Answer
$\frac{dy}{dx} =2xsin^{3}x(sin x + 2xcos x) +cos^{-2}x(1+2xcos^{-1} x \times sin x) $
Work Step by Step
$y= a+b$ where, $a = x^{2} u ^{4 }$ and, $ b = x v^{-2} $
where,
$u = sin x$
$v = cos x$
Now,
$u' = cos x$
$v' = -sin x$
$a' = 2xu^{4} + 4x^{2}u^{3} \times cos x = 2xsin^{4}x + 4x^{2}sin^{3}x cosx = 2xsin^{3}x(sin x + 2xcos x)$
$b' = v^{-2} - 2xv^{-3}(-sin x) = cos^{-2} x + 2xcos^{-3} x \times sin x = cos^{-2}x(1+2xcos^{-1} x \times sin x)$
So now,
$\frac{dy}{dx} = a' +b'$
$\frac{dy}{dx} =2xsin^{3}x(sin x + 2xcos x) +cos^{-2}x(1+2xcos^{-1} x \times sin x) $
