Answer
$\frac{{dy}}{{dx}} = - 40{x^3} - 21{x^2} + 44x + 14$
Work Step by Step
$$\eqalign{
& y = \left( {10{x^2} + 7x - 2} \right)\left( {2 - {x^2}} \right) \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {10{x^2} + 7x - 2} \right)\left( {2 - {x^2}} \right)} \right] \cr
& {\text{Differentiate by using the formula }}\left( {fg} \right)' = fg' + gf' \cr
& {\text{Let }}f = \left( {10{x^2} + 7x - 2} \right){\text{ and }}g = \left( {2 - {x^2}} \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = \left( {10{x^2} + 7x - 2} \right)\left( {2 - {x^2}} \right)' + \left( {2 - {x^2}} \right)\left( {10{x^2} + 7x - 2} \right)' \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = \left( {10{x^2} + 7x - 2} \right)\left( { - 2x} \right) + \left( {2 - {x^2}} \right)\left( {20x + 7} \right) \cr
& {\text{Use the distributive property}} \cr
& \frac{{dy}}{{dx}} = - 20{x^3} - 14{x^2} + 4x + 40x + 14 - 20{x^3} - 7{x^2} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = - 40{x^3} - 21{x^2} + 44x + 14 \cr} $$
