just allowing the limits on both sides to be equal already has a mathematical definition which is that of a functional limit, the function in this case being (f(x) - flim(c))/ (x-c) where flim(c) is the value of a (different) functional limit of f(x): x->c (as f(c) doesn't exist). and yes, by defining a new function with that hole explicitly filled in with a defined value to make it continuous is the typical prescription. It does not imply the derivative exists for the other function as the other post posits.

dwattttt 2y ago

https://en.m.wikipedia.org/wiki/Classification_of_discontinuities https://en.m.wikipedia.org/wiki/Classification_of_discontinu... is responsive and quite accessible. It notes that there doesn't have to be an undefined point for a function to be discontinuous (and that terminology often conflates the two), and matches what I recall of determining that if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists.

bikenaga 2y ago
> ... there doesn't have to be an undefined point for a function to be discontinuous. That's right. In the example f(x) = (x² - 1)/(x - 1) for x ≠ 1, if we further define f(1) = 0, the function is now defined at x = 1, but discontinuous there. > ... if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists. (You probably mean "both sides of the point", since if there's a discontinuity there the derivative can't exist.) Your point that, if the left and right-hand limits both exist and are equal, then the derivative exists (and equals their common value) is true for all limits. Also, there's a difference between the use of the word "continuous" in calc courses and in topology. In calc courses where functions tend to take real numbers to real numbers, a function may be said to be "not continuous" at a point where it isn't defined. So f(x) = 1/(x - 2) is "not continuous at 2". But in topology, you only consider continuity for points in the domain of the function. So since the (natural) domain of f(x) = 1/(x - 2) is x ≠ 2, the function is continuous everywhere (that it's defined).
- dwattttt 2y ago
  I was actually aiming for the situation where a function is defined on all reals but still discontinuous (e.g. the piecewise function in the wiki article for the removable discontinuity). So there's a discontinuity (x=1), however the function is defined everywhere.

smokedetector1 2y ago

The standard definition of a derivative c involves the assumption that f is defined at c. However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard.

deleted 2y ago
[deleted]
JadeNB 2y ago
> However, you could also (probably) define the derivative as lim_{h->0} (f(c+h) - f(c-h))/2h, so without needing f(c) to be defined. But that's not standard. Although this gives the right answer whenever f is differentiable at c, it can wrongly think that a function is differentiable when it isn't, as for the absolute-value function at c = 0.
- smokedetector1 2y ago
  Good point. So this is probably one of the reasons why the version I stated isn't used.
  - JadeNB 2y ago
    It is used, just with the caveat in mind that it may exist when the derivative doesn't. It is usually called the symmetric derivative (https://en.wikipedia.org/wiki/Symmetric_derivative https://en.wikipedia.org/wiki/Symmetric_derivative).
    - deleted 2y ago
      [deleted]