Question

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

• A. $\frac{y+x}{y^2-2x}$
• B. $\frac{y^3-x}{2y^2-2xy-1}$
• C. $\frac{y^3+x}{2y^2-x}$
• D. None of these

Answer 1

Answer: (D)

We have, $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}}$ $\Rightarrow$ $y^2=x+\sqrt{y+\sqrt{x+\sqrt{y+...+\infty }}}$ $\Rightarrow$ $y^2=x+\sqrt{y+y}$ $\Rightarrow$ $(y^2-x^2)=2y$ On differentiating both sides w. r. t . $x$ , we get $2(y^{2-x})\left(2y\frac{dy}{dx}-1\right)=2\frac{dy}{dx}$ $\Rightarrow$ $2y(y^2-x)\frac{dy}{dx}-\frac{dy}{dx}=y^2-x$ $\Rightarrow$ $\frac{dy}{dx}(2y^3-2xy-1)=(y^2-x)$ $\Rightarrow$ $\frac{dy}{dx}=\frac{y^2-x}{(2y^3-2xy-1)}$