If $y = \sqrt{x + \sqrt{x+\sqrt{x+\ldots}}}$ prove that $\frac{dy}{dx}=\frac{1}{2y-1}$

Question

If $$y = \sqrt{x + \sqrt{x+\sqrt{x+\ldots}}}$$

prove that $$\frac{dy}{dx}=\frac{1}{2y-1}$$

Please help me to solve this problem.

Answer 1

Hint: Since $$y = \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}},$$ we can write $$y = \sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

Answer 2

Since

$$ y = \sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

Answer 3

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$\boxed{y' = \frac{1}{2y - 1}}$