ey(x+1)=1ey × 1 + [ey × dy/dx × (x+1)] = 0Differentiating on both sides ⇒ dy/dx = -ey/ey × (x+1) = -1/(x+1)Differentiating again ⇒ d²y/dx² = -[-1/(x+1)²] = 1/(x+1)²⇒ d²y/dx² = 1/(x+1)² = (-1/(x+1))² = (dy/dx)²Hence proved