A possible dumb question about derivative I was solving some differentiation problems when I found the function
$$g(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}.$$
So  I thought: If I define the function $f:\mathbb{R_{x>0}}\to \mathbb{R}$ as
$$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$
What kind of informations have I about the function $f$? Is it continous or differentiable, in some "sense"? If yes, is it correct to say by implicit differentiation that
$$f'(x)=\frac{1}{2f(x)-1}?$$
Thanks so much.
 A: If $$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}$$ then assuming this function actually converges for a particular $x$ (in fact, as it stands, your domain may not be properly restricted for convergence, making your "function" not necessarily well defined.)
Then $$f(x)=\sqrt{x+f(x)}$$
$$[f(x)]^2-f(x)=x$$
Implictly, 
$$2f'(x)f(x)-f'(x)=1$$
$$f'(x)=\frac{1}{2f(x)-1}$$
This entirely assumes your function is, indeed, a function.
You can actually work this out. By the quadratic formula, we have
$$f(x)=\frac{1\pm\sqrt{1+4x}}{2}$$
So clearly, at least, $x\in[-\frac{1}{4},\infty)$. Your chosen domain, $(0,\infty)$, then looks okay. Now to deal with that $\pm$ sign. Your function, in its original form is non-negative, so we can only take the negative branch for
$$\frac{1-\sqrt{1+4x}}{2}\geq0$$
$$\sqrt{1+4x}\leq 1$$
$$1+4x\leq 1$$
$$x\leq 0$$
Which is wonderful, since by our restriction on the domain, we don't need to consider the negative branch. Therefore the function is well defined, and is:
$$f(x)=\frac{1+\sqrt{1+4x}}{2},x>0$$
Which is continuous and differentiable on its domain.
