Suppose a function $F$ is differentiable on an interval $(a,b) \supset [0,1]$. Denote its derivative by $f$, and suppose that $f > 0$ on $[0,1]$.
Question 1: Is it true that $f$ can be bounded away from $0$ on $[0,1]$, i.e. that there exists some $c > 0$ such that $f(x) > c$ for all $x \in [0,1]$? If $f$ is continuous, this is clearly true, as a continuous function attains its minimum on a compact set, and this minimum is $> 0$ by assumption. If $f$ were an arbitrary function (not a derivative), this is clearly false; for instance, consider the function $f(x) = 1$ when $x = 0$ and $f(x) = x$ elsewhere. But this function has a jump discontinuity, and therefore is not the derivative of any function.
Question 2: Is it true that $f$ is bounded on $[0,1]$? Note that if we remove the $f > 0$ requirement, this is not true (for instance, consider $F(x) = x^2 \sin(1/x^2)$, with $F(0) = 0$; $F$ is differentiable, so $f$ exists, but $f$ is not bounded).
This question may be relevant, but it doesn't directly answer the above.