How to Prove the Continuity of a Derivative?

[Treadstone 71]

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?

 

[amcavoy]

Do you mean if f(x) is differentiable at every value of x in [a,b]?

 

[Hurkyl]

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)

 

[devilkidjin]

yes, if is differentiable it has to exists

this can be concluded by the domain being within it

 

[HallsofIvy]

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 [tex]x \neq 0[/itex], f(0)= 0
is differentiable at every point on (-1, 1) but that f' is not continuous at 0.

A lot of people confuse continuity of the derivative with the fact that, if a function is differentiable at a point, the function must be continuous at that point.

 

[Treadstone 71]

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.

 

[Rocketa]

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)..

 

[shmoe]

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)

 

[HallsofIvy]

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.

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
$$lim_{x\rightarrow 0}\frac{x^2sin(\frac{1}{x})}{x}= lim_{x\rightarrow 0}x sin( \frac{1}{x})= 0$$

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.

 

[Treadstone 71]

Ah yes, I was searching for the intermediate value property, thinking it was continuity. Thanks for clearing that up.

 

[mathwonk]

is f' continuous somewhere?

 

[Rocketa]

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
$$lim_{x\rightarrow 0}\frac{x^2sin(\frac{1}{x})}{x}= lim_{x\rightarrow 0}x sin( \frac{1}{x})= 0$$
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!