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I have the following function:

$f(x,y)= \ \begin{cases} (x^2 + y^2)\cdot \sin(\frac{1}{\sqrt{x^2+y^2}}) & (x,y)\neq (0,0) \\ \\0 & (x,y)=(0,0) \end{cases} \\$

Now, I am asked to prove that this function is not continuously differentiable at $(0,0)$, but it is differentiable on $R^2$. I have proven that it is differentiable by taking the first order partial derivatives, using the precise definition. It turns out that:

$\frac{\partial f}{\partial x} at (0,0) = 0$

and $\frac{\partial f}{\partial y} at (0,0) = 0$

My question is how to proceed from here and prove that this function is not twice differentiable at $(0,0)$.

mathcounterexamples.net
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Lazov
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1 Answers1

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You have only shown that $f$ is partially differentiable at $(0,0)$ ! To show that $f$ is differentiable at $(0,0)$ you have to show that

$\frac{f(x,y)-f(0,0)-xf_x(0,0)-yf_y(0,0)}{||(x,y)||} \to 0$ for $(x,y) \to (0,0)$.

Try a proof !

Then you have to show that $f$ is not continuously differentiable at $(0,0)$.

To this end show that the partial derivative $f_x$ is not continuous at $(0,0)$

Fred
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  • I have shown that it is differentiable at $(0,0)$, I just did not write it here, because the solution is long. So, it is enough to take the first-order partial derivative of f (not the one at $(0,0)$) and to show that it is not cont. at $(0,0)$? Because as I have shown the other one $\frac{\partial f}{\partial x} at (0,0) = 0$. – Lazov Dec 12 '16 at 15:29