How do you differentiate g(x)=(1+4x)^5(3+x-x^2)^8?

$g ' \left(x\right) = 8 {\left(4 x + 1\right)}^{5} {\left(- {x}^{2} + x + 3\right)}^{7} \left(- 2 x + 1\right) + 20 {\left(4 x + 1\right)}^{4} {\left(- {x}^{2} + x + 3\right)}^{8}$
$g \left(x\right) = {\left(4 x + 1\right)}^{5} {\left(- {x}^{2} + x + 3\right)}^{8}$
$g ' \left(x\right) = {\left(4 x + 1\right)}^{5} \left(8\right) {\left(- {x}^{2} + x + 3\right)}^{7} \left(- 2 x + 1\right) + 5 {\left(4 x + 1\right)}^{4} \left(4\right) {\left(- {x}^{2} + x + 3\right)}^{8}$
$g ' \left(x\right) = 8 {\left(4 x + 1\right)}^{5} {\left(- {x}^{2} + x + 3\right)}^{7} \left(- 2 x + 1\right) + 20 {\left(4 x + 1\right)}^{4} {\left(- {x}^{2} + x + 3\right)}^{8}$