Question

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

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

Answer 1

The correct option is D $\frac{1}{2y-1}$

$y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots \infty }}}$
$\Rightarrow y=\sqrt{x+y}$
Differentiating both sides wrt $x$,
$\frac{dy}{dx}=\frac{1}{2\sqrt{x+y}}\times (1+\frac{dy}{dx})$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\times (1+\frac{dy}{dx})$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}+\frac{1}{2y}\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}$