What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$? let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :
$\displaystyle \frac{dy}{dx}$ ?.
Note: I have used the definition of derivative of the square root function but i don't succed .
Thank you for any help 

Clarification to solve the question 
The question posed is  calcule Derivative $\sqrt{x+\sqrt{x+\sqrt{x+....}}}$ 
we put $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}} $ 
Before calculating the derivative we simplify the expression $y$
$y^{2}=x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}=x+y$
$y^{2}-y-x=0$
We solve the equation $y^{2}-y-x=0$ 
The solution to this equation is $y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4x}}{2}$
After simplifying the relationship y we calculate $\frac{dy}{dx}$
we've got $\frac{dy}{dx}=\frac{1}{2}.\frac{4}{2\sqrt{1+4x}}=\frac{1}{\sqrt{1+4x}}$
In the general case, if
$y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+...}}}$
$\Rightarrow y^{2}=f(x)+y$
$y^{2}-y-f(x)=0$
$y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4f(x)}}{2}$
$\frac{dy}{dx}=\frac{1}{2}\frac{4\acute{f}(x)}{2\sqrt{1+4f(x)}}=\frac{\acute{f}(x)}{\sqrt{1+4f(x)}}$
