Question

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

• A. $\frac{1}{2y+1}$
• B. $\frac{1}{2y-1}$
• C. $\frac{1}{xy}$
• D. 1

Answer 1

Answer: (B)

Let $y=\sqrt{x+\sqrt{x+\sqrt{x+....\infty }}}$
$\Rightarrow$ $y=\sqrt{x+y}$
$\Rightarrow$ $y^2=x+y$
$\Rightarrow$ $y^2-y-x=0$
On differentiating w.r.t. $x,$ we get
$2y\frac{dy}{dx}-\frac{dy}{dx}-1=0$
$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{(2y-1)}$