y’=3（sin(2x-1)+2〕^2*【sin(2x-1)+2】’
=3（sin(2x-1)+2〕^2*（2cos（2x-1））
=6（sin(2x-1)+2〕^2*cos(2x-1)
y'=(x+1)^4[(x-2)^4]'+[(x+1)^4]'(x-2)^4
=4(x+1)^4(x-2)^3+4(x+1)^3(x-2)^4
=4(x+1)^3(x-2)^3 (2x-1)
1.y=〔sin(2x-1)+2〕^3
1.y=〔sin(2x-1)+2〕^3
2.y=(x+1)^4(x-2)^4
2.y=(x+1)^4(x-2)^4
數(shù)學(xué)人氣：477 ℃時間：2020-03-23 12:34:24
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