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Историк62
@Историк62
July 2022
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найти производную функции y=2 sinx + cos2x - 3 если x0 = -3p/2 объясните подробно плиз
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sharadi
Verified answer
y=2 sinx + cos2x - 3
y' = 2Cosx - 2Sin2x
x
₀= -3π/2
y' = 2Cos(-3π/2) - 2Sin(2*(-3π/2)) = 0 + 0 = 0
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Verified answer
y=2 sinx + cos2x - 3y' = 2Cosx - 2Sin2x
x₀= -3π/2
y' = 2Cos(-3π/2) - 2Sin(2*(-3π/2)) = 0 + 0 = 0