Найти f' (x) и f' (x0)

f (x) = xctgx, x0 = π/4

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Алгебра

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1 мая, 10:18

Найти f' (x) и f' (x0)

f (x) = xctgx, x0 = π/4

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Флор · Answer 1

((uv) '=u'v+uv')

f (x) = xctgx, u=x, v=ctgx

f' (x) = (xctgx) ' = (x') ctgx + x (ctgx) '=ctgx - (x / (sin²x))

f' (π/4) = ctg (π/4) - (π/4) / sin² (π/4) = 1 - (π/4) / (√2/2) ²=1 - (π/4) / (2/4) =

=1 - (π/4) / (1/2) = 1 - (2π/4) = 1 - (π/2) = (2-π) / 2

Найти f' (x) и f' (x0)f (x) = xctgx, x0 = π/4

Найти f' (x) и f' (x0)

f (x) = xctgx, x0 = π/4