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

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.

• Have you seen an example of such a function? – Eric Wofsey Jun 6 '17 at 22:34
• 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.) – Cheerful Parsnip Jun 6 '17 at 22:35
• 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. – Michael Burr Jun 6 '17 at 22:40
• Possibly of interest: Why does existence of directional derivatives not imply differentiability? – Andrew D. Hwang Jun 7 '17 at 0:58
• 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? – user337992 Jun 7 '17 at 8:41

• 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. – DanielWainfleet Jun 7 '17 at 0:56