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denis12541
@denis12541
July 2022
1
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Помогите решить пожалуйста.
Найти тангенс угла наклона касательной к графику функции y=1+2x^2/x в точке x0=2
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moboqe
Y=(1+2x^2)/x
y'=(4x^2-2x^2-1)/x^2
y'=2-(1/x^2)
y'(x0)=2-(1/4)=
7/4
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Answers & Comments
y'=(4x^2-2x^2-1)/x^2
y'=2-(1/x^2)
y'(x0)=2-(1/4)=7/4