Clairauts “equality of mixed partial derivatives” theorem

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SUMMARY

Clairaut's theorem, established in 1743, asserts that if the first partial derivatives of a function are continuous, then the mixed partial derivatives are equal, expressed mathematically as ∂/∂y(∂f/∂x) = ∂/∂x(∂f/∂y). The theorem emphasizes the symmetry of change in multivariable calculus. While visualization of this concept is beneficial, understanding the underlying conditions for its validity is essential for both mathematics and physics students. The discussion highlights the importance of knowing the conditions under which mathematical models operate, particularly in applied fields like physics and engineering.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with multivariable calculus concepts
  • Knowledge of continuity in mathematical functions
  • Basic principles of mathematical modeling in physics
NEXT STEPS
  • Study the implications of Clairaut's theorem in multivariable calculus
  • Explore the concept of continuity and its role in calculus
  • Learn about the significance of mixed partial derivatives in physical models
  • Investigate visualization techniques for understanding multivariable functions
USEFUL FOR

Mathematics students, physics students, engineers, and anyone interested in the application of calculus in real-world scenarios will benefit from this discussion.

davidbenari
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I know how to prove this via limits and I'm okay with that.

What I want to understand is the interpretation of the theorem and specifically a visualisation of why what the theorem states must be the case.

My guess is that this theorem is saying that change is symmetrical. But I don't know if this is only true for second derivatives.

If you don't know this theorem by its name the theorem basically says this:

∂/∂y(∂f/∂x)=∂/∂x(∂f/∂y)

Also, I would like to know if you consider my focus on visualisation to be not worthwhile and that I should instead just trust this theorem.

I thank you in advance.
 
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I am a physics student. I have a question about this "if you are a mathematics student then it is worthwhile" stuff. Would a great physicist, say like Feynman, know why this is so? Even if your answer is speculative, what do you think?
 
If the math is a model of some physical situation, you can often argue that derivatives are continuous, etc, on physical grounds. Usually, a math model only hits the "exceptional" conditions that mathematicians like to understand completely, if it's a poor model of the physics.

IMO it is always worthwhile (not to say essential) to know something about the conditions that make your math valid. But as a physicist or engineer you don't necessarily need to know the most general set of conditions that make it valid, or be able to prove why it is valid.

Of course you can never know "too much" math, but in real life, whether you learn more math or more physics is a time management problem.

(Full disclosure: I've seen both sides of this first hand - I have a math degree, and spent most of my life working on engineering problems).
 
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