I need an example of a function.

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Give an example of a function f:R2-->R , continuous in (0,0), with partial derivatives in (0,0), but not differentiable in (0,0).

I was thinking in something like f(x,y)=IxyI
 
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What is the definition of continuous function at a point like (0, 0) ?
 
limit as (x,y)->(0,0) of If(x,y)-f(0,0)I=0
 
I know what this properties mean, I just can't think of a function with partial derivatives in (0,0) but not differentiable in that point.
 
Actually, I think that the function f (x , y) = x1/3y2/3+y1/3x2/3 works fine, am I right?
 
Yeah that seems to work. I plotted it here.

image.jpg


image.jpg
 
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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