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Discontinuous Derivative

  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations
    n/a

    3. The attempt at a solution
    I know that if you derive x^2sin(1/x)
    you get
    -cos(1/x) + sin(1/x)(2x).
    But what do I do from here? If I use the limit definition, i'll end up getting something like h(sin(1/h)) after evaluating. I still don't understand how the limit definition will show that this exists.
     
  2. jcsd
  3. Oct 26, 2016 #2

    Mark44

    Staff: Mentor

    Please show us what you did in using the definition of the derivative.

    BTW, you don't "derive" x^2 sin(1/x) -- you differentiate it. If you start from a quadratic equation, you can use completing the square to derive the quadratic formula.
     
  4. Oct 26, 2016 #3

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You need to do two things

    First show that ##\lim_{x\to 0}g'(x)## does not exist. That should be easy using the derivative you have calculated above.

    Second, try to calculate ##g'(0)## which is defined as
    $$\lim_{h\to 0}\frac{g(h)-g(0)}{h}$$
    If that limit exists then you are finished.

    The reference to the 'limit definition' in the question is a bit confusing as there are two different limits involved in this question. They are referring to the definition of the derivative as a limit (the second formula I wrote above), not to the limit of ##g'(x)## as ##x\to 0##
     
    Last edited by a moderator: Oct 26, 2016
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