Crazy Calculus 7

Relevant wiki: Implicit Differentiation Problem Solving - Basic

$\begin{aligned} y & = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}} \\ \implies y & = \sqrt{x+y} \\ y^2 & = x+y \\ \implies 2y \frac {dy}{dx} & = 1 + \frac {dy}{dx} \\ (2y - 1) \frac {dy}{dx} & = 1 \\ \implies \frac {dy}{dx} & = \frac 1{2y-1} : P \end{aligned}$

Considering $R$ :

$\begin{aligned} y^2 & = x + y \\ y^2 - y & = x \\ y^2 - y + \frac 14 & = x + \frac 14 \\ \left(y - \frac 12 \right)^2 & = x + \frac 14 \\ y - \frac 12 & = \sqrt{x + \frac 14} \\ 2y - 1 & = 2 \times \frac {\sqrt{1+4x}}2 \\ \implies \frac 1{2y-1} & = \frac 1{\sqrt{1+4x}} : R \end{aligned}$

Considering $S$ :

$\begin{aligned} S: \frac y{2x+y} & = \frac y{2x+2y-y} \\ & = \frac y{2y^2-y} \\ & = \frac 1{2y-1} \end{aligned}$

Considering $Q$ :

To show that $\dfrac x{x+2y} \ne \dfrac 1{2y-1}$ , we can use a case, for example $x=1$ , then from $y^2 = x+y$ , we get $y = \dfrac {1+\sqrt 5}2$ .Substituting into $P: \dfrac 1{2y-1} = \dfrac 1{\sqrt 5}$ , substituting in $Q: \dfrac x{x+2y} = \dfrac 1{2+\sqrt 5} \ne P$ .

Therefore, $\dfrac {dy}{dx} = \boxed{P, R \text{ and }S}$ .