Derivative

What is the derivative of this

y= 1/sin^2x - 2/sinx

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asked by Ken
  1. let z = sin x
    dz/dx = cos x

    then we have
    y = 1/z^2 -2/z

    dy/dz = -2z/z^4 + 2/z^2
    = -2/z^3 +2/z^2

    dy/dx = dy/dz*dz/dx
    = (-2/z^3 +2/z^2)cos x
    = -2cosx/sin^3x + 2 cos x/sin^2x
    =-2 cos x/sin^3x + 2 cos x sin x/sin^3x
    = (2 cos x /sin^3x)(sin x - 1)

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    posted by Damon

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