Nested Radical Calculus

$f(x)=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+f(x)}$ $\implies f(x)^2=f(x)+x\implies f(x)^2-f(x)-x=0$

Solve for $f(x)$ using the quadratic formula:

$\frac{-1+\sqrt{4x+1}}{2}=\frac{1}{2}(\sqrt{4x+1}-1)$

Now take the derivative using the chain rule:

$\frac{d}{dx}(\frac{1}{2}((4x+1)^{\frac{1}{2}}-1)=(4)(\frac{1}{2})(\frac{1}{2\sqrt{4x+1}})=\boxed{\frac{1}{\sqrt{4x+1}}}$

Similar solution with @Andrei Li 's

Let $y$ be the nested radical. Then

$\begin{aligned} y & = \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} & \small \color{#3D99F6} \text{Squaring both sides} \\ y^2 & = x + \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} \\ & = x + y \\ y^2 - y - x & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }y. \\ y & = \frac {1+\sqrt{4x+1}}2 & \small \color{#3D99F6} \text{Note that }y > 0 \end{aligned}$

From

$\begin{aligned} y^2 - y - x & = 0 & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ (2y - 1)\frac {dy}{dx} - 1 & = 0 \\ \frac {dy}{dx} & = \frac 1{{\color{#3D99F6}2y}-1} & \small \color{#3D99F6} \text{Recall }y = \frac {1+\sqrt{4x+1}}2 \\ & = \boxed{\dfrac 1{\sqrt{4x+1}}} \end{aligned}$