摘要：求下列函数在x=x0处的导数. =cosx·sin2x+cos3x.x0=, =.x0=2, =.x0=1. 解 =［cosx(sin2x+cos2x)］′ =′=-sinx, ∴f′()=-. = = =,∴f′(2)=0. =(x)′-x′+(lnx)′=-x-1+, ∴f′(1)=- .

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