(1)∵y==x+x3+, ∴y′=(x)′+(x3)′+(x-2sinx)′ =-x+3x2-2x-3sinx+x-2cosx. (2)方法一 y=(x2+3x+2)(x+3) =x3+6x2+11x+6, ∴y′=3x2+12x+11. 方法二 y′=[(x+1)(x+2)]′(x+3)+(x+1)(x+2)(x+3)′ =[(x+1)′(x+2)+(x+1)(x+2)′](x+3)+(x+1)(x+2) =(x+2+x+1)(x+3)+(x+1)(x+2) =(2x+3)(x+3)+(x+1)(x+2) =3x2+12x+11. (3)∵y=-sin(-cos)=sinx, ∴y′=(sinx) ′= (sinx)′=cosx. (4)y=+==, ∴y′=()′==. |