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Question Number 203910 by Davidtim last updated on 01/Feb/24
if y=(1−x)(2−x)(3−x)∙∙∙(n−x) y^′ =?
$${if}\:\:\:{y}=\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)\centerdot\centerdot\centerdot\left({n}−{x}\right) \\ $$$${y}^{'} =? \\ $$
Answered by mr W last updated on 01/Feb/24
y=Π_(k=1) ^n (k−x) ln y=Σ_(k=1) ^n ln (k−x) ((y′)/y)=−Σ_(k=1) ^n (1/(k−x)) y′=−Π_(k=1) ^n (k−x)Σ_(k=1) ^n (1/(k−x))
$${y}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({k}−{x}\right) \\ $$$$\mathrm{ln}\:{y}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\:\left({k}−{x}\right) \\ $$$$\frac{{y}'}{{y}}=−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}−{x}} \\ $$$${y}'=−\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({k}−{x}\right)\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}−{x}} \\ $$
Answered by witcher3 last updated on 02/Feb/24
f(x)=Π_(i=1) ^n p_i f′(x)=Σ_(i=1) ^n p_i ′.Π_(l=1#i) ^n p_l ;∀n≥2 proof n=2 f(x)=p_1 .p_2 ⇒f′(x)=p_1 ′.p_2 +p_2 ′.p_1 =Σ_(i=1) ^n (p_i )Π_(l=1≠i) ^2 p_l suppose ∀n∈N f′(x)=Σp_i ′Πp_(l ) true shows for f(x)=p_1 ....p_(n+1) f′(x)=Σ_(l=1) ^(n +1) p_l ′Π_(i=1#l) ^(n+1) p_i f(x)=(p_(n+1) ).(p_1 ....p_n )⇒f′(x)=p_(n+1) ′.(p_1 ...p_n ) +p_(n+1) .Σ_(k=l) ^n p_k ′.Π_(l=1#k) ^n p_l =Σ_(k=n+1) ^(n+1) p_k ′.Π_(l=1#k) ^(n+1) p_k +Σ_(k=1) ^n p′_k Π_(l=1#k) ^(n+1) p_l =Σ_(k=1) ^(n+1) p′_k Π_(l=1≠k) ^(n+1) p_l prooved in our exempl p_i =(i−x)⇒p′_i =−1 y′=Σ_(k=1) ^n (−1)Π_(l=1≠k) ^n (l−x)=−Σ_(k=1) ^n (1/(k−x))Π_(l=1) ^n (l−x) =−yΣ_(k=1) ^n (1/(k−x))
$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{p}_{\mathrm{i}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{p}_{\mathrm{i}} '.\underset{\mathrm{l}=\mathrm{1}#\mathrm{i}} {\overset{\mathrm{n}} {\prod}}\mathrm{p}_{\mathrm{l}} ;\forall\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{proof}\:\mathrm{n}=\mathrm{2} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{p}_{\mathrm{1}} .\mathrm{p}_{\mathrm{2}} \Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{p}_{\mathrm{1}} '.\mathrm{p}_{\mathrm{2}} +\mathrm{p}_{\mathrm{2}} '.\mathrm{p}_{\mathrm{1}} =\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{p}_{\mathrm{i}} \right)\underset{\mathrm{l}=\mathrm{1}\neq\mathrm{i}} {\overset{\mathrm{2}} {\prod}}\mathrm{p}_{\mathrm{l}} \\ $$$$\mathrm{suppose}\:\forall\mathrm{n}\in\mathbb{N}\:\:\mathrm{f}'\left(\mathrm{x}\right)=\Sigma\mathrm{p}_{\mathrm{i}} '\Pi\mathrm{p}_{\mathrm{l}\:\:} \mathrm{true} \\ $$$$\mathrm{shows}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{p}_{\mathrm{1}} ….\mathrm{p}_{\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\underset{\mathrm{l}=\mathrm{1}} {\overset{\mathrm{n}\:+\mathrm{1}} {\sum}}\mathrm{p}_{\mathrm{l}} '\underset{\mathrm{i}=\mathrm{1}#\mathrm{l}} {\overset{\mathrm{n}+\mathrm{1}} {\prod}}\mathrm{p}_{\mathrm{i}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{p}_{\mathrm{n}+\mathrm{1}} \right).\left(\mathrm{p}_{\mathrm{1}} ….\mathrm{p}_{\mathrm{n}} \right)\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{p}_{\mathrm{n}+\mathrm{1}} '.\left(\mathrm{p}_{\mathrm{1}} …\mathrm{p}_{\mathrm{n}} \right) \\ $$$$+\mathrm{p}_{\mathrm{n}+\mathrm{1}} .\underset{\mathrm{k}=\mathrm{l}} {\overset{\mathrm{n}} {\sum}}\mathrm{p}_{\mathrm{k}} '.\underset{\mathrm{l}=\mathrm{1}#\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\mathrm{p}_{\mathrm{l}} \\ $$$$=\underset{\mathrm{k}=\mathrm{n}+\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\mathrm{p}_{\mathrm{k}} '.\underset{\mathrm{l}=\mathrm{1}#\mathrm{k}} {\overset{\mathrm{n}+\mathrm{1}} {\prod}}\mathrm{p}_{\mathrm{k}} +\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{p}'_{\mathrm{k}} \underset{\mathrm{l}=\mathrm{1}#\mathrm{k}} {\overset{\mathrm{n}+\mathrm{1}} {\prod}}\mathrm{p}_{\mathrm{l}} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\mathrm{p}'_{\mathrm{k}} \underset{\mathrm{l}=\mathrm{1}\neq\mathrm{k}} {\overset{\mathrm{n}+\mathrm{1}} {\prod}}\mathrm{p}_{\mathrm{l}} \\ $$$$\mathrm{prooved} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{exempl}\:\mathrm{p}_{\mathrm{i}} =\left(\mathrm{i}−\mathrm{x}\right)\Rightarrow\mathrm{p}'_{\mathrm{i}} =−\mathrm{1} \\ $$$$\mathrm{y}'=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(−\mathrm{1}\right)\underset{\mathrm{l}=\mathrm{1}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{l}−\mathrm{x}\right)=−\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}−\mathrm{x}}\underset{\mathrm{l}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{l}−\mathrm{x}\right) \\ $$$$=−\mathrm{y}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}−\mathrm{x}} \\ $$
Commented by York12 last updated on 02/Feb/24
GJ
$$\mathrm{GJ} \\ $$

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