## Calculus: Early Transcendentals (2nd Edition)

$h'(z)=4z^3+9z^2-6z-1$
Product Rule: $h(z) = (z^3+4z^2+z)(z-1)$ $h'(z) = (3z^2+8z+1)(z-1) + (z^3+4z^2+z)(1) =4z^3+9z^2-6z-1$ Expansion Then Power Rule: $h(z) = (z^3+4z^2+z)(z-1) = (z^4+3z^3-3z^2-z)$ $h'(z) = (4)z^{4-1}+3(3)z^{3-1}-3(2)z^{2-1}-(1)z^{1-1} = 4z^3+9z^2-6z-1$