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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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    @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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    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