MHB Is the Function $f(x,y)=\frac{x}{y}+\frac{y}{x}$ Differentiable and $C^1$?

[evinda]

Hello! (Wave)

Suppose that we want to check if $f(x,y)=\frac{x}{y}+\frac{y}{x}$ is differentiable at each point of its domain and if it is $C^1$.

The domain is $D=\{ (x,y) \in \mathbb{R}^2: x \neq 0 \text{ and } y \neq 0\}$.The partial derivatives :

$$\frac{\partial{f}}{\partial{x}}=\frac{x^2-y^2}{x^2 y} \\ \frac{\partial{f}}{\partial{y}}=\frac{-x^2+y^2}{xy^2}$$

are continuous on $D$. So $f$ is $C^1$ and so it is differentiable.

Does this suffice?

Because there is the following definition:

Let $f: \mathbb{R}^2 \to \mathbb{R}$. We say that $f$ is differentiable in $(x_0, y_0)$, if $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ exist in $(x_0, y_0) $ and

$$\frac{f(x,y)-f(x_0, y_0)- \left[ \frac{\partial{f}}{\partial{x}}(x_0, y_0)\right](x-x_0)-\left[ \frac{\partial{f}}{\partial{y}}(x_0, y_0)\right](y-y_0) }{||(x,y)-(x_0, y_0)||} \to 0$$while $(x,y) \to (x_0, y_0)$.Don't we have to show the latter?

 

[I like Serena]

Hello! (Wave)

Suppose that we want to check if $f(x,y)=\frac{x}{y}+\frac{y}{x}$ is differentiable at each point of its domain and if it is $C^1$.

The domain is $D=\{ (x,y) \in \mathbb{R}^2: x \neq 0 \text{ and } y \neq 0\}$.The partial derivatives :

$$\frac{\partial{f}}{\partial{x}}=\frac{x^2-y^2}{x^2 y} \\ \frac{\partial{f}}{\partial{y}}=\frac{-x^2+y^2}{xy^2}$$

are continuous on $D$. So $f$ is $C^1$ and so it is differentiable.

Does this suffice?

Because there is the following definition:

Let $f: \mathbb{R}^2 \to \mathbb{R}$. We say that $f$ is differentiable in $(x_0, y_0)$, if $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ exist in $(x_0, y_0) $ and

$$\frac{f(x,y)-f(x_0, y_0)- \left[ \frac{\partial{f}}{\partial{x}}(x_0, y_0)\right](x-x_0)-\left[ \frac{\partial{f}}{\partial{y}}(x_0, y_0)\right](y-y_0) }{||(x,y)-(x_0, y_0)||} \to 0$$while $(x,y) \to (x_0, y_0)$.Don't we have to show the latter?

Hi evinda! (Smile)

From wiki:

If all the partial derivatives of a function all exist and are continuous in a neighborhood of a point, then the function must be differentiable at that point, and it is of class $C^1$.
​

In this case we know that all the partial derivatives of the function all exist and are continuous in a neighborhood of a point, so it's indeed differentiable.

The definition you mention does not require the partial derivatives in a neighborhood, but only at the point itself.
As a consequence there's more to prove, which is implicitly true if we have the partial derivatives in a neighborhood. (Nerd)

 

[evinda]

Hi evinda! (Smile)

From wiki:

If all the partial derivatives of a function all exist and are continuous in a neighborhood of a point, then the function must be differentiable at that point, and it is of class $C^1$.
​

In this case we know that all the partial derivatives of the function all exist and are continuous in a neighborhood of a point, so it's indeed differentiable.

In this case, the partial derivatives of the function all exist and are continuous in the domain of the function, right?
So can we consider the domain to be a neighborhood of any possible point $\in \mathbb{R}^2$ ?

As a consequence there's more to prove, which is implicitly true if we have the partial derivatives in a neighborhood. (Nerd)

What do you mean?

 

[I like Serena]

In this case, the partial derivatives of the function all exist and are continuous in the domain of the function, right?
So can we consider the domain to be a neighborhood of any possible point $\in \mathbb{R}^2$ ?

Any point except (0,0). (Thinking)

What do you mean?

What you gave is the definition of differentiability at a point.
We took a proposition from wiki. Applying that proposition is easier than trying to apply the definition directly. (Thinking)

 

[evinda]

Any point except (0,0). (Thinking)

(Nod)

What you gave is the definition of differentiability at a point.
We took a proposition from wiki. Applying that proposition is easier than trying to apply the definition directly. (Thinking)

We only apply the definition if we are given a specific point and want to check if the function is differentiable at this point, right?

 

[evinda]

Also, how else could we show that $f$ is differentiable without showing that it is continuously differentiable?

 

[I like Serena]

We only apply the definition if we are given a specific point and want to check if the function is differentiable at this point, right?

Right. (Nod)

Also, how else could we show that $f$ is differentiable without showing that it is continuously differentiable?

I'm not aware of another way. (Thinking)

 

[evinda]

Ok, thank you! (Smile)