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Does $\frac{\partial^2{z} }{\partial{x} \partial{y} }$ always equal $\frac{\partial^2{z} }{\partial{y} \partial{x}}$?

I find myself in situations where using one will be easier and faster than the other.

Orestes Dante
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  • You have a plate in which one candy and one biscuit are available. Eat either first candy then biscuit or vice versa. Final answer will be empty plate! Haha.. sorry – learningstudent Jun 09 '19 at 16:00
  • Quite funny. Thanks for the illustration. But this is maths you know. @learningstudent – Orestes Dante Jun 09 '19 at 16:09
  • @coreyman317 The link you cite points out the answer is "no." The OP asks if they are always equal. – saulspatz Jun 09 '19 at 16:16
  • It's a well-known fact in multivariable calculus that the mixed partials are equal. – Michael Rybkin Jun 09 '19 at 16:16
  • @MichaelRybkin: ...under some conditions! See here for a well-known counterexample: https://math.stackexchange.com/questions/219759/show-that-both-mixed-partial-derivatives-exist-at-the-origin-but-are-not-equal – Hans Lundmark Jun 09 '19 at 16:49
  • @HansLundmark That's absolutely correct. I forgot to mention that. Thank you. – Michael Rybkin Jun 09 '19 at 17:35

1 Answers1

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Since you are probably dealing with elementary functions, this will hold true. But I will point out some caveats. Let's not forget the obvious: the second partial derivatives must exist at the given point you are evaluating the expression.

Additionally, if the partials are continuous at that point, and analogously, if the $n^{\text{th}}$ partials are continuous at the point of evaluation, the $n^{\text{th}}$ degree mixed partials will commute. However, if this is not the case (it is most of the time), then you will have to resort to other methods to determine the truth of this statement.

Here's an example where Clairaut's Theorem fails.

paulinho
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  • Isn't the continuity of second partials only a sufficient condition for mixed partials to be equal, not a necessary one? In other words, is the statement “must be continuous” really true? – Hans Lundmark Jun 09 '19 at 16:51
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    You are correct. I will update the answer. – paulinho Jun 09 '19 at 16:53