(3)y=ln(x+√x2+1) (x+√x2+1 x2+1 I(x)y+(√x2+1)! x+x2+1 1+ x2+1)” x+√x2+12yx2+1 1+ 2x x+√x2+1 2√x2+1 (4 y=tan y=2tan-(tan-) 2 2.sec-() x tan - sec11 ln( 1) 2 y = x + x + 2 2 1 1 y xx ( ) 1 x x ′ = ⋅ + + ′ + + [( ) ( 1) ] 1 1 2 2 ⋅ ′ + + ′ + + = x x x x ( 1) ] 2 1 1 [1 1 1 2 2 2 ⋅ + ′ + ⋅ + + + = x x x x 2 ] 2 1 1 [1 1 1 2 2 x x x x ⋅ + ⋅ + + + = 1 1 2 + = x （3） 2 1 y tan x = 1 2tan t n 1 (a ) x y x ′ = ⋅ ′ ) 1 ( 1 sec 1 2tan 2 = ⋅ ⋅ ′ x x x 2 2 21 1 tan sec xx x =− ⋅ （4）