Take log: lny=xsinx⋅lnx. Differentiating both sides: (1/y)(dy/dx)=(sinx)(lnx)+x(cosx)(lnx)+xsinx⋅(1/x)=sinx(1+lnx)+xcosxlnx. Hence dy/dx=xxsinx[sinx(1+logx)+xlogxcosx].
Note: Option 3 prints sin(1+logx), which is best read as sinx⋅(1+logx) given the standard derivation; we have selected option 3 as the intended answer.