## Calculus: Early Transcendentals (2nd Edition)

$g'(x)=30x^4 - 1$
$g(x) = 6x^5 - x$ Lets use THEOREM 3.5 $g'(x) = (6x^5 -x)' = (6x^5)' + ( -x)'$ $(6x^5)' = 6 \times (x^5)'$ THEOREM 3.4 $(x^4)' = 5x^4$ THEOREM 3.3 therefore $(6x^5)' = 6 \times (x^5)' = 30x^4$ $(-x)' = -1 \times (x)'$ THEOREM 3.4 $(x)' = 1x^0 = 1$ THEOREM 3.3 therefore $(-x)' = -1 \times (x)' = -1$ Therefore $g'(x) = (6x^5 -x)' = (6x^5)' + ( -x)' = 30x^4 - 1$