Differentiate xsinx+(sinx)cosx with respect to x.Answer: xsinx[xsinx+cosxlogx]+(sinx)cosx[cosxcotx−sinxlogsinx]SolutionLet y=u+v with u=xsinx, v=(sinx)cosx.For u: logu=sinxlogx, so dxdu=xsinx[xsinx+cosxlogx].For v: logv=cosxlogsinx, so dxdv=(sinx)cosx[cosxcotx−sinxlogsinx].Add:dxdy=xsinx[xsinx+cosxlogx]+(sinx)cosx[cosxcotx−sinxlogsinx]CBSE Class 12 Board · 2019 · Subjective