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Question Number 108043 by ZiYangLee last updated on 14/Aug/20

$$\mathrm{Given}\mathrm{f}\left(x\right)=x(x+1)(x+2)...(x+n)$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{f}}^{\prime }\left(0\right).$$

Answered by hgrocks last updated on 14/Aug/20

$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\ln (\mathrm{x}+\mathrm{k})$$$$\frac{{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1}+\frac{1}{\mathrm{x}+2}+.......\frac{1}{\mathrm{x}+\mathrm{n}}$$$$\frac{{\mathrm{f}}^{\prime }\left(0\right)}{\mathrm{f}\left(0\right)}=\frac{1}{\mathrm{x}}+1+\frac{1}{2}+\frac{1}{3}+.........\frac{1}{\mathrm{n}}$$$${\mathrm{f}}^{\prime }\left(0\right)=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}{(\mathrm{x}+\mathrm{k})}_{\mathrm{x}=0}+\mathrm{f}\left(0\right)\left({\mathrm{H}}_{\mathrm{n}}\right)$$$$=1.2.3.....\mathrm{n}+0\times {\mathrm{H}}_{\mathrm{n}}$$$$=\mathrm{n}!$$$$$$

Commented by ZiYangLee last updated on 14/Aug/20

$$\mathrm{wow}...$$

Answered by Her_Majesty last updated on 14/Aug/20

$$f\left(x\right)=x(x+1)(x+2)...(x+n)=$$$$=x^{n+1}+(....)x^2+n!x$$$$f^{\prime }\left(x\right)=(n+1)x^n+2(...)x+n!$$$$f^{\prime }\left(0\right)=n!$$

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