1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where is the function differentiable

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    None

    3. 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).##
     
  2. jcsd
  3. Mar 23, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Given that [itex]f(0,0) = 0[/itex] and [itex]f(0,y) = 0[/itex], you have
    [tex]
    \left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{x \to 0} \frac{f(x,0)}{x}
    [/tex]
    but for [itex]y \neq 0[/itex] you have
    [tex]
    \left.\frac{\partial f}{\partial x}\right|_{(0,y)} = \lim_{x \to 0} \frac{f(x,y)}{x}
    [/tex]
    Does
    [tex]
    \lim_{y \to 0} \left.\frac{\partial f}{\partial x}\right|_{(0,y)}[/tex]
    exist, and if it does is it equal to [itex]\left.\dfrac{\partial f}{\partial x}\right|_{(0,0)}?[/itex]
     
  4. Mar 23, 2014 #3
    Hm, lets 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?
     
  5. Mar 23, 2014 #4

    pasmith

    User Avatar
    Homework Helper

    No; you have
    [tex]
    \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)[/tex]
    so you need to take the [itex]x[/itex] limit first. What is
    [tex]
    \lim_{x \to 0} \frac{xy}{x(|x| + |y|)}[/tex]
    when [itex]y \neq 0[/itex]?
     
  6. Mar 23, 2014 #5
    By l'hopital we have that [tex]\lim_{x \to 0} \frac{xy}{x(|x| + |y|)} = \frac{y}{|y|}[/tex]

    Is that correct? The absolute value is throwing me off if I am wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted