Nested radical differentiation

$\begin{aligned} y & = \sqrt{x + \sqrt{x- \color{#3D99F6}{ \sqrt{x + \sqrt{x - \sqrt{x + \cdots }}}}}} \\ y & = \sqrt{x + \sqrt{x - \color{#3D99F6}{y}}} \\ y^2 & = x + \sqrt{x - y} \\ (y^2-x)^2 & = x-y \end{aligned}$

$\begin{aligned} \implies y^4 - 2xy^2 + x^2 -x + y & = 0 & \small \color{#3D99F6}{\text{Putting }x=3} \\ y^4 - 6y^2 + y + 6 & = 0 \\ (y+1)(y^3-y^2 -5y +6) & = 0 \\ (y+1)(y-2)(y^2+y-3) & = 0 \\ \implies y = -1, \ \color{#3D99F6}{2}, \ \frac {\pm \sqrt{13}-1}2 \end{aligned}$

Substituting the four likely solutions of $y$ and $x=3$ in $y = \sqrt{x + \sqrt{x - y}}$ , we find that only $y=2$ is the solution. Now, we have:

$\begin{aligned} y^4 - 2xy^2 + x^2 -x + y & = 0 \\ 4y^3 \frac{dy}{dx} - 2y^2 - 4xy \frac{dy}{dx} + 2x -1+\frac{dy}{dx} & = 0 & \small \color{#3D99F6}{\text{Putting }x=3, \ y= 2} \\ 32 \frac{dy}{dx}\bigg|_{x=3} - 8 - 24 \frac{dy}{dx}\bigg|_{x=3} + 6 -1+\frac{dy}{dx}\bigg|_{x=3} & = 0 \\ 9 \frac{dy}{dx}\bigg|_{x=3} & = 3 \\ \implies \frac{dy}{dx}\bigg|_{x=3} & = \frac 13 \end{aligned}$

$\implies a + b = 1 + 3 = \boxed{4}$