value of $f'(0)$ in composite function 
If $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that
$f(f(x))=1+x. $ Then $f'(0)$ is

what i try: $$f(f(x))=1+x$$
Replace $x\rightarrow f(x)$
$$f(f(f(x)))=1+f(x)$$
$$f(1+x)=1+f(x)$$
How do i solve it , Help me please
 A: This is a partial answer. 
If $x$ is an integer, say $x = n$, the last relation in the OP is a linear difference equation that can be solved to obtain $f(n) = c + n$. Constant $c$ can be computed so that $f(f(n)) = 1 + n$, yielding $f(n) = n + \frac 12$. So, any such function $f$ must be linear, at least over the integers. Naturally, $f(x)=x + \frac 12$ is one possibility. Are there others? The answer is: apparently yes (see here).
It is easy to see that $f$ must be injective and, in fact, increasing. If we assume that $x_1 \ne x_2$ and $f(x_1)=f(x_2)=\alpha$ we obtain a contradiction ($f(\alpha)$ would have to be simultaneously $1+x_1$ and $1+x_2$).
Thinking about how such function would look like (periodic, increasing, coinciding with $x+\frac 12$ over the integers), I'm inclined to conclude that either $f$ is not differentiable at integer points or, being differentiable, its graphic must be tangent to $x+\frac 12$.
My conjecture is that if $f$ is differentiable at $x=0$, the derivative is $f'(0)=1$. 
A: As long as the solution to
$$
f(x+1) = f(x)+1
$$
is 
$$
f(x) = x +\Phi(x)
$$
where $\Phi(x)$ is any periodic function with period  $1$ we have
$$
f'(x) = 1 + \Phi'(0)
$$
assuming $f(x)$ continuous at $x = 0$
