Discussion Forum

If $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+..........\infty}}}}$, then $\frac{dy}{dx}$= 

Answer:

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

  $4 y^{3}\frac{dy}{dx}-2y^{2}-4yx\frac{dy}{dx}-\frac{2dy}{dx}+2x=0$

 $\frac{dy}{dx}(4y^{3}-4xy-2)=-2x+2y^{2}$

$\Rightarrow$      $\frac{dy}{dx}=\frac{y^{2}-x}{2y^{3}-2xy-1}$