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Question Number 33407 by NECx last updated on 15/Apr/18

$$ify=x!finddy/dx$$

Answered by MJS last updated on 15/Apr/18

$$\mathrm{for}x\in \mathbb{N}\mathrm{no}\mathrm{derivate}\mathrm{exists}$$$$\mathrm{if}\mathrm{we}\mathrm{use}$$$$y=x!=\mathrm{\Gamma }\left(x\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^x}{{\mathrm{e}}^t}dt(x\in \mathbb{R})\Rightarrow$$$$\Rightarrow \frac{dy}{dx}={\mathrm{\Gamma }}^{\prime }\left(x\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^x\ln t}{{\mathrm{e}}^t}dt$$$$\mathrm{if}\mathrm{we}\mathrm{use}$$$$y=x!=\mathrm{\Gamma }\left(x\right)=\underset{0}{\stackrel{1}{\int }}{\left(\ln \frac{1}{t}\right)}^{x}dt(x\in \mathbb{R})\Rightarrow$$$$\Rightarrow \frac{dy}{dx}={\mathrm{\Gamma }}^{\prime }\left(x\right)=\underset{0}{\stackrel{1}{\int }}{\left(\ln \frac{1}{t}\right)}^x\ln \left(\ln \frac{1}{t}\right)dt$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$weknowthat\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dtwithx>0$$$$and\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }\left(x\right)soifwenote$$$$x!=\mathrm{\Gamma }(x+1)wegetx!={\int }_0^{\mathrm{\infty }}{t}^x{e}^{-t}dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{xln\left(t\right)}{e}^{-t}dt\Rightarrow \frac{d}{dx}(x!)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial x}\left(e^{xln\left(t\right)}{e}^{-t}\right)dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)t^{x}dt$$

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