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Given the function $$ g(x,y) = \begin{cases} \frac{xy^3}{2x^4+y^4}, \,\,\, \text{if}\,\, (x,y)\neq(0,0)\,\\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{else} \end{cases} $$ check whether $g$ is continuous and differentiable at $(0,0)$ or not.

Considering the two restrictions $y = \pm x$, it is seen $g$ is not continuous at $(0,0)$. How about differentiability? Is it related to continuity? The two partial derivatives are both $0$ at $(0,0)$. Still I'm not sure $g$ is differentiable at $0$, since I should consider any directional derivative.

Thanks for your patience and help.

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  • $\begingroup$ Yes, differentiability is related to continuity by virtue of differentiable functions being continuous. But if you don't know this fact, you shouldn't be able to use it, should you? Do you know the definition of $g$ is differentiable at $(0, 0)$? $\endgroup$ – Git Gud Apr 26 '18 at 13:05
  • $\begingroup$ Guess partial derivatives exist, but $g$ is not differentiable (otherwise it would be continuous).. There is no contradiction. Also, I can say the partial derivates are not (both) continuous at $(0,0)$. Am I right? $\endgroup$ – Francesco Gemma Apr 26 '18 at 16:03
  • $\begingroup$ Yes, because if the partials were continuous, then $g$ would be differentiable. $\endgroup$ – Git Gud Apr 26 '18 at 16:55
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Suppose that $g$ is differentiable at $(0,0)$. Then $g$ is continuous at $(0,0)$.

Conclusion ?

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    $\begingroup$ I think a generous reading of the Question (which is more detailed than your Answer) suggests the OP is concerned with understanding the distinction between the existence of partial derivatives at a point and differentiability at the point. The wording of the Question (perhaps by virtue of being an assignment) lays the groundwork for such a discussion. It would improve your Answer to sketch in more of the proper "Conclusion". $\endgroup$ – hardmath Apr 26 '18 at 13:16

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