Statement Let f(x)=(3x2−1)(x3+2x)f(x)=(3x^2-1)(x^3+2x)f(x)=(3x2−1)(x3+2x). (a) Find f′(x)f'(x)f′(x) using the product rule. (b) Compute f′(1)f'(1)f′(1). Solution (a) Product rule (uv)′=u′v+uv′(uv)'=u'v+uv'(uv)′=u′v+uv′ with u=3x2−1u=3x^2-1u=3x2−1, v=x3+2xv=x^3+2xv=x3+2x: f′(x)=6x(x3+2x)+(3x2−1)(3x2+2)=6x4+12x2+9x4+6x2−3x2−2=15x4+15x2−2.f'(x)=6x(x^3+2x)+(3x^2-1)(3x^2+2)=6x^4+12x^2+9x^4+6x^2-3x^2-2=15x^4+15x^2-2.f′(x)=6x(x3+2x)+(3x2−1)(3x2+2)=6x4+12x2+9x4+6x2−3x2−2=15x4+15x2−2. (b) f′(1)=15+15−2=28f'(1)=15+15-2=28f′(1)=15+15−2=28. f′(x)=15x4+15x2−2;f′(1)=28\boxed{f'(x)=15x^4+15x^2-2;\qquad f'(1)=28}f′(x)=15x4+15x2−2;f′(1)=28