Найдите значение производной. sin^2x x0=pi\12 -x cos2x x0=0 1\2sin2x x0=pi\8

Goshavasa Goshavasa    2   03.09.2019 20:20    2

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Ulyana150806 Ulyana150806  03.08.2020 15:25
Y = sin²x.
u = sinx, v = u²
y' = u'·v' = (sinx)'·2u = 2sinx·cosx = sin2x
y'(x₀) = sin(2·π/12) = sin(π/6) = 1/2

y = -x·cos2x
y' = -(x)'cos2x + -x·(cos2x)' = -cos2x + 2x·sin2x
y'(x₀) = -cos0 + 0 = -1 

y = 1/2·sin2x
y' = 1/2·2cos2x = cos2x
y'(x₀) = cos(2·π/8) = cos(π/4) = √2/2 
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