y=ln√[(1-sin2x)/(1+sin2x)],先用对数性质简化
=(1/2)ln[(1-sin2x)/(1+sin2x)]
=(1/2)ln(1-sin2x) - (1/2)ln(1+sin2x)
y' = (-2cos2x) / [2(1-sin2x)] - (2cos2x) / [2(1+sin2x)],对数公式(lnx)'=1/x,加上链式法则
=(-cos2x) * (1+sin2x+1-sin2x) / [(1-sin2x)(1+sin2x)],分母合并
=(-cos2x) * 2/(1-sin²2x)
=(-cos2x) * 2/cos²2x
=-2/cos2x
=-2sec2x