Question

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

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

Answer 1

Answer: (A)

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