If y = x^{n}logcosx + \frac{xlogx}{sinx^{2}} then dy/dx

Correct Answer is : nx^{n−1}logcosx−x^{n}tanx + \;{sinx^{2}\;(1\;+\;logx)\;−xlogxcos^{2}2x}/{\;(sinx^{2}) ^{2}}

{sinx^{2} (1 + logx) −xlogxcos^{2}2x}/{ (sinx^{2}) ^{2}}

nx^{n−1}logcosx−x^{n}tanx + {sinx^{2} (1 + logx) −xlogxcos^{2}2x}/{ (sinx^{2}) ^{2}}