Question

if $y= \sqrt{sin (x)+\sqrt{sin (x)+\sqrt{sin (x)+....} } }$
till infinity
then $\frac{dy}{dx}=?$

Answer 1

$y= \sqrt{sin x + \sqrt{sin x + \sqrt{sin x + ..} } }$
$y^2-sinx=\sqrt{sin x +\sqrt{sin x +\sqrt{sin x + } .. } }=y$
diff. w.r.t x
$2y \frac{dy}{dx}-cosx= \frac{dy}{dx}$
$\frac{dy}{dx}= \frac{cos x}{2y +1}$

Answer 2

y = undr root= ur = ur of sinx +ur of sinx +ur of sinx.+............
y*y - sinx= "    "    "    "'    "     "
2y = dy/dx -cosx = dy/dx
dy/dx = cosx / 2y+1.
hope this makes u understand.!!!!!