Vì

$\begin{align}  & y'=\frac{\left( \cos x \right)'2{{\sin }^{2}}x-\cos x.\left( 2{{\sin }^{2}}x \right)'}{{{\left( 2{{\sin }^{2}}x \right)}^{2}}} \\  & =\frac{-\sin x.2{{\sin }^{2}}x-\cos x.\left( 2.2\sin x.\cos x \right)}{4.{{\sin }^{4}}x} \\  & =\frac{-2{{\sin }^{3}}x-4\sin x{{\cos }^{2}}x}{4{{\sin }^{4}}x}=-\frac{{{\sin }^{2}}x+2{{\cos }^{2}}x}{2{{\sin }^{3}}x} \\  & =-\frac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+{{\cos }^{2}}x}{2{{\sin }^{3}}x}=-\frac{1+{{\cos }^{2}}x}{2{{\sin }^{3}}x} \\ \end{align}$