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!

Basic bivariate question

  1. Mar 12, 2005 #1
    given f(x,y) = (xy+(x^3))/(x^2+xy)

    I want to calculate df/dx (i.e. first partial derivative wrt x) at the origin.

    f is undefined there, but i can still do it, right?

    so, we get (x^2 + 2xy -y)/(x^2 +2xy +y^2)

    I'm not really sure of where to go with this, though.

    I think the basic definition of the derivative, i.e. lim (h->0) ((x+h)y + (x+h)^3)/((x+h)^2 + (x+h)y) could be the way to go.

    Any advice would be appreciated.

    The only specific techniques i understand are the "sandwich rule", Taylor polynomials, as well as basic differentiation etc, and the 1-var stuff like l'Hopital's rule. So please don't give me something like the epsilon-delta stuff for limits or anything else i won't understand.
     
  2. jcsd
  3. Mar 12, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    1. IF a function is not defined at a point then it does not have partial derivatives there! When yuou talk about "lim (h->0) ((x+h)y + (x+h)^3)/((x+h)^2 + (x+h)y) " you seem to be assuming that f(0,0)= 0. Is that given? If so, then f certainly is defined there!

    2. Note that you want to find the derivative AT (0,0). The partial derivative (with respect to x) is calculated holding y constant so you are really concerned with the derivative of f(x,0)= x3/x2 which is equal to x as long as x is not 0- and, assuming that f(0,0)= 0, is also equal to 0 at x= 0 and so is equal to x for all 1. Surely you can find the derivative of that!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Basic bivariate question
  1. Very basic question (Replies: 2)

  2. Basic Tensor Question (Replies: 4)

Loading...