Compute $\frac{dy}{dx}$ given $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$

Question

I need to know wich answer is right

https://i.stack.imgur.com/QcxdH.jpg

I tried to solve it using recursivity but I didn't get any one of them

• $y_1=\sqrt{x}$
• $y'=\frac{1}{2\sqrt{x}}=\frac{1}{2y_1}$
• $y_2=\sqrt{x+\sqrt{x}}=\sqrt{x+y_1}$ ;
• $y_2'=\frac{1+y_1'}{2\sqrt{x+y_1}}=\frac{1+y_1'}{2y_2}=\frac{1+\frac{1}{2y_1}}{2y_2}=\frac{1+2y_1}{4y_1y_2}$
• $y_3=\sqrt{x+\sqrt{x+\sqrt{x}}}=\sqrt{x+y_2}$
• $y_3'=\frac{1+y_2'}{2y_3}=\frac{1+\frac{1+2y_1}{4y_1y_2}}{2y_3}=\frac{1+2y_1+4y_1y_2}{8y_1y_2y_3}$
• .......
• $y_n=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y_{n-1}}$
• $y_n'=\frac{1+y_{n-1}'}{2y_n}=\frac{1+\sum_{i=1}^{n-1}2^i\prod_{j=1}^iy_j}{2^n\prod_{i=1}^{n}y_i}$

Answer 1

As @WW1 mentioned in his comments, $$y=\sqrt {x+{\sqrt {x+\cdots}}} $$ $$\Rightarrow y=\sqrt {x+y} $$ $$\Rightarrow y^2=x+y $$ Now using the techniques of implicit differentiation, we have $$2y\frac{dy}{dx} = 1+\frac{dy}{dx} $$ Rearranging, we get, $$\Rightarrow \frac{dy}{dx} =\frac {1}{2y-1} $$ Hope it helps.

Answer 2

$y^2=x+y$ now

$2ydy/dx=1+dy/dx$

Answer 3

$$y^2-y=x$$ then $$(2y-1)y'=1.$$