Answer
$$ y' = 24{x^2} - 16{x^3} + 6 - 6x,\,\,\,\,y'' = 48x - 48{x^2} - 6,\,\,\,y''' = 48 - 96x,\,\,\,\,\,\,{y^{\left( 4 \right)}} = - 96,\,\,\,\,\,{y^{\left( 5 \right)}} = 0\,$$
Work Step by Step
$$\eqalign{
& y = \left( {4{x^2} + 3} \right)\left( {2 - x} \right)x \cr
& \cr
& {\text{Multiply by using the distributive property}} \cr
& y = \left( {4{x^2} + 3} \right)\left( {2x - {x^2}} \right) \cr
& y = 4{x^2}\left( {2x - {x^2}} \right) + 3\left( {2x - {x^2}} \right) \cr
& y = 8{x^3} - 4{x^4} + 6x - 3{x^2} \cr
& \cr
& {\text{Find the derivatives of all orders of the function using the rule }}\cr
&\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& y' = \frac{d}{{dx}}\left[ {8{x^3} - 4{x^4} + 6x - 3{x^2}} \right] \cr
& y' = 8\left( {3{x^2}} \right) - 4\left( {4{x^3}} \right) + 6\left( 1 \right) - 3\left( {2x} \right) \cr
& y' = 24{x^2} - 16{x^3} + 6 - 6x \cr
& \cr
& y'' = \frac{d}{{dx}}\left[ {24{x^2} - 16{x^3} + 6 - 6x} \right] \cr
& y'' = 24\left( {2x} \right) - 16\left( {3{x^2}} \right) + 0 - 6\left( 1 \right) \cr
& y'' = 48x - 48{x^2} - 6 \cr
& \cr
& y''' = \frac{d}{{dx}}\left[ {48x - 48{x^2} - 6} \right] \cr
& y''' = 48\left( 1 \right) - 48\left( {2x} \right) - 0 \cr
& y''' = 48 - 96x \cr
& \cr
& {y^{\left( 4 \right)}} = \frac{d}{{dx}}\left[ {48 - 96x} \right] \cr
& {y^{\left( 4 \right)}} = - 96 \cr
& \cr
& {y^{\left( 5 \right)}} = \frac{d}{{dx}}\left[ { - 96} \right] \cr
& {y^{\left( 5 \right)}} = 0 \cr
& \cr
& \cr
& {\text{The derivatives are}} \cr
& y' = 24{x^2} - 16{x^3} + 6 - 6x,\,\,\,\,y'' = 48x - 48{x^2} - 6,\,\,\,y''' = 48 - 96x,\,\,\,\,\,\,{y^{\left( 4 \right)}} = - 96,\,\,\,\,\,{y^{\left( 5 \right)}} = 0\, \cr} $$
