Найти производную

(tgx (cosx+2)) '

((cos2x+sin2x) ^3) '

(5x-2/sinx) '

cos (3x^3+4x+2)) '

(√5-√cosx) '.

Решение

1) (tgx (cosx+2)) ' = 1/cos²[x * (cosx + 2) ] * [ (cosx + 2) - x*sinx] =

= [ (cosx + 2) - x*sinx] / cos²[x * (cosx + 2) ]

2) ((cos2x+sin2x) ^3) ' = 3 * (cos2x + sin2x) ² * (- 2sin2x) * (2cos2x) =

= - 6 * sin4x * (cos2x + sin2x) ²

3) (5x-2/sinx) ' = 5 + 2sinx cosx = 5 + sin2x

cos (3x^3+4x+2)) ' = - sin (3x^3+4x+2) * (9x ² + 4)

(√5-√cosx) ' = - [1 / (2√cosx) ] * (- sinx) = sinx / (2√cosx)