西安毛子律枝大学tan?,求例9J=1解tan?!1tan?小=2ln22xtan?!=2x ln2.2tantanxx tan?!1x ln 2.2(tan =)2).secxtan?!x ln 2.2(tan =) · sec2=22xxx1616 2 1 tan 2 x y = ，求 y  . 解 2 1 tan 2 1 2 ln 2 (tan ) x y x   =  2 1 tan 1 1 2 ln 2 2(tan ) (tan ) x x x =    2 1 tan 1 1 1 2 2 ln 2 2(tan ) sec ( ) x x x x =     2 1 tan 2 2 1 1 1 2 ln 2 2(tan ) sec ( ) x x x x =    − 例9