Prove that the function
f(x) = x^2 * sin (1/x) if x!=0
= 0 if x=0
is differetiable in (-infinity, infinity), but has no second derivative at x=0
Sorry, Captain Black, but this function definitely is differentiable at x= 0.
Use the definition of the derivative:
As long as h is not 0, so .
for all non-zero h while h goes to 0 so the derivative at x= 0 is .
Notice that for x not 0, the derivative is given by the product rule:
.
Now, that has no limit as x goes to 0 so while f is differentiable for all x, its derivative is not continuous at x= 0. That means that f' is not differentiable at x= 0 and so f is not twice differentiable at x= 0.