Question

f(x)=(x-1)(x-2).(x-n)/(x+1)(x+2)...(x+n),求f'(1)的值.

Answer 1

f(x)=(x-1)(x-2).(x-n)(x+1)^(-1)(x+2)^(-1)...(x+n)^(-1)

所以f'(x)=(x-1)’*(x-2).(x-n)(x+1)^(-1)(x+2)^(-1)+(x-1)(x-2)‘.(x-n)(x+1)^(-1)(x+2)^(-1)...(x+n)^(-1)...(x+n)^(-1)+……+(x-1)(x-2).(x-n)[(x+1)^(-1)]'(x+2)^(-1)...(x+n)^(-1)+……+(x-1)(x-2).(x-n)(x+1)^(-1)(x+2)^(-1)...[(x+n)^(-1)]'

=(x-2).(x-n)(x+1)^(-1)(x+2)^(-1)+(x-1)(x-2)‘.(x-n)(x+1)^(-1)(x+2)^(-1)...(x+n)^(-1)...(x+n)^(-1)+……+(x-1)(x-2).(x-n)[(x+1)^(-1)]'(x+2)^(-1)...(x+n)^(-1)+……+(x-1)(x-2).(x-n)(x+1)^(-1)(x+2)^(-1)...[(x+n)^(-1)]'

除了第一项以外,其他每项都有x-1,所以x=1时都等于0

所以f'(1)=(1-2).(1-n)=(-1)^(n-1)*(n-1)!