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is $\exp \left( - 1 \over x^2 \right ) $ differentiate at $x=0$. Wolframlapha says it is. But is it continuous since we have $1 \over x^2$ and $x=0$? Can we really do this $f(0) = \exp \left( - {1 \over 0} \right)$? I hope I'm making any sense.

EDIT:: Does differentiability necessitate continuity? Is above function continuous at $x=0$.
Also a function $f(x) = {|x| \over x}$ also seems to be differentiable at $x=0$ as mentioned below in comment.

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    It is a common misconception that continuity has something to do with evaluating at a point. It is nothing to do with this, it is about how the function behaves "approaching" a point.2012-09-20
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    @fretty doesn't differentiability necessiate contuinity?2012-09-20
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    Yes but you show in your post that you do not understand continuity...2012-09-20
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    The fact that you cannot evaluate $f$ at $0$ does not imply discontinuity.2012-09-20
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    Depends on how strict your definition of differentiable is. You can just add into the definition $f(0)=0$ and everything will be fine by every definition.2012-09-20
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    @NickAlger no ... we are not defining $f(0) = 0$2012-09-20
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    @fretty could you please be more elaborate. I indeed thought that it would be discontinuous at $f(0) = 0$2012-09-20
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    Well then you just need to go back to the exact definition of differentiability you are using, and see if it satisfies those properties. (eg, what is the function space $f$ is defined to live in, are you allowed to redefine the function on a set of measure zero, etc)2012-09-20
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    hmmm @NickAlger we are defining the domain of $f(x)$ on real numbers.2012-09-20
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    What is the space that $f$ maps to, and what is the definition of differentiability you are using?2012-09-20
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    @NickAlger this is just normal real-to-real function. I'm using the usual definition of differentiability. Just there seem [to be a theorem](http://mathforum.org/library/drmath/view/53637.html) that says the differentiability implies contuinity at that interval. I don't understand why this function defined from real to real ... is not continuous but differentiable at x=02012-09-20
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    Normally the definition of differentiability of a function in a point requires that the function be defined in that point. Would you consider the function $x\mapsto\frac{|x|}x$ on $\mathbb R\setminus\{0\}$ to be differentiable at $0$?2012-09-20
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    @MarcvanLeeuwen amazing I didn't really know that ... could you add a bit of explanation? why this is differentiable at x=0 but not continuous?2012-09-20
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    Under normal definitions, $f$ as defined is not actually a function $\mathbb R \rightarrow \mathbb R$, so again under normal definitions there is no meaning to the statement "derivative of $f$ at 0". There are other definitions where it's OK since the "problems" with the function are trivially fixed. What's important is that you understand the definitions you are using, and communicate them clearly whenever confusion might arise.2012-09-20
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    @MonkeyD.Luffy: I think you misunderstood my rhethoric question: I'm saying that the mentioned function is _not_ differentiable at $0$ because it is not defined in $0$. The "normal" definition of differentiability of $f$ at $a$ is that $\lim_{h\to0}\frac{f(a+h)-f(a)}h$ exists, and at the very least this supposes that $f(a)$ is defined. Are you using another definition?2012-09-20

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The function $$ x \mapsto e^{-\frac{1}{x^2}} $$ can be extended in a continuous way to $x=0$. This extended function turns out to be differentiable (and even $C^\infty$). Strictly speaking, the original function is undefined at $x=0$, so that it is perfectly meaningless to ask whether it is differentiable.

About you edited question: yes, in the common definition of derivative, the function must be defined at the point where you are computing the derivative. And the common definition implies that a differentiable function is always continuous. Hence, no hope to differentiate a discontinuous function! I know that it is customary to confuse a disconinuous function with its extension by continuity, if this extension exists. But, as we see in this discussion, it is a dangerous abuse, at least for beginners.

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    what is this *extended function* ?2012-09-20
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    @MonkeyD.Luffy In this case, you simply put $f(0)=0$.2012-09-20
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    can we treat all jump discontinuities this way?2012-09-20
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    @MonkeyD.Luffy Jump discontinuities are never removable!2012-09-20
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    woops sorry!! i mean the removable discontinuity (where limit exits but $f(x) \neq L$ )? also could you add another answer regrading [this](http://www.wolframalpha.com/input/?i=differentiate+abs%28x%29%2Fx+at+x%3D0) and [this](http://www.wolframalpha.com/input/?i=plot+abs%28x%29%2Fx+from+-1+to+1)2012-09-20
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    From a mathematical viewpoint, the sign function $x \mapsto |x|/x$ is discontinuous at $x=0$, and there is no way to define it at zero so that it becomes continuous.2012-09-20
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    so this is not differentiable at $x=0$?? right??2012-09-20
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    Actually, you *could* define a derivative which also is well defined at isolated undefined points, provided that the function can be extended to an ordinarily differentiable function there: Just use $\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ instead of the standard definition. Indeed, with that definition, you could even derive non-continuous functions, as long as they only differ from continuous ones in isolated points (this is similar to the Lebesgue integral which also doesn't care about localized changes, although there you've got even more freedom).2012-09-20
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    thank you @celtschk I will remember this for future.2012-09-20
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    In some settings there's good reason to want the derivative of a jump to exist and be some sort of "impulse" ("delta function"), though doing this rigorously involves the theory of distributions.2012-09-20