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Determined to finish off my assignment

  1. Sep 21, 2006 #1
    Determined to finish off my assignment!!

    Firstly, it's 2:50 am here and I'm NOT going to sleep until I finish this assignment off!! I have few remaining ones and I'd be glad if someone could help me out so I can go to sleep...

    Please consider this question(and the one I posted below);
    f:R^2-->R is defined by f(x,y)=lxyl where lxyl is absolute value of xy

    (a)Show that f is differentiable at (0,0)
    For this, I think using definition of 'differentiable' is almost impossible, so I'm thinking should I use the theorem which says the function is differentiable if all of partial derivatives exist, and each entry in jacobian matrix exists in a neighbourhood of the point and continuous at that point.

    But... partial derivative of something inside absolute value?... also to show each entry in jacobian is continuous... do i use epsilon delta argument??

    (b)Show that there is no disc B((0,0);delta) throughout which f has all of its directional derivatives
    This one scares me off...
  2. jcsd
  3. Sep 21, 2006 #2
    I'm a bit confused about (b) because I think there is a theorem which states that if function f:R^m-->R^n is differentiable then directional derivatives Dvf(a) for all v a vector in R^m exists and is equal to f'(a)(v)? This seems to contradict to (a) since we are proving that it 'is' differentiable right?
  4. Sep 21, 2006 #3
    What is the definition of differentiability that you have? From what I can remember, the definition of differentiability I know can be readily applied to this question.

    In fact, if you were given the definitions of partial derivatives and differentiability, your first question is actually quite straight forward. By inspection (using the definition of a partial derivative), the partials wrtx and wrty evaluated at (x,y)=(0,0) are clearly zero meaning that the Jacobian is the 2 dimensional zero vector since f:R^2->R.

    f(0,0) = 0 clearly. So if you're using the same definition of differentiability as the one I am using then this comes down to showing that [tex]\frac{{\left| {xy} \right|}}{{\sqrt {x^2 + y^2 } }} \to 0[/tex] as (x,y) goes to (0,0). This is also straight forward.

    Edit: As for part (b) I think you have got it the other way around. That is, if all of the partial derivatives of f are continuous in a neighbourhood of the point of interest then f is differentiable.
    Last edited: Sep 21, 2006
  5. Sep 22, 2006 #4


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    From Calculus 9th edition by Sala, Hille, and Etgen, page 872

    "Theorem 15.2.4
    If f is differentiable at x, then f has a directional derivative at x in every direction u, where us is a unit vector"

    However, it appears that while |xy| is differentiable at (0,0), it is not differentiable for any (x,0) with x not 0, or (0,y) with y not 0. So the point of (b) is not directional derivatives at (0,0) but rather at (x, 0) and (0,y).
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