Properties of a derivative on a compact interval 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.

No.  For instance, let $f$ be piecewise linear and positive on $[0,1)$ such that on an infinite sequence of intervals approaching $1$, $f$ alternates between jumping down to values approaching $0$, jumping up to values approaching $\infty$, and jumping back down to $1$ and remaining constant with value $1$.  Define $F(x)=\int_0^xf(t)\,dt$.  If we choose the intervals where $f$ takes values other than $1$ to be sufficiently small and sparse (so as $t\to 1$, $f(t)=1$ for a quickly increasingly large proportion of the time), the integrals of $f$ over these intervals will have a negligible effect on the limiting behavior of $F(x)$ as $x\to 1$.  So, $F$ will extend continuously to $1$ with $F'(x)=1$.  We can then extend $F$ to be differentiable on an open interval containing $[0,1]$ (e.g., by making its derivative $1$ outside of $[0,1]$).
