(2)y′=
=.
(3)法一:∵y=(4x-x)(ex+1)=4xex+4x-xex-x,
∴y′=(4xex+4x-xex-x)′
=(4x)′ex+4x(ex)′+(4x)′-[x′ex+x(ex)′]-x′
=ex4xln 4+4xex+4xln 4-ex-xex-1
=ex(4xln 4+4x-1-x)+4xln 4-1.
法二:y′=(4x-x)′(ex+1)+(4x-x)(ex+1)′
=(4xln 4-1)(ex+1)+(4x-x)ex
=ex(4xln 4+4x-1-x)+4xln 4-1.
(4)y′=(x)′=x′+x′
= +
=.
(5)y′=(sin3x+sin x3)′=(sin3x)′+(sin x3)′
=3sin2xcos x+cos x3·3x2
=3sin2xcos x+3x2cos x3.
8.解:∵y′=(e2x)′·cos 3x+e2x·(cos 3x)′
=2e2x·cos 3x-3e2x·sin 3x,
∴f′(0)=2,∴曲线在点(0,1)处的切线方程为
y-1=2(x-0),即y=2x+1.
设直线l的方程为y=2x+b,
根据题意,得=,解得b=6或-4.
∴直线l的方程为y=2x+6或y=2x-4.