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Homework Help: Product rule

  1. Nov 15, 2003 #1
    Can the product rule be applied if one of the functions is not differentiable? For example,
    f(x)={g(x)sin(1/x), x not =0
    =0, x=0
    where g(0)=g'(0)=0.
    f'(0)=g'(0)sin(1/0) + g(0)dsin(1/x)/dx
    =0sin1/0+0dsin(1/x)/dx=0?
    applying the limit definition, I get
    f'(0)=g'(0)lim sin(1/h) where h-->0
    is this zero?
     
  2. jcsd
  3. Nov 15, 2003 #2
    It turns out that in order to find f'(0) I had to go back to the e-d definition of limit. Anyone see an easier way?
     
  4. Nov 15, 2003 #3

    NateTG

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    Science Advisor
    Homework Helper

    The limit you're describing does not exist.a

    Product rule:
    f(x)g(x)=f'(x)g(x)+f(x)g'(x)

    You need derivatives of both.

    Now we have
    g(x)sin(1/x)

    the derivative of sin(1/x) is
    x-2cos(1/x)
    and not defined at zero (no limit at zero either)
    you'd need g(x) to grow at better than x2 to have a potential derivative there.
     
  5. Nov 15, 2003 #4
    Hi NateTG, I have found the limit. See attached.
    I had to go back to epsilons and deltas. I was wondering if anyone knows an easier way to find the derivative.
     

    Attached Files:

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