0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Michael Burr Sep 25 '17 at 8:52
  • $\begingroup$ possibly related. $\endgroup$ – Michael Burr Sep 25 '17 at 8:54
1
$\begingroup$

Example:

$f(x,y)=\frac{xy^2}{x^2+y^4}$ 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 $x \to 0+$)

The directional derivative only measures the derivative along straight lines ! But not along curves.

$\endgroup$
0
$\begingroup$

Infact you can prove that if partial derivative exist and they are bounded then function is continuous. You can try that it is simple excercise of epsilon delta and triangle inequality and mean value theorem for partial derivative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.