y=sin^3（2x）求导
\u6c42\u5bfc\u6570\uff0cy\uff1dsin^2X(sin²x)' = 2sinx(sinx)'
= 2sinxcosx
= sin2x
\u6216\uff1a
(sin²x)' = [(1-cos2x)/2]'
= [1/2 - (cos2x)/2]'
= 0 - ½(-sin2x)(2x)'
= ½(sin2x)\u00d72
= sin2x
y'=2sin(2x+\u03c0/3)*[sin(2x+\u03c0/3)]'
=2sin(2x+\u03c0/3)*cos(2x+\u03c0/3)*(2x+\u03c0/3)'
=2sin(2x+\u03c0/3)*cos(2x+\u03c0/3)*2
=4sin(2x+\u03c0/3)*cos(2x+\u03c0/3)
=2sin(4x+2\u03c0/3)
\u671b\u91c7\u7eb3\uff0c\u5982\u6709\u4e0d\u59a5\u8bf7\u56de\u590d\u3002
=3D（sin^2（2x））
=3sin^2（2x））cos（2x）D（2x）
=6 Cos[2 x] Sin[2 x]^2
y/=3sin^2(2x)cos(2x)*2=3sin(4x)sin(2x)
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