Assume is a real-valued function defined on an open set of and all partial derivatives of exist on . Is it possible that there is some and such that the directional derivative of in the direction of at does not exist?
Thanks!
Assume is a real-valued function defined on an open set of and all partial derivatives of exist on . Is it possible that there is some and such that the directional derivative of in the direction of at does not exist?
Thanks!
If f is differentiable at a point, then the directional derivative in all directions exists. But, as Opalg showed, it may happen that has partial derivatives at a point but is not "differentiable" there.
Another example is
At (0,0) both partial derivatives exist and are 0. But the function is not differentiable and directional derivatives do not exist except in the directions of the axes.
Umm, I hate to disagree, but that example does not have partial derivatives at the origin:
(ouch!)
I think you probably meant But no, on second thoughts that won't work either, because that function doesn't have partial derivatives on either of the axes (except at the origin). I have a feeling that the example I gave previously can't really be simplified.
Thank you! Now I have a further question: Assume the conditions in my first post still hold, that is, is a real-valued function defined on an open set of and all partial derivatives of exist on . If the directional derivative of in the direction of at a point does exist, is it always true that the value of the directional derivative equals the inner product ? Thanks!
Ouch! You are right. That is the example I meant to give. It does have partial derivatives at (0,0) but is not differentiable there. exists at (x, 0) but not and vice-versa at (0, y).
That is not a good example because the initial post asked for an example where the partial derivatives exist on some open set. Thank you.