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I'm having trouble with the concept of continuous partial derivative and differentiability.

Let

$$f(x,y) =\begin{cases} 1& \text{when } xy = 0\\0& \text{when } xy \neq 0\end{cases}$$

then clearly, df/dx = 0 and df/dy = 0.

So both Partial Derivative are continuous and thus Differentiable everywhere?

using definition of Differentiability however, $\lim\limits_{(x,y) \to (0,0)} [f(x,y) - f(0,0) - 0 - 0]/[(x^2 + y^2)^{0.5}]$, the limit does not exist and therefore is not differentiable at (0,0).

So which is right?

and in the case, is $f(x,y)=\sin(x)\sin(x+y)\sin(x-y)$ differentiable at (0,0)

Thanks for all the help. Cheers!

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1 Answer 1

up vote 3 down vote accepted

Check again: $\partial f/\partial x$ isn't defined when $x=0\ne y$, and $\partial f/\partial y$ isn't defined when $y=0\ne x$, because you get $(0-1)/\Delta$ as $\Delta\to0$ when trying to evaluate the limit definition. Also, this is Math.SE; if you posted this in Overflow you'd get your question closed very quickly and redirected here.


$\sin(x)\sin(x+y)\sin(x-y)$ is differentiable because all of the partials exist and are continuous.

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Brilliant answer! Thanks a lot. Yea, i was redirected here... Best of luck to you. –  adsisco Sep 14 '11 at 6:23
    
@adsisco: No problem. Best of luck to you too. –  anon Sep 14 '11 at 6:26
    
Thanks J.M. for the great edit. btw, I toggled the tick. –  adsisco Sep 14 '11 at 6:32

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