Does a nondecreasing, differentiable function have continuous derivative? Are the following statements true? How to prove or disprove?
(1). Let $f$ be a nondecreasing, differentiable function on $[0,1]$. Then $f$ is absolutely continuous?
To be stronger,
(2).   Let $f$ be a nondecreasing, differentiable function on $[0,1]$. Then $f'$ is bounded on $[0,1]$?
 A: Perhaps a sledehammer:
The following holds: 
If 
$\ \ \ $ 1) $f$ is continuous on $[a,b]$, 
$\ \ \ $ 2) $f'$ exists and is finite for all but a countable subset of $(a,b)$, 
and 
$\ \ \ $ 3) $f'$ is Lebesgue integrable, 
then
$$
f(x)-f(a)=\int_a^x f'(t)\,dt
$$
for all $a\le x\le b$.
In particular, $f$ is absolutely continuous.
This is Excercise 18.41 in Hewitt and Stromberg's Real and Abstract Analysis. The absolute contunuity of $f$ follows from the integral representation above and is proved in 18.17 of Hewitt and Stromberg.
Your condition that $f$ is monotone insures $f'$ is integrable; since $f'\ge0$ and $\int_a^b f'\le f(b)-f(a)$  (c.f., Theorem 18.14 of Hewitt and Stromberg).
So the answer to your first question is "yes".

The answer to your second question is "no". Given any interval $[x,y]$, an integer $n$, and any two numbers $d>c$, one can construct an increasing differentiable function $g$ with $g(x)=c$, $g(y)=d$, $g'(c)=g'(d)=0$ and $g'((c+d)/2)\ge n$.
Use this to contruct a counterexample $f$ with $0\le f(x)\le x^2$ for all $x$ (on the interval $I_n=[1/(n+1), 1/n]$ construct a $g$ as above with $\sup_{x\in I_n}g'(x)\ge n$). Defining $f(0)=0$ will insure $f$ is differentiable at $0$. 
A: The answer to the second question is no, as $f(x)=\sqrt x$ shows.
