Question

已知：f(x)=(x-1)(x-2)(x-3)(x-4)(x-5). 求：f '(0)= ?

Answer 1

f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)
f′(x)=(x-1)′(x-2)(x-3)(x-4)(x-5)+(x-1)(x-2)′(x-3)(x-4)(x-5)+(x-1)(x-2)(x-3)′(x-4)(x-5)+(x-1)(x-2)(x-3)(x-4)′(x-5)+(x-1)(x-2)(x-3)(x-4)(x-5)′
=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-1)(x-2)(x-4)(x-5)+(x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-3)(x-4)

f′(0)=(-2)(-3)(-4)(-5)+(-1)(-3)(-4)(-5)+(-1)(-2)(-4)(-5)+(-1)(-2)(-3)(-5)+(-1)(-2)(-3)(-4)
=120+60+40+30+24
=274

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Answer 2

f(x)=(x-1)(x-2)(x-3)(x-4)(x-5).
这个导数可以不求出的
f'(0)=2*3*4*5+1*3*4*5+1*2*4*5+1*2*3*5+1*2*3*4

Answer 3

f'(x)=(x-1)(x-2)(x-3)(x-4)+(x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-2)(x-3)(x-4)(x-5)
x=0代入上式 得

f '(0)= 2x3x4 + 2x3x5 + 2x4x5 + 3x4x5 + 2x3x4x5
= 24+30 + 40 + 60 + 120
= 274