differentiate y= (x) cosx + (cosx)sinx Share with your friends Share 1 Deepali Gupta answered this y = xcosx + cosxsinxlet u= xcosx and let v = cosxsinxy = u+vdydx = dudx +dvdxu= xcosxtaking logarithm on both sideslog u = cosx log (x)differentiating w.r.t x1ududx = cosx ddxlog (x) + log (x)ddxcosx1ududx = cosx 1x + log (x)-sinxdudx = xcosx cosx 1x + log (x)-sinxdudx = cosx . xcosx-1 - xcosx. log (x) . sinxlet v = cosxsinxtaking logarithm on both sdeslog v = sinx log (cosx)differentiating w.r.t x1vdvdx = sinx ddxlog (cosx) + log (cosx)ddxsinx1vdvdx = sinx 1cosx-sinx + log (cosx)cosxdvdx = cosxsinx sinx 1cosx-sinx + log (cosx)cosxdvdx = -sin2x . cosxsinx-1 + cosxsinx+1 . log (cosx)From original equationdydx = dudx +dvdxdydx = cosx . xcosx-1 - xcosx. log (x) . sinx - sin2x . cosxsinx-1 + cosxsinx+1 . log (cosx) 4 View Full Answer
