Question Number 147402 by vvvv last updated on 20/Jul/21
$$\boldsymbol{{f}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{2}\right)….\left(\boldsymbol{{x}}+\mathrm{2018}\right) \\ $$$$\boldsymbol{{f}}'\left(\mathrm{0}\right)=? \\ $$
Answered by SEKRET last updated on 21/Jul/21
$$\:\:\boldsymbol{\mathrm{y}}=\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)\centerdot….\left(\boldsymbol{\mathrm{x}}+\mathrm{2018}\right) \\ $$$$\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)+\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)+..+\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{2018}\right) \\ $$$$\:\:\frac{\boldsymbol{\mathrm{y}}'}{\boldsymbol{\mathrm{y}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{2}}+…+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{2018}} \\ $$$$\:\boldsymbol{\mathrm{y}}\:'\:=\:\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)\centerdot..\left(\boldsymbol{\mathrm{x}}+\mathrm{2018}\right)\centerdot\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{2}}+…+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{2018}}\right) \\ $$$$\:\boldsymbol{\mathrm{y}}\:'\:\left(\mathrm{0}\right)=\:\mathrm{2018}!\centerdot\left(\frac{\mathrm{1}}{\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+..+\frac{\mathrm{1}}{\mathrm{2018}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\:\boldsymbol{\mathrm{ABDUVALIYEV}}\: \\ $$
Answered by Olaf_Thorendsen last updated on 20/Jul/21
$$\mathrm{I}\:\mathrm{suppose}\:\mathrm{you}\:\mathrm{mean} \\ $$$${f}\left({x}\right)\:=\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+\mathrm{2018}\right) \\ $$$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2018}} {\prod}}\left({x}+{k}\right) \\ $$$$\mathrm{ln}{f}\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}\mathrm{ln}\left({x}+{k}\right) \\ $$$$\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}\frac{\mathrm{1}}{{x}+{k}} \\ $$$$\frac{{f}'\left(\mathrm{0}\right)}{{f}\left(\mathrm{0}\right)}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2018}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\frac{{f}'\left(\mathrm{0}\right)}{\mathrm{2018}!}\:=\:{H}_{\mathrm{2018}} \\ $$$${f}'\left(\mathrm{0}\right)\:=\:\mathrm{2018}!×{H}_{\mathrm{2018}} \\ $$