# If f is differentiable, is f ' continuous?

1. Oct 6, 2009

### l'Hôpital

So, a certain discussion occured in class today...

If f is differentiable, is f ' continuous?

At first sight, there seems no reason to think so. However, we couldn't think any counterexample. It also seems logical that f' is continuous since otherwise f wouldn't be differentiable.

For example, suppose f(x) = ln x, for x > 0

Then f'(x) = 1/x. Yes, this is discontinuous, but it's not for the domain x > 0.

So, the question remains:

If f is differentiable, is f ' continuous?

2. Oct 6, 2009

### Office_Shredder

Staff Emeritus
Re: Continuity.

x2sin(1/x) if x is not zero
0 if x is zero

3. Oct 6, 2009

### l'Hôpital

Re: Continuity.

^Haha, yes, I was just talking about that with a friend.

You mean, f(x) = x2 sin(1/x) for x =/= 0, 0 otherwise, right?

EDIT:

However, wouldn't that make f not differentiable since it's not differentiable at x = 0?

For example, f(x) = l x l , it's not differentiable at x = 0, thus we deem it "not differentiable". Wouldn't the same apply here?

EDIT #2: Nevermind, just an error. Alright. I think I got it for sure now! Haha.

Last edited: Oct 6, 2009
4. Oct 7, 2009

### Werg22

Re: Continuity.

If a function is differentiable at a point, it is necessarily continuous at this point. To see this, recall the definition of a limit:

lim h->0 f(x+h) - f(x) / h

Since it presumably exists, and the denominator goes to 0, lim h->0 f(x+h) - f(x) = 0. From this, it's clear the function is continuous at x.

So if a function is differentiable (everywhere), it's continuous.

5. Oct 7, 2009

### chislam

Re: Continuity.

Remember though, that although the above is true, "if a function is continuous, then it is differentiable" is not necessarily true. Consider y = |x| at x = 0.

6. Oct 8, 2009

### HallsofIvy

Re: Continuity.

That is true but was not the original question. The question was, if a function is differentiable at x= a is the derivative continuous there.

Office Shredder's answer was "no, the derivative is not necessarily continuous".

And, l'Hopital, it is not the same situation as |x|.

The difference quotient, at x= 0, is
$$\frac{h^2 sin(1/h)}{h}= h sin(1/h)$$
for $h\ne 0$ and that goes to 0 as h goes to 0. Unlike |x|, the derivative at x=0 does exist and is 0.

That derivative is not continuous at x=0 because, for $x\ne 0$, the derivative is $2xcos(1/x)+ sin(1/x)$ and that has no limit as x goes to 0.

It is, however, true that the derivative must satisfy the "intermediate value property". That is, if f' exists on [a, b], then it must take all values between f'(a) and f'(b). That means, in particular, that if the two one sided limits, $\lim_{x\to a^+}f'(x)$ and $\lim_{x\to a^-}f'(x)$, exist, they must be equal and equal to the value of the derivative at a (in which case, f' is continuous there. The only way a derivative can exist at a point and not be continuous there is if the two one sided limits themselves do not exist, as in Office Shredder's example.