If a function is differentiable, does that mean its higher order derivatives are also differentiable?

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- October 18th 2010, 09:29 PMJskiddiferntialbility of a function
If a function is differentiable, does that mean its higher order derivatives are also differentiable?

- October 18th 2010, 09:41 PMJskid
I guess not since a polynomial's derivative will (eventually) have a removable discontinuity at x=0

- October 19th 2010, 04:53 AMHallsofIvy
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 , which is not continuous at x=0, gives g(x)= -x if x< 0 and g(x)= x if (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 if x< 0 and if . 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. - October 19th 2010, 10:12 AMJskid
That answers my initial question but I have trouble with something else.

Let then

In general but at x=0 for n=1 the function is undefined.

So how would ? - October 21st 2010, 11:32 AMJskid
Am I right? Since but at x=0 is undefined.