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Thread: 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 $\displaystyle x\ge 0$, which is not continuous at x=0, gives g(x)= -x if x< 0 and g(x)= x if $\displaystyle x\ge 0$ (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 $\displaystyle h(x)= -x^2/2$ if x< 0 and $\displaystyle h(x)= x^2/2$ if $\displaystyle x\ge 0$. 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 $\displaystyle f(x)=x^2$ then $\displaystyle f'(x)= 2x$
    In general $\displaystyle \frac{dx}{xy}=nx^{n-1}$ but at x=0 for n=1 the function is undefined.
    So how would $\displaystyle f''(x)=0$?
    Last edited by Jskid; Oct 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 $\displaystyle \frac{dy}{dx} 2x = 2(1)x^{1-1}$ but at x=0 $\displaystyle 2(0)^0$ is undefined.
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