Please help. Thanks.?

　
y = √[(cos 2x)/(1 + sin 2x)]

dy/dx = 1/2 [(cos 2x)/(1 + sin 2x)]^(−1/2) * d/dx (cos 2x)/(1 + sin 2x)
dy/dx = 1/2 √[(1 + sin 2x)/cos 2x] * [−2sin 2x(1+sin 2x) − cos 2x(2 cos 2x)] / (1 + sin 2x)²
dy/dx = 1/2 √[(1 + sin 2x)/cos 2x] * [−2sin 2x − 2sin² 2x − 2cos² 2x] / (1 + sin 2x)²
dy/dx = 1/2 √[(1 + sin 2x)/cos 2x] * [−2sin 2x − 2] / (1 + sin 2x)²
dy/dx = −√[(1 + sin 2x)/cos 2x] * (sin 2x + 1) / (1 + sin 2x)²
dy/dx = −√[(1 + sin 2x)/cos 2x] / (1 + sin 2x)
dy/dx = −1 / √[cos 2x (1 + sin 2x)]

y = √[cos2x / (1 + sin2x)]
y = (√cos2x) / [√(1 + sin2x)]

dy/dx
= d{(√cos2x) / [√(1 + sin2x)]}/dx
= {[√(1 + sin2x)]•d(√cos2x)/dx - (√cos2x)•d[√(1 + sin2x)]/dx} / [√(1 + sin2x)]²
= {[√(1 + sin2x)]•d(√cos2x)/d(cos2x)•d(cos2x)/dx - (√cos2x)•d[√(1 + sin2x)]}/d(1 + sin2x)•d(1 + sin2x)/dx} / (1 + sin2x)
= {[√(1 + sin2x)]•(1/2√cos2x)•(-2sin2x) - (√cos2x)•[1/2√(1 + sin2x)]•(2sin2x)} / (1 + sin2x)
= {-[√(1 + sin2x)]²•(sin2x) - (√cos2x)²•(sin2x)} / (1 + sin2x)√[(cos2x)(1 + sin2x)]
= {-(1 + sin2x)•(sin2x) - (cos2x)•(sin2x)} / (1 + sin2x)√[(cos2x)(1 + sin2x)]
= -{1 + sin²2x + cos²2x} / (1 + sin2x)√[(cos2x)(1 + sin2x)]
= -2 / (1 + sin2x)√[(cos2x)(1 + sin2x)]
= -2√[(cos2x)(1 + sin2x)] / (1 + sin2x){√[(cos2x)(1 + sin2x)]}²
= -2√[(cos2x)(1 + sin2x)] / (1 + sin2x)²(cos2x)

ln(y) = (1/2)[ln(cos2x) - ln(1+sin2x)]
y'/y = -tan(2x) - [cos2x/(1+sin2x)]
...

sqrt(u) -> u'/2sqrt(u)
u' = (v/w)' = (v'w-vw')/w²

Here u = cos(2x)/(1+sin(2x))
u' = [-2sin(2x)(1+sin(2x)) - 2cos²(2x)] / (1+sin(2x))² = -2(1+sin(2x))/(1+sin(2x))² = -2/(1+sin(2x))

y' = -2/(1+sin(2x)) / 2sqrt[(cos2x)/(1 + sin2x)]
= -1/(1+sin2x) * sqrt[(1+sin2x)/cos(2x)]
= -1/sqrt[(1+sin2x)*cos(2x)]