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!

[Analysis] Derivative in Two Dimensions.

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f : ℝ2 -> ℝ be some function that is defined on a neighborhood of a point c in ℝ2. If D1f (the derivative of f in the direction of e1) exists and is continuous on a neighborhood of c, and D2f exists at c, prove that f is differentiable at c.

    2. Relevant equations

    The only sufficient condition I have for a function to be differentiable at a point is that the partial derivatives exist and are continuous at that point.

    3. The attempt at a solution

    If I could show that D2f is continuous at c, then I would be done. To show that D2f is continuous at c, I have to be sure that

    |D2f(x) - D2f(c)| < ε

    when ||x - c|| is small (do I know that D2f(x) even exists for x ≠ c?). If f were continuous at c, then I think that I could argue that, since D2f(x) is close to some difference quotient of f at x and D2f(c) is close to some difference quotient of f at c (and these difference quotients can be made close to each other), then D2f(x) is close to D2f(c).

    So now I am thinking of how to show that f is continuous. I know that if the partial derivatives are bounded on some region of c, then f is continuous at c. I think that the partial derivatives are bounded near c, but I am not totally sure. Could someone tell me if I am going in the right direction?
     
  2. jcsd
  3. Nov 4, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    f(x, y) = x2 sin(1/x) for x >0, 0 otherwise.
    fy exists and is continuous in a neighbourhood of (0, 0).
    fx exists but is not continuous in a neighbourhood of (0, 0).
    Does this help to narrow down the possible approaches?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: [Analysis] Derivative in Two Dimensions.
  1. A two dimension limit (Replies: 1)

Loading...