Question

If $y=\sqrt{[\sin x+\sqrt{\sin x+\sqrt{\sin x+\sqrt{(\sin x+\cdot \cdot \cdot \infty )}}}]}$, then

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

Answer 1

The correct option is B $\frac{dy}{dx}=\frac{\cos x}{2y-1}.$

Given, $y=\sqrt{[\sin x+\sqrt{\sin x+\sqrt{\sin x+\sqrt{(\sin x+\cdot \cdot \cdot \infty )}}}]}$

$y=\sqrt{\sin x+y}$

$\therefore y^2=\sin x+y$

$\therefore y^2-y=\sin x$

$\therefore (2y-1)\frac{dy}{dx}=\cos x$

$\therefore \frac{dy}{dx}=\frac{\cos x}{2y-1}$