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I stumbled upon a question involving discontinuous derivative but I couldn't understand how could it happen that the derivative of a function (let's say $f(x)$) is defined at point (let's suppose $c$) but then it is discontinuous (or undefined) at the same point $c$. Is there a problem in calculus?

(The function $f(x)$ is also continuous at $c$)

Arctic Char
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Adidas10
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  • Your question has a contradiction, when you say that $f'$ is both defined and undefined at $c$. – José Carlos Santos Nov 01 '20 at 18:53
  • "Discontinuous" doesn't mean the same thing as "undefined." I don't really understand what you're asking. – saulspatz Nov 01 '20 at 18:55
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    $f(x) = x^2\sin(1/x)$, with value $f(0)=0$, is differentiable everywhere (in particular, $f'(0)=0$) but its derivative is not continuous at $x=0$. – Greg Martin Nov 01 '20 at 18:58
  • The literal answer to your question in its current form is: "They exist because [insert proof of the comment by @GregMartin]". And that proof will be perfectly valid, therefore there is no problem in calculus. So perhaps this is still not quite what you wanted to ask? – Lee Mosher Nov 01 '20 at 19:04
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    The answers to this question have a great deal of information on such functions. – Brian M. Scott Nov 01 '20 at 19:10
  • @saulspatz Why do some continuous functions need to have discontinuous derivative function ? If I imagine the derivative at a point as a tangent to the curve and the derivative being discontinuous (assumed) at some point $c$, the slope of tangent will jump nearby $c$ but because our function was continuous it shouldn't happen like that so maybe the only possibility left is that the derivative is undefined. – Adidas10 Nov 03 '20 at 13:02

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The function $$ f(x) = \begin{cases} x^2\sin(1/x), &\text{if } x\ne0, \\ 0, &\text{if } x=0 \end{cases} $$ is differentiable everywhere (in particular, $f'(0)=0$) but its derivative is not continuous at $x=0$.

Whether this is a "problem" in calculus is in the eye of the beholder! (By way of analogy, the school of Pythagoras discovered that a real number could be defined but could not be written as a quotient of two integers. To them that might have seemed like a problem, but today we're pretty comfortable with the idea of irrational numbers.)

Certainly the drive to formalize these concepts in analysis in the 19th century was motivated in large part by "pathological" examples like this one. In the end, the definitions of continuity and differentiability we settled one do permit functions to have derivatives that are defined everywhere but not continuous everywhere, just as they permit functions that are themselves defined everywhere but not continuous everywhere.

Mathematical definitions develop in order to reflect our intuitions as much as possible within mathematics. But once we have settled on mathematical definitions, if ever their consequences run counter to our intuitions, we must be prepared to update our intuitions.

Greg Martin
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  • When we use limits to evalute the derivative of the function we get $\lim_{x \to 0} xsin(1/x) = 0$ but when we differentiate it the other way we get $cos(1/x)$ as one of the terms in its derivative function which is clearly undefined at x approaching '0' so where am I wrong. – Adidas10 Nov 03 '20 at 12:41
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Remember that it’s entirely possible for a function to be continuous at a point where the derivative is not. Moreover, take care not to confuse continuity with existence (or being well defined for that matter). There is no prerequisite for differentiability of a function based on the differentiability of its derivative. A thorough discussion can be found here: Discontinuous derivative.

AJAX
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