Question Number 105113 by yahyajan last updated on 26/Jul/20

Answered by Dwaipayan Shikari last updated on 26/Jul/20

(d/dx)(x!)=y  logx+log(x−1)+log(x−2)+....=logy  (1/x)+(1/(x−1))+....=(1/y) (dy/dx)  x!((1/x)+(1/(x−1))+....)=(dy/dx)

Answered by OlafThorendsen last updated on 26/Jul/20

x! = Γ(x+1)  (d/dx)x! = Γ′(x+1) = Γ(x+1)ψ_0 (x+1)  with by definition Γ(x+1) = xΓ(x)  and ψ_0 (x+1) = ψ_0 (x)+(1/x)  (d/dx)x! = xΓ(x)[ψ_0 (x)+(1/x)]  (d/dx)x! = Γ(x)[xψ_0 (x)+1]  Γ : Gamma function  ψ_0  : digamma function  (d/dx)x! = x!ψ_0 (x)+(x−1)!