Differentiate `(x)/(sinx)` w.r.t . `sinx`.

Answer

N/a

Explanation

Let `u = (x)/(sinx)` and `v = sinx`
`:. (du)/(dx) = (sinx.(d)/(dx)x-x.(d)/(dx)sinx)/((sinx)^(2))`
`=(sinx-xcosx)/(sin^(2)x)"....."(i)`
and `(dv)/(dx) = d/(dx)sinx = cosx "...(ii)`
`:. du = (du//dx)/(dv//dx) = (sinx-xcosx//sin^(2)x)/(cosx)`
`=(sinx-xcosx)/(sin^(2)xcosx) = ((sinx-xcosx)/(cosx))/((sin^(2)xcosx)/(cosx))`
[dividing by cos x iin both numerator and denominator]
`= (tanx-x)/(sin^(2)x)`