20060625, 16:09  #12 
May 2005
Argentina
2·3·31 Posts 
Differentiable but not C1
In Marsden's Vector Calculus book, it says that a function is called C1 at a point A, if its partial derivatives exist and are continious in the neighborhood of A.
It also says that if a function is C1 at A, then it is differentiable at A. But that a function been differentiable not necesarily is C1. Has anybody a counterexample for showing this? I mean, a function been differentiable at a point, but not been C1 there. Because by reading "upside down" the proof of C1 imply dif, I think it also demostrates that dif implies C1, so C1 and dif are equal. Thanks. 
20060625, 21:59  #13 
Dec 2003
Hopefully Near M48
3336_{8} Posts 
To answer your question Damian, just because a function is differentiable doesn't mean that the derivative is continuous. If I am not mistaken, this is equivalent to saying that there exist discontinuous functions that have an antiderivative.
Last fiddled with by jinydu on 20060625 at 22:04 
20060626, 15:14  #14 
May 2005
Argentina
2·3·31 Posts 
Can you give me an example of a function that is differentiable in the neighborhood of a point, but whose derivative isn't continious there?

20060701, 22:26  #15  
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
Quote:
This is easily differentiable, but it's derivative is not continuous where cos(x) = 0 

20060701, 22:32  #16  
Jun 2005
2·191 Posts 
Quote:
How about abs(x) at x=0, instead. 

20060702, 23:48  #17 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Take your favorite piecewiesecontinuous (but not continuous) function and integrate it. That should do the trick.

20060704, 13:06  #18 
May 2005
Argentina
2×3×31 Posts 
Thanks for all your answers.
Jinydu, I think your example does the trick. Anyway I found a good example of a function differentiable but with derivative not continious here: http://pirate.shu.edu/projects/reals/cont/fp_c1.html 
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