Question

高数求导化简

多项乘除的函数在某点上的导数，要求用导数的定义来求

Answer 1

(〃•⊖•〃)

Answer 2

f(x) = (x-1)(x-2)....(x-n)/[(x+1)(x+2)...(x+n)]
lnf(x)= ln(x-1)+ln(x-2)+...+ln(x-n) -ln(x+1)-ln(x+2)-...-ln(x+n)
f'(x)/f(x) = 1/(x-1)+1/(x-2)+...+1/(x-n) - 1/(x+1)-1/(x+2)-...-1/(x+n)
f'(x)
=[1/(x-1)+1/(x-2)+...+1/(x-n) - 1/(x+1)-1/(x+2)-...-1/(x+n) ] .f(x)
=[1/(x-1)+1/(x-2)+...+1/(x-n) - 1/(x+1)-1/(x+2)-...-1/(x+n) ] .
{ (x-1)(x-2)....(x-n)/[(x+1)(x+2)...(x+n)] }
f'(1)
=(1-2)(1-3)....(1-n)/[(1+1)(1+2)...(1+n)]
=(-1)(-2)....(-(n-1))/ (n+1)!
= (-1)^(n-1) . (n-1)! / (n+1)!
= (-1)^(n-1) / [n(n+1)]

Answer 3

如下图所示