Question

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

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

Answer 1

Answer: (D)

$y=\sqrt{x+y}\Rightarrow y^2=x+y\Rightarrow 2y\frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow \left(2y-1\right)\frac{dy}{dx}=1\therefore \frac{dy}{dx}=\frac{1}{2y-1}$