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Question Number 128069 by 0731619177 last updated on 04/Jan/21

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$\frac{d}{dx}(x!)={\mathrm{\Gamma }}^{\prime }(x+1)=\mathrm{\Gamma }(x+1)\psi (x+1)=x!(-\gamma +\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+x})$$

Answered by A8;15: last updated on 04/Jan/21

$$\frac{\mathrm{d}(\mathrm{x}!!)}{\mathrm{dx}}=\frac{1}{4}\cdot \mathrm{x}!!\cdot (2{\psi }^{\left(0\right)}(\frac{\mathrm{x}}{2}+1)+\pi \cdot \log \left(\frac{2}{\pi }\right)\cdot \sin \left(\pi \mathrm{x}\right)+\log \left(4\right))$$

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