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!

Is this function differentiable?

  1. Oct 11, 2011 #1
    Hey guys, I'm just wondering if I got this question right:

    Discuss where the following in [itex]R^{2}[/itex] is differentiable:
    [itex]f(x,y)=(x^{3}+y^{3})^{2/3}[/itex]

    So I take the partial derivative:

    [itex]f_{x}(x,y)=\frac{2x^{2}}{(x^{3}+y^{3})^{1/3}}[/itex]

    and see that f(x,y) might not be differentiable at (0,0), so I used the definition of a partial derivative to see if the partial derivatives exist at (0,0):

    [itex]f_{x}(0,0)=lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=0[/itex]

    I thought this was the end of the question, since it shows that the partial derivative at (0,0) does exist, and thus it's continuous for all (x,y) (showing that f(x,y) is differentiable)
    But under the question there's a hint, it says:

    Hint: Verify the handy inequality [itex]|x^{3}+y^{3}| \leq 2(x^{2}+y^{2})^{3/2}[/itex]

    Did I miss something?
     
  2. jcsd
  3. Oct 11, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Look at f(x, y) in the neighborhood of the line y=-x .

    Look at fx(x, y) in the neighborhood of the line y=-x .
     
    Last edited: Oct 11, 2011
  4. Oct 11, 2011 #3
    So I'm not done the question? What am I looking for at the moment to show that it's differentiable?

    Sorry, just trying to understand the question better! :)
     
  5. Oct 11, 2011 #4
    Ohh, I see what you're saying, when y=-x, the denominator of fx = 0, so would I just say that f(x,y) isn't differentiable when y=-x? Or do I have to go back to limits and use squeeze theorem somehow? (Just a cause cause the hint looks like squeeze theorem :P)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is this function differentiable?
Loading...