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Why some functions $f(x,y)$ can be discontinuous but its partial derivatives still could exist?

I am slightly confused,... the relationship between continuity, limits, partial derivatives and differentation. I don't understand that from the definition very well.

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  • $\begingroup$ Have you seen an example of such a function? $\endgroup$ – Eric Wofsey Jun 6 '17 at 22:34
  • $\begingroup$ Jason's function at math.stackexchange.com/questions/1749822/… is an example of a function whose partial derivatives exist, but it's not continuous at the origin. (Exercise.) $\endgroup$ – Cheerful Parsnip Jun 6 '17 at 22:35
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    $\begingroup$ Directional derivatives correspond to just looking at a single path. On the other hand, discontinuities consider the entire disk around a point. So, one is looking at a one-dimensional path while the other is looking at a two-dimensional region. $\endgroup$ – Michael Burr Jun 6 '17 at 22:40
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    $\begingroup$ Possibly of interest: Why does existence of directional derivatives not imply differentiability? $\endgroup$ – Andrew D. Hwang Jun 7 '17 at 0:58
  • $\begingroup$ So if I differentiate my function I differentiate it whole (all directions, so the whole function must be continuous) but if I just take the partial derivatives I do it by pieces in some fixed direction I want? $\endgroup$ – user337992 Jun 7 '17 at 8:41
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The loose, hand-wavy answer to all of this is that once you move beyond functions of a single variable, the number of possible ways to go wrong explodes.

Let’s take limits and continuity. For single-variable functions, the limit exists at a point if you get the same value approaching that point from the left or the right. If this value matches the function’s value at that point, it’s continuous as well. With a multivariable function, not only is there an infinite number of directions that all have to agree, but the value you get along any path that approaches the point has to be consistent as well. So, for instance, you might have a function of two variables for which the limit as you approach a point along any straight line is the same, but you get a completely different answer if you follow a parabolic path.

Differentiation is more of the same in that the linear approximation to the change in the function’s value has to be consistent for all ways that you can approach the point. Partial derivatives are really directional derivatives in the directions of the positive coordinate axes. So, when you compute a function’s partial derivatives, out of all of the infinite possible paths that approach a point you’re only looking at how the function behaves along a specific finite set of lines through the point. That’s not nearly enough to guarantee that the function is continuous or differentiable, since it could be doing something wildly different along all of those other paths that you haven’t examined. A relative of the function I posited in the previous paragraph not only has partial derivatives at the point, but every directional derivative exists there, and yet it’s not even continuous, let alone differentiable.

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    $\begingroup$ Nicely phrased...... For the O.P.: Consider $f(x,y)=xy/(x^2+y^2)$ when $x^2+y^2\ne 0,$ and $ f(0,0)=0.$ If $x\ne 0$ then $f(x,x)=1/2$ but $f(x,0)=0 $, so $f$ is discontinuous at $(0,0).$ But $f$ has continuous partial derivatives at every point. $\endgroup$ – DanielWainfleet Jun 7 '17 at 0:56

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