How to Prove the Continuity of a Derivative?

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Discussion Overview

The discussion revolves around the relationship between the differentiability of a function and the continuity of its derivative. Participants explore whether the differentiability of a function on an interval implies that its derivative is continuous on that interval, examining definitions and providing counterexamples.

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants question if differentiability of a function on an interval implies the continuity of its derivative.
  • One participant suggests that the right approach is to rewrite the question using definitions related to limits, continuity, and derivatives.
  • A counterexample is provided: the function f(x) = x²sin(1/x) for x ≠ 0 and f(0) = 0, which is differentiable everywhere on (-1, 1) but has a derivative that is not continuous at 0.
  • Another participant asserts that if a function is differentiable on an interval, then its derivative must be continuous, which is challenged by the counterexample.
  • Discussion includes the intermediate value property of derivatives, noting that if f is differentiable on [a,b], then f' takes on every value between f'(a) and f'(b).
  • Clarifications are made regarding the original question, emphasizing the distinction between the continuity of the function and the continuity of its derivative.

Areas of Agreement / Disagreement

Participants express differing views on whether the differentiability of a function guarantees the continuity of its derivative. Multiple competing perspectives remain unresolved.

Contextual Notes

Some participants express confusion over terminology and definitions, particularly regarding the implications of differentiability and the conditions under which continuity of the derivative may fail.

Who May Find This Useful

This discussion may be of interest to students and educators in calculus and real analysis, particularly those exploring the properties of derivatives and continuity.

Treadstone 71
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If f is differentiable on (a,b), does it imply that f' is continuous on (a,b)? If so, is there a way of proving it?
 
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Do you mean if f(x) is differentiable at every value of x in [a,b]?
 
The nice thing about analysis is that, unless you have a clever trick to apply, the right first step is almost always the same: rewrite the question in terms of the definitions. (i.e. continuity and derivatives are defined in terms of limits, and limits have an epsilon-delta definition)
 
yes, if is differentiable it has to exists

this can be concluded by the domain being within it
 
devilkidjin said:
yes, if is differentiable it has to exists
this can be concluded by the domain being within it
I have no idea what this means! "domain being within it"? What does "it" refer to? In fact, what does "it has to exist" refer to? This doesn't appear to have anything to do with the original question- whether the existence of f' on an interval implies that f' must be continuous on that interval.

The answer to that question is "no"- for example the function
f(x)= x^2sin(\frac{1}{x}) if x \neq 0[/itex], f(0)= 0<br /> is differentiable at every point on (-1, 1) but that f&#039; is not continuous at 0.<br /> <br /> A lot of people confuse continuity of the <b>derivative</b> with the fact that, if a function is differentiable at a point, the <b>function</b> must be continuous at that point.
 
What's the derivative of your said function at 0?

I'm pretty sure if f is differentiable on I, then f' is continuous on I.
 
Last edited:
yes.

lim(x->x0) {[f(x)-f(x0)]/(x-x0)}*lim(x->x0){x-x0}=0

Therefore, lim(x->x0){f(x)-f(x0)}=0, lim(x->x0) {f(x)}=f(x0)..
 
Treadstone 71 said:
What's the derivative of your said function at 0?

For Hall's example, f'(0)=0, you can work it out using the definition of the derivative. The limit of f' as you approach zero doesn't exist and your assertion that f' must be continuous is false.

Derivatives do have an intermediate value property though. If f is differentiable on [a,b] and f'(a)<c<f'(b) then f'(x)=c for some x in (a,b)
 
No, treadstone and rocketa, the question was NOT "if f is differentiable is f continuous?"

The question was "If f is differentiable is f '(x) continuous?". without a space between the f and the ' it's a little bit hard to see but if you look closely it's there! If f is differentiable on an interval, it's derivative is NOT necessarily differentiable there. The example I gave earlier:
f(x)= x2sin(x) if x is not 0
f(0)= 0
is differentiable on for all x but its derivative is not continuous at x= 0.

treadstone said:
What's the derivative of your said function at 0?
If that was a reference to my example, the derivative at 0 is, of course
lim_{x\rightarrow 0}\frac{x^2sin(\frac{1}{x})}{x}= lim_{x\rightarrow 0}x sin(<br /> \frac{1}{x})= 0

For x not equal to 0, the derivative is
2x sin(\frac{1}{x})- cos(\frac{1}{x})
which does not have a limit as x goes to 0.

That is, the derivative exists for all x but is not continuous at x=0.
 
Last edited by a moderator:
  • #10
Ah yes, I was searching for the intermediate value property, thinking it was continuity. Thanks for clearing that up.
 
  • #11
is f' continuous somewhere?
 
  • #12
HallsofIvy said:
No, treadstone and rocketa, the question was NOT "if f is differentiable is f continuous?"
The question was "If f is differentiable is f '(x) continuous?". without a space between the f and the ' it's a little bit hard to see but if you look closely it's there! If f is differentiable on an interval, it's derivative is NOT necessarily differentiable there. The example I gave earlier:
f(x)= x2sin(x) if x is not 0
f(0)= 0
is differentiable on for all x but its derivative is not continuous at x= 0.
If that was a reference to my example, the derivative at 0 is, of course
lim_{x\rightarrow 0}\frac{x^2sin(\frac{1}{x})}{x}= lim_{x\rightarrow 0}x sin(<br /> \frac{1}{x})= 0
For x not equal to 0, the derivative is
2x sin(\frac{1}{x})- cos(\frac{1}{x})
which does not have a limit as x goes to 0.
That is, the derivative exists for all x but is not continuous at x=0.

Nod, Nod.

I misunderstood the question. Sorry!
 

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