F(x)= ln 〖tg〗^22x ; f`(П/24)

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вопрос

To find the derivative of the function at x = π/24, we first find the derivative of the function F(x) = ln(tan^2(2x)).

Using the chain rule and properties of logarithms, we have:

F'(x) = 1/(tan^2(2x)) 2tan(2x) sec^2(2x)
F'(x) = 2tan(2x) * sec^2(2x) / tan^2(2x)
F'(x) = 2sec^2(2x)

Now we can find the value of the derivative at x = π/24:

F'(π/24) = 2sec^2(2π/24)
F'(π/24) = 2sec^2(π/12)
F'(π/24) = 2/(cos^2(π/12))
F'(π/24) = 2/(cos^2(15°))

Using trigonometric identities, we can simplify further:

cos(15°) = cos(45° - 30°) = cos(45°)cos(30°) + sin(45°)sin(30°)
cos(15°) = (sqrt(2)/2 sqrt(3)/2) + (sqrt(2)/2 1/2)
cos(15°) = (sqrt(6) + sqrt(2)) / 4

Therefore, F'(π/24) = 2 / ((sqrt(6) + sqrt(2))^2 / 16)
F'(π/24) = 32 / (6 + 2 + 2sqrt(12) + 2sqrt(2))
F'(π/24) = 32 / (8 + 2sqrt(3) + 2sqrt(2))
F'(π/24) = 32 / (8 + 2sqrt(3) + 2sqrt(2)) (8 - 2sqrt(3) - 2sqrt(2)) / (8 - 2sqrt(3) - 2sqrt(2))
F'(π/24) = (32(8 - 2sqrt(3) - 2sqrt(2))) / (64 - 12 - 16)
F'(π/24) = (256 - 64sqrt(3) - 64sqrt(2)) / 36
F'(π/24) = 128/18 - 8sqrt(3)/3 - 8sqrt(2)/3

Therefore, F'(π/24) = 64/9 - 8sqrt(3)/3 - 8sqrt(2)/3.