作业帮设f(x)=(x-2)(x-1)x(x+1)(x+2),则f′(1)的值为
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设f(x)=(x-2)(x-1)x(x+1)(x+2),则f′(1)的值为
设f(x)=(x-2)(x-1)x(x+1)(x+2),则f′(1)的值为
设f(x)=(x-2)(x-1)x(x+1)(x+2),则f′(1)的值为
答：
f(x)=(x-2)(x-1)x(x+1)(x+2)
f'(x)=(x-2)x(x+1)(x+2)+(x-1)[(x-2)x(x+1)(x+2)]'
f'(1)=(-1)*1*2*3+0
f'(1)=-6
f'(x)=(x-1)x(x+1)(x+2)+(x-2)x(x+1)(x+2)+(X-2)(X-1)(X+1)(X+2)+(x-2)(x-1)x(x+2)+(x-2)(x-1)x(x+1)
所以f'(1)=(1-2)1(1+1)(1+2)=-6