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Calculus

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Find the d^3z/dydx^2 partial derivative of:

ln[sin(x-y)]

The answer is supposed to be 1/2; however, I cannot figure out how to get this answer. Any help is appreciated.

  • Calculus - ,

    I got the dz/dy derivative to be:

    cos(x-y)/sin(x-y)

    I got the d^2z/dydx derivative to be:

    -(sin(x-y))^2 - (cos(x-y))^2/ (sin(x-y))^4

  • Calculus - ,

    I think there's a minus sign missing there.

    ∂z/∂y = -cos(x-y)/sin(x-y) = -cot(x-y)

    ∂^2z/∂y∂x = csc^2(x-y)
    ∂^3z/∂y∂x^2 = 2csc(x-y) (-csc(x-y)cot(x-y))
    = -2csc^2(x-y)cot(x-y)

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