## Calculus: Early Transcendentals (2nd Edition)

$f'(x)=3x^{2}-6x+4$
Original Function $f(x)=(x-3)(x^{2}+4)$ Apply the product rule: $f(x)g'(x)+g(x)f'(x)$ $f'(x)=(x-3)(2x)+(x^{2}+4)(1)$ Distributed $f'(x)=2x^{2}-6x+x^{2}+4$ Simply and you get the answer $f'(x)=3x^{2}-6x+4$