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Question Number 203910 by Davidtim last updated on 01/Feb/24

$$ify=(1-x)(2-x)(3-x)\cdot \cdot \cdot (n-x)$$$$y^{{}^{\prime }}=?$$

Answered by mr W last updated on 01/Feb/24

$$y=\underset{k=1}{\prod ^n}(k-x)$$$$\ln y=\underset{k=1}{\sum ^n}\ln (k-x)$$$$\frac{y^{\prime }}{y}=-\underset{k=1}{\sum ^n}\frac{1}{k-x}$$$$y^{\prime }=-\underset{k=1}{\prod ^n}(k-x)\underset{k=1}{\sum ^n}\frac{1}{k-x}$$

Answered by witcher3 last updated on 02/Feb/24

$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{i}=1}{\prod ^{\mathrm{n}}}{\mathrm{p}}_{\mathrm{i}}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{proof}\mathrm{n}=2$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{p}}_1.{\mathrm{p}}_2\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{p}}_1^{\prime }.{\mathrm{p}}_2+{\mathrm{p}}_2^{\prime }.{\mathrm{p}}_1=\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\left({\mathrm{p}}_{\mathrm{i}}\right)\underset{\mathrm{l}=1\ne \mathrm{i}}{\prod ^2}{\mathrm{p}}_{\mathrm{l}}$$$$\mathrm{suppose}\mathrm{\forall }\mathrm{n}\in \mathbb{N}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\mathrm{\Sigma }{\mathrm{p}}_{\mathrm{i}}^{\prime }\mathrm{\Pi }{\mathrm{p}}_{\mathrm{l}}\mathrm{true}$$$$\mathrm{shows}\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{p}}_1....{\mathrm{p}}_{\mathrm{n}+1}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{f}\left(\mathrm{x}\right)=\left({\mathrm{p}}_{\mathrm{n}+1}\right).({\mathrm{p}}_1....{\mathrm{p}}_{\mathrm{n}})\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{p}}_{\mathrm{n}+1}^{\prime }.({\mathrm{p}}_1...{\mathrm{p}}_{\mathrm{n}})$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}{\mathrm{p}}_{\mathrm{k}}^{\prime }\underset{\mathrm{l}=1\ne \mathrm{k}}{\prod ^{\mathrm{n}+1}}{\mathrm{p}}_{\mathrm{l}}$$$$\mathrm{prooved}$$$$\mathrm{in}\mathrm{our}\mathrm{exempl}{\mathrm{p}}_{\mathrm{i}}=(\mathrm{i}-\mathrm{x})\Rightarrow {\mathrm{p}}_{\mathrm{i}}^{\prime }=-1$$$${\mathrm{y}}^{\prime }=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(-1)\underset{\mathrm{l}=1\ne \mathrm{k}}{\prod ^{\mathrm{n}}}(\mathrm{l}-\mathrm{x})=-\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}-\mathrm{x}}\underset{\mathrm{l}=1}{\prod ^{\mathrm{n}}}(\mathrm{l}-\mathrm{x})$$$$=-\mathrm{y}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}-\mathrm{x}}$$

Commented by York12 last updated on 02/Feb/24

$$\mathrm{GJ}$$

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