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Calculus

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Need help to find dy/dx:

1) y = (sin x) ^-1/3

answer dy/dx = - 1/ 3(sinx) ^4/3

2) y = sqrt 8 + sin(6x)
answer - lost

  • Calculus - ,

    1) close, you left out the derivative of the base, so
    dy/dx = - 1/ 3(sinx) ^4/3(cosx)

    for 2)
    rewrite it as
    y = (8+sin(6x))^1/2
    dy/dx = (1/2)((8+sin(6x))^-1/2)(cos(6x))(6)
    = 3cos(6x)((8+sin(6x))^-1/2)

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