✔ 最佳答案
f(x) = {x(1+x)(2+x)(3+x).....(n+x)} / {(1-x)(2-x)(3-x)....(n-x)}
設分母項為G(x),分子項為Q(x)
G(x) =(1-x)(2-x)(3-x)....(n-x)
Q(x) =x(1+x)(2+x)(3+x).....(n+x) = {x } {(1+x)(2+x)(3+x)....(n+x) }
Q'(x) = (x)' {(1+x)(2+x)(3+x)....(n+x) } +(x) {(1+x)(2+x)(3+x)....(n+x) } '
=(1+x)(2+x)(3+x) ....(n+x) + (x) {(1+x)(2+x)(3+x)....(n+x) } '
f(x) =Q(x)/G(x)
f '(x) ={ Q '(x) G(x)- Q(x)G'(x) }/ {G(x) } ^2
f '(0) = { Q '(0) G(0)- Q(0)G'(0) }/ {G(0) } ^2
= { (1×2×3×....×n)(1×2×3×....×n)- 0×G'(x) }/(1×2×3×....×n)
=1×2×3×....×n =n!
2014-10-19 20:51:11 補充:
更正 f '(0)
f '(0) = { Q '(0) G(0)- Q(0)G'(0) }/ {G(0) } ^2
= { (1×2×3×....×n)(1×2×3×....×n)- 0×G'(0) }/(1×2×3×....×n)^2
=1