(1)y′=(
x5)′-(1 5
x3)′+(3x2)′+(4 3
)′
2
=x4-4x2+6x.
(2)法一:∵y=(3x3-4x)(2x+1)=6x4+3x3-8x2-4x,
∴y′=24x3+9x2-16x-4.
法二:y′=(3x3-4x)′(2x+1)+(3x3-4x)(2x+1)′
=(9x2-4)(2x+1)+(3x3-4x)?2
=24x3+9x2-16x-4.
(3)y′=
x′(1?x+x2)?x(1?x+x2)′ (1?x+x2)2
=
=(1?x+x2)?x(?1+2x) (1?x+x2)2
.1?x2
(1?x+x2)2