Can the Subdifferential of a Non-Convex Function at a Point be Computed?

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Define f:R->R by
f(x) = { x^2 sin(1/x) x!=0, 0 x=0
Compute subdifferential f(0)

I went through my notes on subdifferentiation and still do not have a clue how to do this process, is there a formula to directly do this? any help would be greatly appreciated.
 
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On the entire of the world wide web, the first result on "subdifferentiation" comes from this thread. What exactly is subdifferentiation?
 
Wikipedia says: http://en.wikipedia.org/wiki/Subderivative

(I know what people are going to say; don't link to wiki; but there doesn't really seem to be another website on this this is openly accessible!)
 
Ok with that definition:

What makes you (the OP) think that we can find the subderiavative at zero? Is the function convex...
 
the function is not convex. but it is an assignment question so it must be doable some how.
so i looked at the wiki you sent me and saw the example. but then i guess it is not the same
as my question as that function is convex and mine is not.
Example i am referring to:
Consider the function f(x)=|x| which is convex. Then, the subdifferential at the origin is the interval [−1, 1].

But with that said, I graphed the x^2sin(1/x) function to try to see the behaviour around 0, and both sides are approaching 0 and the slope seems to be also 0.. but i don't know if this is how to actually do it.

what are your thoughts after that wiki read?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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