Answer
$g'(x)=30x^4 - 1$
Work Step by Step
$g(x) = 6x^5 - x$
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
$g'(x) = (6x^5 -x)' = (6x^5)' + ( -x)' = 30x^4 - 1$
