已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为

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已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为已知f(x)=x(x-1)(x-2)

已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为
已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为

已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为
由f‘(a)=lim(x→0) [f(a+x)-f(a)] / x
f‘(0)=lim(x→0) x(x-1)(x-2)(x-3)(x-4)(x-5)/x
=lim(x→0) (x-1)(x-2)(x-3)(x-4)(x-5)
=-1×2×3×4×5
=-120

f‘(0)=lim(x→0)x(x-1)(x-2)(x-3)(x-4)(x-5)/x
=lim(x→0)(x-1)(x-2)(x-3)(x-4)(x-5)
=-1×2×3×4×5
=-120