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Show f is differentiable but partial derivatives are not continuous

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Define f: Rn --------> R as
    f(x) = (||x||^2)*sin (1/||x||) for ||x|| ≠ 0
    f(x) = 0 for ||x|| = 0
    Show that f is differentiable everywhere but that the partial derivatives are not continuous.


    2. Relevant equations



    3. The attempt at a solution
    Showing that it is differentiable everywhere
    if it is differentiable at the origin then

    lim h---> 0 of ( f(0+h) - f(0) - c•h )/ ||h|| = 0


    f(o+h) = ||h||^2sin(1/||h||)
    f(0) = 0
    c = gradient of f at 0

    breaking it up into individual limits we have
    lim h--->o of f(0+h)/||h|| + lim h--->0 of f(o)/||h|| + lim h--->0 of c•h/||h||

    f(o+h)/||h|| = ||h||sin(1/||h||) which goes to zero since the absolute value of sin(1/||h||) is bounded by 1
    f(0) / ||h|| goes to zero since f(0) = 0

    but what does c•h/||h|| go to as h ---> 0
    my intuition tells me it goes to c....but if the function is diffable at 0 then it has to go to zero as well...so i'm not sure how to show its differentiable everywhere

    as for showing that the partial derivatives are not continuous

    to calculate the jth partial derivatives (the derivative of f with respect to xj) i rewrite the function as (∑(xn^2))sin (1/root[∑(xn^2)])

    df / dxj (using the product rule) = 2xj * sin (1/root[∑(xn^2)]) + cos(1/root[∑(xn^2)])* -0.5(∑(xn^2))^-1.5 * 2xj * [∑(xn^2)]

    now...i can see that the partical derivatives are not continuous at ||x|| = 0 since the term (∑(xn^2))^-1.5 would be undefined if x1 to xn are all equal to zero

    but am i also supposed to show that the limit as ||x|| approaches zero from the 2 sides of the partical derivatives is not equal? or what?
     
  2. jcsd
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