a,
$y=\dfrac{\sin^2x}{(1+\cos x)^2}$
$=\dfrac{\sin^2x}{1+\cos^2x+2\cos x}$
$y'=\dfrac{(\sin^2x)'(1+\cos^2x+2\cos x)-\sin^2x (1+\cos^2x+2\cos x)'}{(1+\cos x)^4}$
$=\dfrac{\sin2x(1+\cos^2x+2\cos x)-\sin^2x(-2\sin x\cos x-2\sin x)}{(1+\cos x)^4}$
$=\dfrac{\sin2x+\cos^2x.\sin2x+2\sin2x.\cos x +\sin2x.\sin^2x-2\sin^3x}{(1+\cos x)^4}$
b,
$y'=\dfrac{1}{2\sqrt{\cot x}}(\cot x)'$
$=\dfrac{-1}{2\sin^2x\sqrt{\cot x}}$
c,
$y'=3(2+\sin^22x)^2(2+\sin^22x)'$
$=3(2+\sin^22x)^2.(2\sin 2x(\sin 2x)')$
$=3(2+\sin^22x)^2.4\sin2x\cos2x$
$=6(2+\sin^22x)^2\sin4x$
