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设f(x)=x(x-1)(x-2)(x-3)(x-4),则f'(4)=
人气:271 ℃ 时间:2020-07-10 23:50:22
解答
f(x)=x(x-1)(x-2)(x-3)(x-4),
f'(x)=[x(x-1)(x-2)(x-3)]'(x-4)+x(x-1)(x-2)(x-3)[(x-4)']
=[x(x-1)(x-2)(x-3)]'(x-4)+x(x-1)(x-2)(x-3)
f'(4)==0+4×3×2×1=24
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