Question

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

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

Answer 1

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

We have
$y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\cdots to\infty }}}$
The given series may be written as
$y=\sqrt{\sin x+y}$
Squaring both the sides, we get
$y^2=\sin x+y$
Differentiating both sides w.r.t. $x$
$2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$
$2y\frac{dy}{dx}-\frac{dy}{dx}=\cos x$
$\frac{dy}{dx}(2y-1)=\cos x$
$\frac{dy}{dx}=\frac{\cos x}{2y-1}$