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

In summary, we discuss the differentiability and $C^1$-differentiability of a function $f(x,y)=\frac{x}{y}+\frac{y}{x}$ in its domain $D=\{ (x,y) \in \mathbb{R}^2: x \neq 0 \text{ and } y \neq 0\}$ and conclude that it is differentiable and $C^1$-differentiable because its partial derivatives are continuous in $D$. We also consider the definition of differentiability at a point and use a proposition from wiki to show that the function is differentiable in its domain. Finally, we discuss other ways to show differentiability and conclude that showing continuous
  • #1
evinda
Gold Member
MHB
3,836
0
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?
 
Physics news on Phys.org
  • #2
evinda said:
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)
 
  • #3
I like Serena said:
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 $\in \mathbb{R}^2$ ?

I like Serena said:
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?
 
  • #4
evinda said:
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)
 
  • #5
I like Serena said:
Any point except (0,0). (Thinking)

(Nod)

I like Serena said:
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?
 
  • #6
Also, how else could we show that $f$ is differentiable without showing that it is continuously differentiable?
 
  • #7
evinda said:
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)
evinda said:
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)
 
  • #8
Ok, thank you! (Smile)
 

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

What is a differentiable function?

A differentiable function is a mathematical function that can be differentiated at every point in its domain. This means that the function's derivative exists and is continuous at every point.

How is differentiability related to continuity?

A function is differentiable if and only if it is continuous at every point in its domain. This means that a function must be continuous in order to be differentiable, but not all continuous functions are differentiable.

What is the difference between a differentiable function and a non-differentiable function?

A differentiable function is one that has a well-defined derivative at every point in its domain, while a non-differentiable function does not have a well-defined derivative at one or more points in its domain. This can be due to the function being discontinuous or having sharp corners, breaks, or jumps in its graph.

How is the derivative of a differentiable function calculated?

The derivative of a differentiable function is calculated using the limit definition of the derivative. This involves finding the slope of a tangent line at a specific point on the function's graph, which is equivalent to finding the function's instantaneous rate of change at that point.

What are some real-world applications of differentiable functions?

Differentiable functions are used extensively in fields such as physics, engineering, economics, and statistics to model and solve various problems. They are also used in optimization and machine learning algorithms to find the maximum or minimum value of a function. Examples of real-world applications include calculating velocity and acceleration in physics, finding the optimal production level in economics, and predicting stock prices in finance.

Similar threads

Replies
5
Views
1K
Replies
4
Views
3K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
10
Views
3K
Replies
1
Views
2K
Back
Top