# Thread: condition for differentiability

1. ## condition for differentiability

I have to determine if $\displaystyle f(x,y) = \frac{x}{y} + \frac{y}{x}$ is differentiable at all points in its domain and of class $\displaystyle C^1$

$\displaystyle \frac{\partial f}{\partial x} = \frac{1}{y} - \frac{y}{x^2}$
$\displaystyle \frac{\partial f}{\partial y} = \frac{1}{x} - \frac{x}{y^2}$

seems that if $\displaystyle (x,y)=(0,0)$ then its not continuous or differentiable

and if $\displaystyle x=0$ or $\displaystyle y=0$ then one or both partials fail to be continuous or differentiable

so no, its not of class $\displaystyle C^1$

correct?

2. ## Re: condition for differentiability

$C^1$ is the class of functions who's first derivatives are continuous at all points their domain.

$(0,0)$ is not in the domain of $f(x,y)$

Are there any points in the domain of $f(x,y)$ where the first partials are discontinuous?

3. ## Re: condition for differentiability

Originally Posted by romsek
$C^1$ is the class of functions who's first derivatives are continuous at all points their domain.

$(0,0)$ is not in the domain of $f(x,y)$

Are there any points in the domain of $f(x,y)$ where the first partials are discontinuous?
if (0,0) is not in the domain then theres no points where the partials are discontinuous so it would be $\displaystyle C^1$. and (0,0) is not in it because it is not defined there, duh haha.

4. ## Re: condition for differentiability

More to the point: none of the points (x, y) with either x or y zero, are in the domain.