若y=x(x-1)(x-2)(x-3),则y′(0)=
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若y=x(x-1)(x-2)(x-3),则y′(0)=若y=x(x-1)(x-2)(x-3),则y′(0)=若y=x(x-1)(x-2)(x-3),则y′(0)=把y=x(x-1)(x-2)(x-3)
若y=x(x-1)(x-2)(x-3),则y′(0)=
若y=x(x-1)(x-2)(x-3),则y′(0)=
若y=x(x-1)(x-2)(x-3),则y′(0)=
把y=x(x-1)(x-2)(x-3)看成y=x*(x-1)(x-2)(x-3) y′=(x-1)(x-2)(x-3)+x[(x-1)(x-2)(x-3)]′ y′(0)=-6+0=-6
y′(x)=(x-1)(x-2)(x-3)+x(x-2)(x-3)+x(x-1)(x-3)+x(x-1)(x-2) 所以y′(0)=(-1)*(-2)*(-3)=-6
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若y=x(x-1)(x-2)(x-3),则y′(0)=
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