Are the following two derivatives same?

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Discussion Overview

The discussion centers on the equality of mixed partial derivatives, specifically whether the derivatives fxy and fyx are equal in general. Participants explore the conditions under which this equality holds, referencing continuity and differentiability criteria.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that if the mixed partial derivatives fxy and fyx exist and are continuous, then they are equal, referencing the concept known as Clairaut's theorem.
  • Others clarify that continuity of fxy and fyx in a neighborhood of a point is sufficient for their equality at that point.
  • A later reply elaborates on the conditions required for the equality, stating that if certain differentiability and continuity conditions are met, then the mixed partial derivatives are equal at that point.

Areas of Agreement / Disagreement

Participants generally agree on the conditions under which the mixed partial derivatives are equal, though the discussion includes varying levels of detail regarding those conditions.

Contextual Notes

The discussion does not resolve potential limitations related to the assumptions of continuity and differentiability, nor does it address specific cases where these conditions might not hold.

sgsawant
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1. fxy
2. fyx


Are the above 2 derivatives equal, in general. Please explain if you know the answer.

Regards,

-sgsawant
 
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Yes, in general if these two derivatives exist and are continuous, then they are equal. This is called "equality of mixed partial derivatives". I have also seen it called Clairaut's theorem, although supposedly it was first proved by Euler (like so much of the rest of mathematics).
 
As long as [itex]f_{xy}[/itex] and [itex]f_{yx}[/itex] are continuous in some neighborhood of a point, then, at that point, they are equal.
 
Even better:

Let U in R^2 be open, [itex]f:U\to\matbb{R}[/itex] partial differentiable w.r.t. both variables, and [itex]D_1f[/itex] partial differentiable w.r.t the second variable. Suppose further that (x,y) is in V, and [itex]D_2D_1f[/itex] is continuous at (x,y). Then [itex]D_2f[/itex] is partial differentiable w.r.t the first variable at (x,y), and

[tex]D_1D_2f(x,y)=D_2D_1f(x,y)[/tex].

i.e. we only need [itex]D_2D_1f[/itex] to exist and be continuous at some point in the interioir, this already implies that [itex]D_1D_2f[/itex] exists at that point and the two are equal.
 
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