No, that's impossible. Saying that the function is differentiable on [0, 1] means that it has a finite derivative at each point on [0, 1]. Since [0, 1] is a closed and bounded set, that derivative must be bounded.
Question: Give an example of a function f that is differentiable on [0,1] but its derivative is not bounded on [0,1].
Ok, I know that the derivative f' cannot be continuous, because then it would be bounded on [0,1]. I also know that it cannot be increasing with jump discontinuities, because the derivative has to have the intermediate value property. I also do not think (but am not 100 percent sure) that it can have any infinite discontinuities, because I think that would make f unbounded, which it cannot be on [0,1] if its differentiable on [0,1].
Any ideas?
Halls, what about when x isn't 0, and f(0) = 0.
f is differentiable at all non-zero points, and note that . Thus . Thus f is also differentiable at 0 and f(0) = 0.
Note that on (0, 1], , which goes to -inf as x goes to 0 through points of (0, 1], thus f' is not bounded on [0,1]