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dara1999uers
@dara1999uers
July 2022
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(tgx(cosx+2))'
((cos2x+sin2x)^3)'
(5x-2/sinx)'
cos(3x^3+4x+2))'
(√5-√cosx)'.
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Answers & Comments
nKrynka
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)
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Answers & Comments
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) * (9x² + 4)
(√5-√cosx)' = - [1 / (2√cosx)] * (- sinx) = sinx / (2√cosx)