Question

If $y=\sqrt{\sin x-\sqrt{\sin x-\sqrt{\sin x-.....\infty }}}$ , then $\frac{dy}{dx}=$

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

Answer 1

The correct option is A $\frac{dy}{dx}=\frac{\cos x}{2y+1}$

$y=\sqrt{\sin x-\sqrt{\sin x-\sqrt{\sin x-.....\infty }}}{y}^2=\sin x-\sqrt{\sin x-\sqrt{\sin x.....\infty }}$
$y^2=\sin x-y$
$2y\frac{dy}{dx}=\cos x-\frac{dy}{dx}\frac{dy}{dx}=\frac{\cos x}{2y+1}$