# directional derivatives and continuity

FOR A TWO VARIABLE FUNCTION $f(x,y)$,

I understood that the directional derivative at a point $(x_0,y_0)$ along direction of $\vec u$ is the derivative of the 2D graph that we obtain on the plane kept perpendicular to $xy$ plane passing through the point $(x_0,y_0)$ and parallel to u.

But for 2D graphs to be differentiable , it must be continuos.which means if directional derivative exist along some direction, then the function should be continuous in that direction.

Then how this statement is true:"a fun may be discontinuous even if all directional derivative exists"??

IT WOULD BE REALLY HELPFUL IF U CAN EXPLAIN WITH MY APPROACH

• The basic idea is that the directional derivative only measures the derivative along straight lines, but the surface could be pinched at a point so that the derivative exists along straight lines, but not along curves. – Michael Burr Sep 25 '17 at 8:52
• – Michael Burr Sep 25 '17 at 8:54

$f(x,y)=\frac{xy^2}{x^2+y^4}$ if $(x,y) \ne (0,0)$ and $f(0,0)=0$
Prove that the directional derivative at the point $(0,0)$ along any direction exists.
But $f$ is in $(0,0)$ not continuous (to this end show that $f(x,\sqrt{x}) \to 1/2$ for $x \to 0+$)