Is the Function \( f(x,y) = \frac{x}{y} + \frac{y}{x} \) Class \( C^{1} \)?

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SUMMARY

The function \( f(x,y) = \frac{x}{y} + \frac{y}{x} \) is classified as \( C^{1} \) because its partial derivatives exist and are continuous within its domain. The domain of \( f \) is defined as \( A = \{(x,y) \in \mathbb{R}^{2} | (x \neq 0 \text{ and } y \neq 0)\} \). The confusion regarding the continuity of the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) is clarified by confirming that continuity is checked within the domain of \( f \), not the derivatives. Thus, the function meets the criteria for being of class \( C^{1} \).

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with the concept of continuity in multivariable calculus
  • Knowledge of set notation for defining domains
  • Basic principles of vector calculus as outlined in "Vector Calculus" by Marsden and Tromba
NEXT STEPS
  • Study the properties of \( C^{1} \) functions in multivariable calculus
  • Learn how to derive and analyze partial derivatives of functions of two variables
  • Explore the implications of continuity in the context of rational functions
  • Review examples of functions that are not \( C^{1} \) to understand the boundaries of this classification
USEFUL FOR

Students studying multivariable calculus, particularly those focusing on differentiability and continuity of functions, as well as educators looking to clarify concepts related to class \( C^{1} \) functions.

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Homework Statement


Decide if the function is C^{1} or just differentiable

<br /> f(x,y) = \frac{x}{y} + \frac{y}{x}<br />



Homework Equations


"A function whose partial derivatives exist and are continuous is said to be of the class C^{1}." Vector Calculus Marsden and Tromba p. 114



The Attempt at a Solution


Something I'm confused about is I would like to see the above statement be modified to "A function with domain A whose partials exist and are continuous in the domain A are said to be of class C^{1}". Is this right?

So to answer the question to task is to:
1. find the domain of f
2. find the partials and see if they exist and are continuous in the domain of f.

So as stated above one of the places I'm feeling confused is: Am I checking if df/dx is continuous in dom(f) or dom(df/dx)?

I suspect it's in the dom(f) which I would describe using set notation as:
<br /> \text{the domain }A = \{(x,y) \in R^{2} | (x \neq 0 \text{ and } y \neq 0\}<br />

So now another place where I'm confused. Attached is the solution to this question as found in the Vector Calc study guide. Looking at their solution it seems like they suggest that the only place where point not in either dom(f) or dom(df/dx) is (0,0. But I don't think this is right, I think that it should be what I wrote above. Which is right?

In this case it doesn't seem to matter too much because either way the partials are rational functions which are continuous and defined on whatever domain is right, so the function is C^{1}. But still I would like to clear up in my mind the points above.
IMG_20140117_181400.jpg
 
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mbigras said:

Homework Statement


Decide if the function is C^{1} or just differentiable

<br /> f(x,y) = \frac{x}{y} + \frac{y}{x}<br />

Homework Equations


"A function whose partial derivatives exist and are continuous is said to be of the class C^{1}." Vector Calculus Marsden and Tromba p. 114

The Attempt at a Solution


Something I'm confused about is I would like to see the above statement be modified to "A function with domain A whose partials exist and are continuous in the domain A are said to be of class C^{1}". Is this right?

So to answer the question to task is to:
1. find the domain of f
2. find the partials and see if they exist and are continuous in the domain of f.

So as stated above one of the places I'm feeling confused is: Am I checking if df/dx is continuous in dom(f) or dom(df/dx)?

I suspect it's in the dom(f) which I would describe using set notation as:
<br /> \text{the domain }A = \{(x,y) \in R^{2} | (x \neq 0 \text{ and } y \neq 0\}<br />

So now another place where I'm confused. Attached is the solution to this question as found in the Vector Calc study guide. Looking at their solution it seems like they suggest that the only place where point not in either dom(f) or dom(df/dx) is (0,0. But I don't think this is right, I think that it should be what I wrote above. Which is right?

In this case it doesn't seem to matter too much because either way the partials are rational functions which are continuous and defined on whatever domain is right, so the function is C^{1}. But still I would like to clear up in my mind the points above.
IMG_20140117_181400.jpg

I think your description of the domain is correct. Neither x nor y can be 0. It's not just ##\left( x,y \right) \ne \left( 0,0 \right)##. Though maybe they meant that and are just being really sloppy.
 
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