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I need to check whether following function is differentiable at (0,0) $$ F(x,y) = \begin{cases} \exp\left(-\frac{1}{x^2+y^2}\right) & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases} $$ My attempt:

I checked that it is continuous at origin. Also to check differentiability I checked if $$\lim \frac{f (0+h,0+k)-f (0,0)-dz}{\sqrt{h^2+k^2}}=0$$ hence differentiable.

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  • $\begingroup$ The differential at $(0,0)$ is also $0$ (a $0$ linear functional, or if you want, a vector). I assume you also checked $\frac{ f(h,k)}{\sqrt{h^2 + k^2}} \to 0$. $\endgroup$ – Orest Bucicovschi Feb 4 '15 at 7:47
  • $\begingroup$ Ya I checked its 0 $\endgroup$ – singularity Feb 4 '15 at 7:48
  • $\begingroup$ What is the $dz$ in your last limit? $\endgroup$ – 5xum Feb 4 '15 at 7:53
  • $\begingroup$ dz comes to be zero $\endgroup$ – singularity Feb 4 '15 at 8:01
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The function $$f(t):=\left\{\eqalign{e^{-1/t}\quad&(t>0) \cr 0\quad&(t\leq0)\cr}\right.$$ is differentiable on all of ${\Bbb R}$ since $$\lim_{t\to0+}{e^{-1/t}-0\over t}=\lim_{\tau\to\infty}\tau\>e^{-\tau}=0\ .$$ The function $$g:\quad{\mathbb R}^2\to{\mathbb R},\qquad (x,y)\mapsto x^2+y^2$$ is differentiable; therefore the composition $F=f\circ g$ is differentiable on all of ${\Bbb R}^2$ as well.

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