What is the derivative of this
y= 1/sin^2x - 2/sinx
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)