Найти производную функции (3x^2+2x-1 )/(2x+1) в точке х0 = 2. Created by ayyyyylmao. algebra-ru

(2x+1) в точке х0 = 2...

ayyyyylmao @ayyyyylmao

August 2022 1 12 Report

Найти производную функции (3x^2+2x-1 )/(2x+1) в точке х0 = 2

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Universalka

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$(\frac{3x^{2}+2x-1 }{2x+1})'=\frac{(3x^{2}+2x-1)'*(2x+1)-(3x^{2}+2x-1)*(2x+1)'}{(2x+1)^{2}}=\frac{(6x+2)*(2x+1)-(3x^{2}+2x-1)*2}{(2x+1)^{2}}=\\\\=\frac{12x^{2}+6x+4x+2-6x^{2}-4x+2}{(2x+1)^{2}}=\frac{6x^{2}+6x+4}{(2x+1)^{2}}$

$\frac{6*2^{2}+6*2+4 }{(2*2+1)^{2}}=\frac{24+12+4}{5^{2}}=\frac{40}{25}=1,6\\\\Otvet:\boxed{1,6}$

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ayyyyylmao спасибо большое, а можете еще на этот ответить? https://znanija.com/task/38477289

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