What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$?

I am going to assume that the given problem is over $\mathbb{R}^+$, otherwise the definition of $f$ makes no sense. Over $\mathbb{R}^+$, the given function is differentiable by the concavity of $g(x)=\sqrt{x}$.

Such function fulfills $f(x)^2 = x+f(x)$, hence by termwise differentiation $$ 2\,f'(x)\,f(x) = 1 + f'(x) $$ and: $$ \frac{d}{dx}\,f(x) = \frac{1}{2\,f(x)-1}.$$

$ y = \sqrt{x+\left(\sqrt{{x}+\sqrt{x+\cdots}}\right)}$

$y = \sqrt{x + y}$

Then $y^2 = x + y$

Now find derivative.

$2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$

$2y\frac{dy}{dx} - \frac{dy}{dx} = 1$

$(2y - 1)\frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{2y - 1}$