Existence and Mixed derivatives

In summary, Clairaut's theorem states that for any function f there exists a function g such that d/dx(gf/gy) = d/dy(gf/h). This result is called Clairaut's theorem, and it merely requires that all the second partial derivatives are continuous.
  • #1
Physics_wiz
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I remember before reading bits and pieces about how if we have a function of two variables, say f = f(x,y), then it must be true that d/dx(df/dy) = d/dy(df/dx), where the "d"'s are partials.

Can anyone guide me to what this theorem is called or to its implications? Also, does it work in reverse? i.e. if it is true that d/dx(df/dy) = d/dy(df/dx) for some function f, then does f necessarily exist?
 
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  • #2
This result is called Clairaut's theorem, and it merely requires that all the second partial derivatives are continuous. The reciprocal of this theorem is not true, since there is at least one function with a pair of 2nd partial derivatives equal at a point while at least one of the 2nd derivatives is not continuous at that point.
 
  • #3
Physics_wiz said:
I remember before reading bits and pieces about how if we have a function of two variables, say f = f(x,y), then it must be true that d/dx(df/dy) = d/dy(df/dx), where the "d"'s are partials.
Provided the second partials are continuous.

Can anyone guide me to what this theorem is called or to its implications? Also, does it work in reverse? i.e. if it is true that d/dx(df/dy) = d/dy(df/dx) for some function f, then does f necessarily exist?
If f does not exist then what in the world would you mean by "some function f"? Have you miswritten?
 
  • #4
Yes, I see now how I wrote doesn't make sense. I was trying to use this fact to solve the problem in my last post of the "Expressing multi-variable functions" Thread, but I guess I can't use this fact to check for whether a function exists or not.
 

1. What is the concept of "existence" in terms of mixed derivatives?

The concept of "existence" in terms of mixed derivatives refers to the existence of a function that satisfies certain conditions in a given domain. In other words, it is the existence of a solution to a partial differential equation that involves mixed derivatives.

2. What are mixed derivatives and how are they different from ordinary derivatives?

Mixed derivatives are derivatives of a function with respect to multiple variables. They are different from ordinary derivatives in that they involve taking the derivative with respect to more than one independent variable at a time.

3. Why are mixed derivatives important in mathematics?

Mixed derivatives are important in mathematics because they allow us to study the rate of change of a function in multiple directions. This is particularly useful in fields such as physics and engineering, where problems often involve multiple variables.

4. Can mixed derivatives be calculated using the traditional derivative rules?

No, mixed derivatives cannot be calculated using the traditional derivative rules. Instead, they require the use of more advanced techniques, such as the chain rule or the product rule, depending on the specific problem at hand.

5. How are mixed derivatives used in real-world applications?

Mixed derivatives are used in a variety of real-world applications, including physics, engineering, economics, and statistics. For example, they can be used to model the flow of fluids, calculate the rate of change of temperature in a room, or analyze the relationship between multiple variables in a statistical study.

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