Find $f'(1)$ if $f$ is continuous and such that $f (f (x))=1+x$ for every $x$ 
If $ f $ is a continuous function satisfying $f (f (x))=1+x$ find $f'(1)$.

I just guessed $f (x)=x+1/2$.Any formal method for this sum ?

Take the $\lim_{x\rightarrow 1}{{f(f(x))-f(f(1))}\over {x-1}}
=\lim_{x\rightarrow 1}{{x+1-2}\over{ x-1}}=1.$ 
So $f'(f(1))=1$. 
We now derivate for x:
$${d\over {dx}}[f(f(x))]={d\over dx}(1+x)$$
$$f'(x)\cdot f'(f(x))=1.$$ Hence for $x=1:$ $$f'(1)\cdot f'(f(1))=1$$
$$f'(1)\cdot 1=1$$
$$f'(1)=1.$$
