Question

вычислите f'(п/6),если f(x)=2sinx+cosx-ctgx

Answer 1

f(x)=2sinx + cosx - ctgx

f'(x)=2cosx - sinx - 1/sin²x

f'(π/6)=2·cos(π/6) - sin(π/6) - 1/sin²(π/6) =

= 2·0.5√3 - 0.5 - 1/(1/4) = √3 - 0.5 - 4 = √3 - 4.5

Примерно: √3 - 4.5 ≈ 1,73 - 4,5 ≈ - 2,77