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Question Number 91622 by Ar Brandon last updated on 02/May/20

$$\frac{\mathrm{d}(\mathrm{x}!)}{\mathrm{dx}}=$$

Commented by MJS last updated on 02/May/20

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{several}\mathrm{times}\mathrm{before}$$$$x!\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{N}\Rightarrow \mathrm{no}\mathrm{derivate}\mathrm{exists}$$$$\mathrm{if}\mathrm{you}\mathrm{mean}\mathrm{\Gamma }\left(x\right)\mathrm{you}\mathrm{must}\mathrm{say}\mathrm{it}...$$

Answered by  M±th+et+s last updated on 02/May/20

$$iamnotsureifmysolutioniscorrect$$$$\frac{d}{dx}\mathrm{\Gamma }\left(x\right)=\mathrm{\Gamma }\left(x\right){\mathrm{\varnothing }}_0\left(x\right)$$$${\mathrm{\varnothing }}_0\left(x\right)=\underset{k=1}{\sum ^{x-1}}\frac{1}{k}-\gamma$$$$x!=\mathrm{\Gamma }(x+1)$$$$so\frac{d(x!)}{dx}=\frac{d(\mathrm{\Gamma }(x+1)}{dx}=\mathrm{\Gamma }(x+1){\mathrm{\varnothing }}_0(x+1)$$$$$$

Answered by Ar Brandon last updated on 02/May/20

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