Question

y=[1-sin2x /1+sin2x] ^1/2

Answer 1

1st take log on both sides,
then differentiate the those function
logy=1/2log (1_sin2x/1+sin2x)
use product rule as well as chain rule
dy/dx×1/y=1/2d/dx (1_sin2x/1+sin2x)(1_sin2x/1+sin2x)
differentiate the above function using quotient rule
dy/dx=1_sin2x/1+sin2x)^1/2(cos2x_cos2x×sin2x)
I hope this ans will be right