# Directional derivatives at critical points

x*abs(y)*(y+x^2+x)=f(x,y)
so, on normal points they are tangent vectors on some point in the chosen direction. how about in critical points, where there shouldnt be any on a geometrical standpoint? can i say they exist if i get them with the definition? or the result i get has no value?
like for example on f(0,0), i get 0 for any derivative. do i say they exist and their value is 0 or what?
when a directional derivative dont exist?

fresh_42
Mentor
x*abs(y)*(y+x^2+x)=f(x,y)
so, on normal points they are tangent vectors on some point in the chosen direction. how about in critical points, where there shouldnt be any on a geometrical standpoint? can i say they exist if i get them with the definition? or the result i get has no value?
like for example on f(0,0), i get 0 for any derivative. do i say they exist and their value is 0 or what?
when a directional derivative dont exist?
I'm not sure they don't exist, the graphic on wolframalpha looks pretty smooth. But anyway, you could have chosen another example. If you have more than one limit, aka possible tangent, then you have more than one directional derivative. It always depends on the path on which you approach your point, critical or not. It's the total derivative which creates the problems or if you get different values in the same direction, depending on left and right.

WWGD
Gold Member
It seems the abs|y| would give you a problem with ##f_y## at 0. But it seems you can check using the product rule can't you?

Last edited by a moderator:
I'm not sure they don't exist, the graphic on wolframalpha looks pretty smooth. But anyway, you could have chosen another example. If you have more than one limit, aka possible tangent, then you have more than one directional derivative. It always depends on the path on which you approach your point, critical or not. It's the total derivative which creates the problems or if you get different values in the same direction, depending on left and right.
what do you mean with more then one limit?
It seems the abs|y| would give you a problem with ##f_y## at 0. But it seems you can check using the product rule can't you?
im not familiar with the terminology, im not a native english speaker, you mean the "gradient formula"?

Last edited by a moderator:
Ray Vickson
Homework Helper
Dearly Missed
x*abs(y)*(y+x^2+x)=f(x,y)
so, on normal points they are tangent vectors on some point in the chosen direction. how about in critical points, where there shouldnt be any on a geometrical standpoint? can i say they exist if i get them with the definition? or the result i get has no value?
like for example on f(0,0), i get 0 for any derivative. do i say they exist and their value is 0 or what?
when a directional derivative dont exist?

I cannot parse exactly what your question is, but in this example we have:
(1) the function is differentiable at ##(0,0)##; all its directional derivatives equal zero at that point.
(2) the function is not differentiable at points of the form ##(x,0), x \neq 0.## At such points there are directional derivatives in all directions. However, the directional derivative in the direction ##(k_x,k_y)## is not a linear function of ##(k_x,k_y),## and that is why ##f## is not differentiable at such points.
(3) The function is continuously differentiable at all points ##(x,y) , y \neq 0.##

Kenneth1997 and fresh_42
WWGD
Gold Member
I cannot parse exactly what your question is, but in this example we have:
(1) the function is differentiable at ##(0,0)##; all its directional derivatives equal zero at that point.
(2) the function is not differentiable at points of the form ##(x,0), x \neq 0.## At such points there are directional derivatives in all directions. However, the directional derivative in the direction ##(k_x,k_y)## is not a linear function of ##(k_x,k_y),## and that is why ##f## is not differentiable at such points.
(3) The function is continuously differentiable at all points ##(x,y) , y \neq 0.##
What do you mean by direction ##(k_x,k_y)##?

Last edited by a moderator:
fresh_42
Mentor
what do you mean with more than one limit?
As in ##x \longmapsto |x|## where you have ##-1## from the left and ##+1## from the right; both in ##x-##direction.

WWGD
Gold Member
As in ##x \longmapsto |x|## where you have ##-1## from the left and ##+1## from the right; both in ##x-##direction.
Oesnt this just mean that the limit does not exist?

fresh_42
Mentor
Oesnt this just mean that the limit does not exist?
In this case yes. I actually had in mind a curve crossing itself, but didn't remember the formula.

i explained myself horribly. what im asking is: there is a way to know when the directional derivatives dont exist?

What do you mean by direction #(k_x,k_y)#?
i think he just mean a generic direction

Ray Vickson
I mean that when you go away from point ##(x_0,y_0)## along a "step" vector ##(k_x,k_y)## you will be looking at points of the form ##(x_0 + k_x t, y_0 + k_y t)##, where ##t## is a real scalar. The distance moved would be ##|t| \sqrt{k_x^2 + k_y^2}## and the actual direction would be given by the unit vector $$\frac{(k_x,k_y)}{\sqrt{k_x^2+k_y^2}}$$