Set of zeroes of the derivative of a pathological function For a continuous function $f : [0,1] \to {\mathbb R}$, let us set 
$$
X_f=\lbrace x \in [0,1] \bigg| f'(x)=0 \rbrace
$$
(for a $x\not\in X_f$, $f'(x)$ may be a nonzero value or undefined).
There are well-known "Cantor staircase" examples where $X_f$ is a dense open set in
$[0,1]$. Is there a continuous $f$ with $X_f={\mathbb Q}\cap [0,1]\,$ ? Is there a continuous $f$ with $X_f=[0,1] \setminus {\mathbb Q}$ ?
UPDATE 06/02/2012 Since $X_f$ is dense in $[0,1]$, the continuity set of $f'$ is included in $X_f$. Since $X_f$ has empty interior, the continuity set of $f'$ cannot be a $G_{\delta}$ in any subinterval of $[0,1]$, so $f$ cannot be everywhere differentiable on any subinterval of $[0,1]$. 
All the links proposed so far in the comments are about everywhere differentiable functions, so they do not suffice to answer my question.
An interesting sub-question is obtained if, in addition, we also require $f$ to be increasing (so that $f$ will be a homeomorphism from $[0,1]$ onto some other interval).
It is easy enough to construct a $f$ and control the behaviour $f'$ on a countable set, by the usual step-by-step procedure. But it seems very hard to say anything at all on the behaviour of $f'$ on the other points of $[0,1]$.
SECOND UPDATE 06/02/2012 As noted in the link provided in a comment below, it follows from Cousin's lemma that if $f$ is a continuous function such that $f'=0$ everywhere except 
for a countable set, then $f$ is constant.
So there is no $f$ such that $X_f=[0,1] \setminus {\mathbb Q}$ : the answer to my second question is NO. My first question remains open however.
 A: The answer to the first question is YES. In fact, one can show the following stronger statement : 
THEOREM.  There exists an increasing homeomorphism $f: [0,1] \to [0,1]$, such that $f'(q)=0$ for all $q\in {\mathbb Q}$ and any irrational $x\in [0,1]$ is contained in arbitrary small intervals $[u,v]$ with $\frac{f(v)-f(u)}{v-u} \geq 1$ (so that $f'(x)$, if defined, is $\geq 1$).
We use a familiar "inbreeding", piecewise and iterative construction, enumerating the rationals and imposing $f'(q)=0$ for each $q$ one by one, keeping at the same time a tight net of values $x,y$ such that $\frac{f(y)-f(x)}{y-x} \geq 1$. Formally : 
DEFINITION.  Let $f,g$ be two function $[a,b] \to {\mathbb R}$. We say that $(f,g)$ is a claw on $[a,b]$ when 
(1) $f(a)=g(a),f(b)=g(b),f'(a)=g'(a)=f'(b)=g'(b)=0$ and 
(2) $f(x)<g(x)$ when $a<x<b$. 
If, in addition, one has 
(3) For any $x\in ]a,b[$, there are $u<v$ in $]a,b[$ with $x\in [u,v]$, $\frac{f(v)-g(u)}{v-u} \geq 1$,
then we say that $(f,g)$ is a tight claw.
FUNDAMENTAL LEMMA.  If $(f,g)$ is any claw on $[a,b]$, then there is another claw $(F,G)$ with $f \leq F \leq G \leq g$ on $[a,b]$ and $(F,G)$ is tight. We may further impose that the supremum norm $||F-G||_{\infty}$ be $\lt \epsilon$ for $\epsilon$ as small as we want.
{\bf Proof of theorem using fundamental lemma}. We construct by induction two sequences $f_n,g_n$ of homeomorphisms $[0,1] \to [0,1]$ such that : 
(i) $f_n \leq g_n$ on $[0,1]$ 
(ii) $(f_n,g_n)$ is a tight claw on each interval $I_{n,k}=[\frac{k}{n!},\frac{k+1}{n!}]$ for $0 \leq k\ \leq n!$. 
(iii) $||f_n-g_n||_{\infty} \leq \frac{1}{n}$. 
We start with $f_0(x)=4x^3-3x^4$ and $g_0(x)=3x^2-2x^3$. Suppose now that $(f_n,g_n)$ have already been constructed. The two maps $f_n$ and $g_n$ coincide on
the finite set $X_n=\lbrace \frac{k}{n!} | 0  \leq k \leq n! \rbrace$. There is an increasing map $\phi : X_{n+1} \to [0,1]$, extending $f_{|X_n}=g_{|X_n}$, such that  $f_n \lt  \phi \lt g_n $ on $X_{n+1} \setminus X_n$.
On each interval $I_{n+1,k}$, there is a claw $(p_k,q_k)$ with $f_n \leq p_k \leq q_k \leq g_n$, and $p_k$ nad $q_k$ coincide with $\phi$ on the extremities of the interval. By fundamental lemma, there is a tight claw $(r_k,s_k)$ on $I_{n+1,k}$ with $p_k \leq r_k \leq s_k \leq q_k$, and $||r_k-s_k||_{\infty} \leq \frac{1}{n}$.
We may then take $f_{n+1}$ ($g_{n+1}$, respectively) to be the unique function that coincides with $r_k$ ($s_k$, respectively) on $I_{n+1,k}$. This completes the inductive construction.
Then by (i) and (iii) we have ${\sf sup}(f_n)={\sf inf}(g_n)$ ; let us call $f$ this function. Then $(f_n)$ nad $(g_n)$ converge uniformly towards $f$, and $f$ is a homeomorphism $[0,1] \to [0,1] $. We have for $x \in I_{n,k}$ and $q=\frac{k}{n!}$, and any $m\geq n$,
$$
f_m(q)=g_m(q)=f(q), \ \frac{f_n(x)-f(q)}{x-q} \leq  \frac{f(x)-f(q)}{x-q} \leq \frac{g_n(x)-f(q)}{x-q}
$$
whence $f'(q)=0$ on the right. Similarly, $f'(q)=0$ on the left. 
The fundamental lemma can be shown from the following :
LEMMA 1. Let $(f,g)$ be a claw on $[a,b]$. Then there are three increasing sequences $(x_k)_{k\in \mathbb Z},(y_{1,k})_{k \in \mathbb Z}$ and $(y_{2,k})_{k \in \mathbb Z}$, with
$$
{\sf lim}_{k \to -\infty} x_k=a, \
{\sf lim}_{k \to +\infty} x_k=b, \
$$
$$
{\sf lim}_{k \to -\infty} y_{1,k}={\sf lim}_{k \to -\infty} y_{2,k}=f(a)=g(a),\
{\sf lim}_{k \to +\infty} y_{1,k}={\sf lim}_{k \to +\infty} y_{2,k}=f(b)=g(b)
$$
and
$$
\frac{y_{1,k+2}-y_{2,k}}{x_{k+2}-x_k} \leq 1, f(x_k) \lt y_{1,k} \lt y_{2,k} <g(x_k),  |y_{2,k}-y_{1,k}| \lt \epsilon
$$
{\bf Proof of fundamental lemma from lemma 1.} Set $F(x_k)=y_{1,k}, G(x_k)=y_{2,k}$ for all $k\in \mathbb Z$ and interpolate in-between (take $F$ and $G$ affine on $[x_k,x_{k+1}]$, for instance).
Finally, lemma 1 follows from the iteration of 
LEMMA 2. Let $(f,g)$ be a claw on $[a,b]$, and let $a',b'$ such that $a<a'<b'<b$. Then there are three increasing finite sequences $(x_k)_{|k| \leq M},(y_{1,k})_{|k| \leq M}$ and $(y_{2,k})_{|k| \leq M}$, with
$$
x_{-M} \leq a', \ b' \leq x_{M}
$$
$$
\frac{y_{1,k+2}-y_{2,k}}{x_{k+2}-x_k} \leq 1, f(x_k) \lt y_{1,k} \lt y_{2,k} <g(x_k),  |y_{2,k}-y_{1,k}| \lt \epsilon
$$
