find the dervative of Xsinx from 1st principle Share with your friends Share 23 Vipin Verma answered this y =xsinxLet δy be the increment in y , corresponding to increase in δx.y + δy = (x+δx) sin(x+δx)δy =(x+δx) sin(x+δx) - xsinxDividing throughout by δx, we getδyδx = (x+δx) sin(x+δx) - xsinxδxOr dydx=limδx→0δyδx= limδx→0(x+δx) sin(x+δx) - xsinxδx= limδx→0[x{sin(x+δx)-sinx}δx+sin(x+δx)]=limδx→0[x{sin(x+δx)-sinx}δx] +limδx→0sin((x+δx))=limδx→0[x{sin(x+δx)-sinx}δx] +sinx=limδx→0[2xcos(x+δx2)sinδx2δx] + sinx= {x×limδx→0cos(x+δx2) ×limδx→0sinδx2δx2}+sinx=xcosx+ sinx (Ans) 27 View Full Answer