# Where is the function differentiable

• Lee33
In summary: By l'hopital we have that \lim_{x \to 0} \frac{xy}{x(|x| + |y|)} = \frac{y}{|y|}Is that correct? The absolute value is throwing me off if I am wrong.Yes, that is correct. Now, what is the limit as y \to 0 of \frac{y}{|y|} ?In summary, the function f:E^2\to\mathbb{R} given by f(x,y)=\begin{cases}\frac{xy}{|x|+|y|} & , \ \text{if} \ (x,y)\ne(0,0)\\0
Lee33

## Homework Statement

Where is the function ##f:E^2\to\mathbb{R}## given by ##f(x,y)=\begin{cases}\frac{xy}{|x|+|y|} & , \ \text{if} \ (x,y)\ne(0,0)\\
0 & , \ \text{if} \ (x,y)=(0,0) \end{cases}## differentiable?

None

## The Attempt at a Solution

The function is continuous so the partials exists, thus we have ##\frac{\partial f}{\partial x} = \begin{cases} \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x>0\\
\frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x<0. \end{cases}##

##\frac{\partial f}{\partial y} = \begin{cases} \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y>0\\
\frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y<0. \end{cases}##

For ##(x,y)\ne (0,0),## the partials ##\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}## exists and are continuous so ##F## is differentiable at any ##(x,y)\ne (0,0). ##

For ##(x,y)=(0,0),## I am having trouble showing why the partials are not continuous at ##(0,0).##

Lee33 said:

## Homework Statement

Where is the function ##f:E^2\to\mathbb{R}## given by ##f(x,y)=\begin{cases}\frac{xy}{|x|+|y|} & , \ \text{if} \ (x,y)\ne(0,0)\\
0 & , \ \text{if} \ (x,y)=(0,0) \end{cases}## differentiable?

None

## The Attempt at a Solution

The function is continuous so the partials exists, thus we have ##\frac{\partial f}{\partial x} = \begin{cases} \frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x>0\\
\frac{y(|x|+|y|)\pm xy|y|}{(|x|+|y|)^2}, & x<0. \end{cases}##

##\frac{\partial f}{\partial y} = \begin{cases} \frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y>0\\
\frac{x(|x|+|y|)\pm xy|x|}{(|x|+|y|)^2}, & y<0. \end{cases}##

For ##(x,y)\ne (0,0),## the partials ##\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}## exists and are continuous so ##F## is differentiable at any ##(x,y)\ne (0,0). ##

For ##(x,y)=(0,0),## I am having trouble showing why the partials are not continuous at ##(0,0).##

Given that $f(0,0) = 0$ and $f(0,y) = 0$, you have
$$\left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{x \to 0} \frac{f(x,0)}{x}$$
but for $y \neq 0$ you have
$$\left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{x \to 0} \frac{f(x,y)}{x}$$
Does
$$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)}$$
exist, and if it does is it equal to $\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}?$

pasmith said:
Given that $f(0,0) = 0$ and $f(0,y) = 0$, you have
$$\left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{x \to 0} \frac{f(x,0)}{x}$$
but for $y \neq 0$ you have
$$\left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{x \to 0} \frac{f(x,y)}{x}$$
Does
$$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)}$$
exist, and if it does is it equal to $\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}?$

Hm, let's see if I understand your reasoning because I am a bit confused.

##\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \frac{f(x,y)}{x}## which does equal ##\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}##.

Am I correct on that?

Lee33 said:
Hm, let's see if I understand your reasoning because I am a bit confused.

##\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \frac{f(x,y)}{x}## which does equal ##\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}##.

Am I correct on that?

No; you have
$$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \left(\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}\right)$$
so you need to take the $x$ limit first. What is
$$\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}$$
when $y \neq 0$?

pasmith said:
No; you have
$$\lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{y \to 0} \left(\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}\right)$$
so you need to take the $x$ limit first. What is
$$\lim_{x \to 0} \frac{xy}{x(|x| + |y|)}$$
when $y \neq 0$?

By l'hopital we have that $$\lim_{x \to 0} \frac{xy}{x(|x| + |y|)} = \frac{y}{|y|}$$

Is that correct? The absolute value is throwing me off if I am wrong.

## 1. What is the definition of differentiability?

Differentiability is a mathematical concept that describes the smoothness of a function. A function is said to be differentiable at a point if it has a well-defined derivative at that point. In simpler terms, this means that the function is continuous and has a well-defined slope at that point.

## 2. How do I determine if a function is differentiable at a specific point?

To determine if a function is differentiable at a specific point, you can use the definition of differentiability. This means checking if the function is continuous at that point and if the limit of the slope of the function exists at that point. If both of these conditions are met, then the function is differentiable at that point.

## 3. Can a function be differentiable but not continuous?

No, a function cannot be differentiable but not continuous. Differentiability requires continuity, so if a function is differentiable at a certain point, it must also be continuous at that point.

## 4. What types of functions are always differentiable?

Polynomial functions, rational functions, exponential functions, and trigonometric functions are all examples of functions that are always differentiable. However, there are some types of functions, such as piecewise functions, that may have points of non-differentiability.

## 5. How does differentiability relate to the smoothness of a function?

Differentiability is directly related to the smoothness of a function. A function that is differentiable at a point is also infinitely smooth at that point. This means that the function has no sharp corners or breaks and can be drawn without lifting the pen from the paper.

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