If f(x)=(x2+1)(x−3)f(x) = (x^2 + 1)(x - 3)f(x)=(x2+1)(x−3), then f′(x)=f'(x) =f′(x)=Ax2+1+(x−3)(2x)x^2 + 1 + (x - 3)(2x)x2+1+(x−3)(2x)B2x+12x + 12x+1C2x(x−3)2x(x-3)2x(x−3)D3x2−6x+13x^2 - 6x + 13x2−6x+1check_circleExplanationUse product rule: 2x(x−3)+(x2+1)(1)=2x2−6x+x2+1=3x2−6x+12x(x-3) + (x^2+1)(1) = 2x^2 - 6x + x^2 + 1 = 3x^2 - 6x + 12x(x−3)+(x2+1)(1)=2x2−6x+x2+1=3x2−6x+1.