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