Question

$y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+-\infty }}}$ then $\frac{dy}{dx}$ equals:$(\sin x>0)$

• A. $\frac{\cos x}{2y-1}$
• B. $\frac{y}{2\tan x+y\sec x}$
• C. $\frac{1}{\sqrt{1+4\sin x}}$
• D. $\frac{2\cos x}{\sin x+2y}$

Answer 1

Answer: (A)

$\frac{\cos x}{2y-1}$

$y=\sqrt{\sin x\sqrt{\sin x\sqrt{\sin x+-\infty }}}$
Taking square on both the sides, we get
$y^2=\sin x+\sqrt{\sin x\sqrt{\sin x\sqrt{\sin x+-\infty }}}$
$y^2=\sin x+y$ ............ Since $y=\sqrt{\sin x\sqrt{\sin x\sqrt{\sin x+-\infty }}}$
$2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{\cos x}{2y-1}$