Decide if a function is C1

1. Jan 17, 2014

mbigras

1. The problem statement, all variables and given/known data
Decide if the function is $C^{1}$ or just differentiable

$$f(x,y) = \frac{x}{y} + \frac{y}{x}$$

2. Relevant 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

3. 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?

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:
$$\text{the domain }A = \{(x,y) \in R^{2} | (x \neq 0 \text{ and } y \neq 0\}$$

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.

2. Jan 17, 2014

Dick

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.

Last edited: Jan 17, 2014