Is the Derivative of a Differentiable Function Always Continuous?

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The discussion centers on whether the derivative of a differentiable function is always continuous. It establishes that if a function f is differentiable at a point, it is also continuous at that point, but the continuity of the derivative is not guaranteed. Participants explore examples of functions that are continuous and differentiable yet have discontinuous derivatives, such as the Weierstrass function and f(x) = (x^2) * sin(1/x). The conversation highlights the complexity of the relationship between differentiability and the continuity of derivatives, emphasizing that while the first two statements are true, the third is not. The exploration of these concepts illustrates the nuanced nature of calculus and the behavior of functions.
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Here is a problem that I am having trouble with.

Let f be a function such that:

Limit as h -->0 [f(2 + h) - f(2)] / h = 5

(essentially that: f ' (2) = 5 ).


It then asks: which of the following must be true?

1) f is continuous at x = 2.

2) f is differentiable at x = 2.

3) The derivative of f is continuous at x = 2.


THE ANSWERS: The first two are true, but the last one is false.

I understand the first two statements and how they must be true, but I also think the third statement must also be true. I can't think of an example of a function that disproves number 3. Can you?

I've tried drawing some possible functions for f ' that has a discontinuity (asymptote, jump, point) at x = 2 but whose antiderivative f is still continuous and differentiable at x = 2, but I can't come up with an example that works!

Thanks for any help
 
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Unless I'm missing something, this doesn't seem like a simple problem. Consider a function that has derivatives only at 0, and points of the form 1/n, where n is a non-zero integer. Suppose its derivative at 0 is 0, and is 1 at all those other points. Then such a function has a derivative at 0, but it is discontinuous there, since any open set around 0 will contain some of the points of the form 1/n, but the derivatives there will always be 1, so the limit does not exist. Now, can a continuous function have such a weird "derivative"? Well, the Weierstrass Function is continuous but only has derivatives on a set of measure 0. The set consisting of 0 and 1/n for all non-zero integers n is a set of measure 0 (this is not a definition of measure 0, just an example of one). So there is a continuous function that has a derivative that is weird in some related sense, but that doesn't mean that it is "weird enough", and the Weierstrass function may still have a continuous derivative. Also, the discontinous derivative I suggested is one way it may be discontinuous, there may be other ways (like a jump discontinuity, etc.).
 
I googled this up to the surface:

An example of a function which is differentiable at 0, but whose derivative is not continuous at 0

f(x) = (x^2) * sin(1/x), with f(0) = 0

I haven't worked out the details. Perhaps someone else would like to try.
 
Last edited:

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