Directional derivatives at critical points

  • #1
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?
 

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  • #2
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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.
 
  • #3
WWGD
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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?
 
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  • #4
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"?
 
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  • #5
Ray Vickson
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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.##
 
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  • #6
WWGD
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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)##?
 
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  • #7
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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.
 
  • #8
WWGD
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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?
 
  • #9
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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.
 
  • #10
i explained myself horribly. what im asking is: there is a way to know when the directional derivatives dont exist?
 
  • #11
What do you mean by direction #(k_x,k_y)#?
i think he just mean a generic direction
 
  • #12
Ray Vickson
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What do you mean by direction #(k_x,k_y)#?
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 $$ \frac{(k_x,k_y)}{\sqrt{k_x^2+k_y^2}}$$

Different sources seem to disagree about whether you need to have a unit vector ##\vec{k}## when definining the directional derivative. I have seen some authors say your directional derivative should, in fact, depend only on the direction, so they would demand use of a unit vector. I have seen other authors define the directional derivative with respect to the whole vector ##\vec{k}##, so for them the directional derivative depends on both the direction and the magnitude of the step ##\vec{k}##. Personally, I tend to favor the latter, but it is easy to pass back and forth between the two.
 
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