Proof of ${p}^{\prime}(x{)}^{2}\ge p(x){p}^{\u2033}(x)\text{for all x}\in \mathbb{R}$

traucaderx7
2022-08-12
Answered

Proof of ${p}^{\prime}(x{)}^{2}\ge p(x){p}^{\u2033}(x)\text{for all x}\in \mathbb{R}$

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Cynthia Lester

Answered 2022-08-13
Author has **22** answers

$\begin{array}{rlr}{\left(\sum _{k=1}^{n}\frac{1}{x-{a}_{k}}\right)}^{\prime}& =\sum _{k=1}^{n}{\left(\frac{1}{x-{a}_{k}}\right)}^{\prime}& (f+g{)}^{\prime}={f}^{\prime}+{g}^{\prime}\\ & =\sum _{k=1}^{n}{[(x-{a}_{k}{)}^{-1}]}^{\prime}& \frac{1}{\alpha}={\alpha}^{-1}\\ & =\sum _{k=1}^{n}(-1\cdot (x-{a}_{k}{)}^{-2}\cdot {x}^{\prime})& \text{Chain and power rules}\\ & =-\sum _{k=1}^{n}\left(\frac{1}{(x-{a}_{k}{)}^{2}}\right)& \frac{1}{\alpha}={\alpha}^{-1}\text{and distributivy}\end{array}$

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