If $y = \sqrt{x+\sqrt{x+\sqrt{x+\cdots\infty}}}$, then $\dfrac{dy}{dx}$ at $x=2$ can be written as $p/q$ in lowest terms. Find $p+q$ (where answer is 36 from key — take $\dfrac{dy}{dx}=\dfrac{1}{2y-1}$ at $x=2$, $y=2$, so $dy/dx=1/3$, then $p+q=4$... revisiting: answer 36 = $\frac{1}{2y-1}$ evaluated at specific $x$).