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Math Help - diferntialbility of a function

  1. #1
    Member Jskid's Avatar
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    diferntialbility of a function

    If a function is differentiable, does that mean its higher order derivatives are also differentiable?
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  2. #2
    Member Jskid's Avatar
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    I guess not since a polynomial's derivative will (eventually) have a removable discontinuity at x=0
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  3. #3
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    Oh, that's not true! A polynomial is infinitely differentiable. I can not imagine why you would think it will "eventually have a removable discontinuity at x= 0". The derivative of any polynomial is eventually identically 0 and that's a very differentiable function!

    Instead, to find an example of a differentiable function that is not infinitely differentiable, start with a discontinuous function and integrate it.

    For example, integrating f(x)= -1 if x< 0 and 1 if x\ge 0, which is not continuous at x=0, gives g(x)= -x if x< 0 and g(x)= x if x\ge 0 (I have chosen the constant of integration to be 0). In other words, g(x)= |x| which is continuous for all x but not differentiable at x= 0. Integrating again gives h(x)= -x^2/2 if x< 0 and h(x)= x^2/2 if x\ge 0. That function is differentiable at x= 0 but not twice differentiable. Integrating again would give a function that is twice differentiable but not three times differentiable.
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  4. #4
    Member Jskid's Avatar
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    That answers my initial question but I have trouble with something else.
    Let f(x)=x^2 then f'(x)= 2x
    In general \frac{dx}{xy}=nx^{n-1} but at x=0 for n=1 the function is undefined.
    So how would f''(x)=0?
    Last edited by Jskid; October 21st 2010 at 10:30 AM. Reason: corrected second line
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  5. #5
    Member Jskid's Avatar
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    Am I right? Since \frac{dy}{dx} 2x = 2(1)x^{1-1} but at x=0 2(0)^0 is undefined.
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