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If f is differentiable, is f ' continuous?

  1. Oct 6, 2009 #1
    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. jcsd
  3. Oct 6, 2009 #2


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    Re: Continuity.

    x2sin(1/x) if x is not zero
    0 if x is zero
  4. Oct 6, 2009 #3
    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?


    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
  5. Oct 7, 2009 #4
    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.
  6. Oct 7, 2009 #5
    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.
  7. Oct 8, 2009 #6


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    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
    [tex]\frac{h^2 sin(1/h)}{h}= h sin(1/h)[/tex]
    for [itex]h\ne 0[/itex] 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 [itex]x\ne 0[/itex], the derivative is [itex]2xcos(1/x)+ sin(1/x)[/itex] 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, [itex]\lim_{x\to a^+}f'(x)[/itex] and [itex]\lim_{x\to a^-}f'(x)[/itex], 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.
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