Infinitely nested differential

$y\quad =\quad \sqrt { x\quad +\quad \sqrt { x\quad +\quad \sqrt { x\quad +\quad ... } } } \\ y\quad =\quad \sqrt { x\quad +\quad y } \\ { y }^{ 2 }\quad =\quad x\quad +\quad y\\ { y }^{ 2 }\quad -\quad y\quad =\quad x\\ \frac { d }{ dx } (y^{ 2 }-y)\quad =\quad \frac { dx }{ dx } \\ 2y\frac { dy }{ dx } \quad -\quad \frac { dy }{ dx } \quad =\quad 1\\ \frac { dy }{ dx } (2y\quad -\quad 1)\quad =\quad 1\\ \frac { dy }{ dx } \quad =\quad \frac { 1 }{ 2y\quad -\quad 1 }$ Substitute $4.5$ for $y$ , you'll get: $\frac { dy }{ dx } \quad =\quad \frac { 1 }{ 2(4.5)\quad -\quad 1 } \\ \frac { d y}{ dx } \quad =\quad \frac { 1 }{ 8 } \quad =\quad \boxed { 0.125 }$

it is similar with $y^2 = x + y$

then $2y dy/dx = 1 + dy/dx$

subst $y = 4.5$

$9dy/dx = 1 + dy/dx$

$dy/dx = 0.125$

yay :D

Squaring

y^2 = x + y

2 y dy/dx = 1 + dy/dx

( 2 y - 1 ) dy/dx = 1

dy/dx = 1/( 2 y - 1 )

y = 4.5

dy/dx = 1/8 = 0.125