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77rtemnikovdo
@77rtemnikovdo
August 2022
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Народ, помогите, пожалуйста с алгеброй срочно
Найдите производную функции
y=2sinx-4x
в точке х0 = п/3
если можно с объяснениями.
Кто решит верно, тому респеки
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nKrynka
Решение
y=2sinx-4x
y` = 2cosx - 4
y`(π/3) = 2cos(π/3) - 4 = 2*(1/2) - 4 = 1 - 4 = - 3
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Answers & Comments
y=2sinx-4x
y` = 2cosx - 4
y`(π/3) = 2cos(π/3) - 4 = 2*(1/2) - 4 = 1 - 4 = - 3