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Question Number 105113 by yahyajan last updated on 26/Jul/20
Answered by Dwaipayan Shikari last updated on 26/Jul/20
ddx(x!)=ylogx+log(x−1)+log(x−2)+....=logy1x+1x−1+....=1ydydxx!(1x+1x−1+....)=dydx
Answered by OlafThorendsen last updated on 26/Jul/20
x!=Γ(x+1)ddxx!=Γ′(x+1)=Γ(x+1)ψ0(x+1)withbydefinitionΓ(x+1)=xΓ(x)andψ0(x+1)=ψ0(x)+1xddxx!=xΓ(x)[ψ0(x)+1x]ddxx!=Γ(x)[xψ0(x)+1]Γ:Gammafunctionψ0:digammafunctionddxx!=x!ψ0(x)+(x−1)!
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