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Question Number 151884 by mathdanisur last updated on 23/Aug/21
f(x)=(x-1)(x-2)...(x-2021) f^′ (2021) = ?
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)…\left(\mathrm{x}-\mathrm{2021}\right) \\ $$$$\mathrm{f}\:^{'} \left(\mathrm{2021}\right)\:=\:? \\ $$
Answered by mr W last updated on 24/Aug/21
ln y=ln (x−1)+ln (x−2)+...+ln (x−2021) ((y′)/y)=(1/(x−1))+(1/(x−2))+...+(1/(x−2021)) y′=(x−1)(x−2)...(x−2021)[(1/(x−1))+(1/(x−2))+...+(1/(x−2021))] y′∣_(x=2021) =2020×2019×...×2×1=2020!
$$\mathrm{ln}\:{y}=\mathrm{ln}\:\left({x}−\mathrm{1}\right)+\mathrm{ln}\:\left({x}−\mathrm{2}\right)+…+\mathrm{ln}\:\left({x}−\mathrm{2021}\right) \\ $$$$\frac{{y}'}{{y}}=\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+…+\frac{\mathrm{1}}{{x}−\mathrm{2021}} \\ $$$${y}'=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\left({x}−\mathrm{2021}\right)\left[\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+…+\frac{\mathrm{1}}{{x}−\mathrm{2021}}\right] \\ $$$${y}'\mid_{{x}=\mathrm{2021}} =\mathrm{2020}×\mathrm{2019}×…×\mathrm{2}×\mathrm{1}=\mathrm{2020}! \\ $$
Commented by mathdanisur last updated on 24/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by Olaf_Thorendsen last updated on 23/Aug/21
f(x) = Π_(k=1) ^(2021) (x−k) f(x) = (x−2021)Π_(k=1) ^(2020) (x−k) Let g(x) = Π_(k=1) ^(2020) (x−k) f(x) = (x−2021)g(x) f′(x) = (x−2021)g′(x)+g(x) f′(2021) = g(2021) = Π_(k=1) ^(2020) (2021−k) f′(2021) = Π_(k=1) ^(2020) k = 2020!
$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2021}} {\prod}}\left({x}−{k}\right) \\ $$$${f}\left({x}\right)\:=\:\left({x}−\mathrm{2021}\right)\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\prod}}\left({x}−{k}\right) \\ $$$$\mathrm{Let}\:{g}\left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\prod}}\left({x}−{k}\right) \\ $$$${f}\left({x}\right)\:=\:\left({x}−\mathrm{2021}\right){g}\left({x}\right) \\ $$$${f}'\left({x}\right)\:=\:\left({x}−\mathrm{2021}\right){g}'\left({x}\right)+{g}\left({x}\right) \\ $$$${f}'\left(\mathrm{2021}\right)\:=\:{g}\left(\mathrm{2021}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\prod}}\left(\mathrm{2021}−{k}\right) \\ $$$${f}'\left(\mathrm{2021}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\prod}}{k}\:=\:\mathrm{2020}! \\ $$
Commented by mathdanisur last updated on 23/Aug/21
Thank You Ser
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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