Answer:
Option D
Explanation:
We have,
$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+..........\infty}}}}$
$y=\sqrt{x+\sqrt{y+y}}\Rightarrow y^{2}=x+\sqrt{2y}$
$\Rightarrow$ $ y^{2}-x=\sqrt{2y}\Rightarrow (y^{2}-x)^{2}=2y$
$\Rightarrow$ $y^{4}-2xy^{2}+x^{2}$=2y
$\Rightarrow$ $y^{4} -2xy^{2}-2y+x^{2}$=0
On differentiating w.r.t to x , we get
$4 y^{3}\frac{dy}{dx}-2y^{2}-4yx\frac{dy}{dx}-\frac{2dy}{dx}+2x=0$
$\frac{dy}{dx}(4y^{3}-4xy-2)=-2x+2y^{2}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{y^{2}-x}{2y^{3}-2xy-1}$
