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Question Number 81226 by 20092104 last updated on 10/Feb/20

$$\frac{d}{dx}(x!)and\frac{d}{dx}(1+2+3+...+x)$$

Answered by MJS last updated on 10/Feb/20

$$\frac{d}{dx}[x!]{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}\mathrm{because}x!\mathrm{is}\mathrm{not}\mathrm{continuous}$$$$1+2+3+...+x=\frac{x(x+1)}{2}$$$$\frac{d}{dx}\left[\frac{x(x+1)}{2}\right]=x+\frac{1}{2}$$

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