## Calculus: Early Transcendentals (2nd Edition)

$f'(x) = 6x+1$
Using Product Rule: $f(x) = (x-1)(3x+4)$ $f'(x) = (1)(3x+4)+(x-1)(3) = 3x+4+3x-3=6x+1$ Expansion First and then Power Rule: $f(x) = (x-1)(3x+4) = 3x^2+x-4$ $f'(x)=3(2)x^{2-1}+(1)x^{1-1}-4(0)x^{0-1} = 6x+1$