Question Number 130496 by Adel last updated on 26/Jan/21
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=? \\ $$
Answered by MJS_new last updated on 26/Jan/21
![x! is defined for x∈N ⇒ no derivate exists if you mean x!=Γ (x+1) (d/dx)[Γ (x+1)]=∫_0 ^∞ t^x e^(−t) ln t dt](https://www.tinkutara.com/question/Q130499.png)
$${x}!\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{no}\:\mathrm{derivate}\:\mathrm{exists} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:{x}!=\Gamma\:\left({x}+\mathrm{1}\right) \\ $$$$\frac{{d}}{{dx}}\left[\Gamma\:\left({x}+\mathrm{1}\right)\right]=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}} \mathrm{e}^{−{t}} \mathrm{ln}\:{t}\:{dt} \\ $$
Commented by Ar Brandon last updated on 26/Jan/21
Uhhh ! You are less nervous today Sir wrt the question. Haha !
Commented by MJS_new last updated on 26/Jan/21
$$\mathrm{o}\:\mathrm{tempora}\:\mathrm{o}\:\mathrm{mores} \\ $$
Commented by MJS_new last updated on 26/Jan/21
![s (x) := { ((−1; x=−(π/2)+2nπ)),((0; x=nπ)),((1; x=(π/2)+2nπ)) :}∀n∈Z (d/dx)[s (x)]=? same procedure as above s (x) =sin x ⇒ (d/dx)[s (x)]=cos x this is not how we should work I′m still of the opinion but I give up explaining by now](https://www.tinkutara.com/question/Q130507.png)
$$\mathrm{s}\:\left({x}\right)\::=\begin{cases}{−\mathrm{1};\:{x}=−\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi}\\{\mathrm{0};\:{x}={n}\pi}\\{\mathrm{1};\:{x}=\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi}\end{cases}\forall{n}\in\mathbb{Z} \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{s}\:\left({x}\right)\right]=? \\ $$$$\mathrm{same}\:\mathrm{procedure}\:\mathrm{as}\:\mathrm{above} \\ $$$$\mathrm{s}\:\left({x}\right)\:=\mathrm{sin}\:{x}\:\Rightarrow\:\frac{{d}}{{dx}}\left[\mathrm{s}\:\left({x}\right)\right]=\mathrm{cos}\:{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{how}\:\mathrm{we}\:\mathrm{should}\:\mathrm{work} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{still}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opinion}\:\mathrm{but}\:\mathrm{I}\:\mathrm{give}\:\mathrm{up}\:\mathrm{explaining} \\ $$$$\mathrm{by}\:\mathrm{now} \\ $$
Answered by Dwaipayan Shikari last updated on 26/Jan/21
$$\Gamma\left({x}+\mathrm{1}\right)={x}!={y} \\ $$$${y}'=\Gamma'\left({x}+\mathrm{1}\right)=\Gamma\left({x}+\mathrm{1}\right)\psi\left({x}+\mathrm{1}\right) \\ $$$${Example} \\ $$$$\frac{{dx}!}{{dx}}\mid_{{x}=\mathrm{1}} =\Gamma\left(\mathrm{2}\right)\psi\left(\mathrm{2}\right)=\mathrm{1}−\gamma\:\:\left(\gamma=\mathcal{E}{ulerian}\:{Constant}=\mathrm{0}.\mathrm{57721}\right) \\ $$
Commented by Ar Brandon last updated on 26/Jan/21
$$\Gamma\left(\mathrm{m}\right)\psi\left(\mathrm{n}\right)=\:? \\ $$
Commented by Ar Brandon last updated on 26/Jan/21
$$\psi'\left(\mathrm{x}+\mathrm{1}\right)= \\ $$
Commented by Dwaipayan Shikari last updated on 26/Jan/21
Answered by Olaf last updated on 26/Jan/21
$$\frac{{d}}{{dx}}{x}!\:=\:\frac{{d}\Gamma}{{dx}}\:\left({x}+\mathrm{1}\right)\:=\:\Gamma\left({x}+\mathrm{1}\right)\psi_{\mathrm{0}} \left({x}+\mathrm{1}\right) \\ $$$$\frac{{d}}{{dx}}{x}!\:=\:{x}\Gamma\left({x}\right)\psi_{\mathrm{0}} \left({x}+\mathrm{1}\right) \\ $$
Commented by Adel last updated on 26/Jan/21
$$\mathrm{whats}\:\mathrm{the} \\ $$$$\psi_{\mathrm{0}} \left(\mathrm{x}+\mathrm{1}\right)=? \\ $$
Commented by Dwaipayan Shikari last updated on 26/Jan/21
$${Digamma}\:{function} \\ $$$$\frac{\Gamma'\left({x}\right)}{\Gamma\left({x}\right)}=\psi\left({x}\right) \\ $$$$\psi\left({x}\right)=−\gamma+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+{x}}\:\:\:\:\left(\gamma=\mathrm{0}.\mathrm{57721}\:{Eulerian}\:{Constant}\right) \\ $$