Suppose f(x,y) is continuous for all $(x,y) \neq (a,b)$, (not continuous at (a,b)), can the partial derivative with respect to x (or y) at (a,b) still exist?
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where at (a,b)?If it is so, It will not exist at (a,b). – Ri-Li Sep 26 '14 at 04:19
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Yes, look at this example:
$f$ is not continuous at $(0,0)$, though the partial derivatives exist, as you can show
$f(x,y) = \begin{cases} \frac{xy}{x^2-y^2} &, x\neq \pm y \\ 0 &, x = \pm y \end{cases}$
The Jacobian Matrix exists at $(0,0)$, (use definition to show that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist).
Aaron Maroja
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"(not continuous at (a,b))", that's what user110617 says. You are right though. – Aaron Maroja Sep 26 '14 at 06:11
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1What I meant is that the question starts out with "Suppose $f(x,y)$ is continuous for all $(x,y) \neq (a,b)$", and this requirement is not satisfied by your example. – Hans Lundmark Sep 26 '14 at 11:52
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Just to show that the partials derivatives exist, that was the first example that came to mind, as I said you are right. – Aaron Maroja Sep 26 '14 at 13:31
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Another example is $$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & \text{ if}\ (x,y)\neq (0,0) \\ 0 & \text{ if}\ (x,y)=(0,0) \end{cases}$$
orangeskid
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