Answer
$\frac{{dy}}{{dx}} = 24{x^2} + 40x + 6$
Work Step by Step
$$\eqalign{
& y = \left( {4{x^2} + 3} \right)\left( {2x + 5} \right) \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {4{x^2} + 3} \right)\left( {2x + 5} \right)} \right] \cr
& {\text{Differentiate by using the formula }}\left( {fg} \right)' = fg' + gf' \cr
& {\text{Let }}f = \left( {4{x^2} + 3} \right){\text{ and }}g = \left( {2x + 5} \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = \left( {4{x^2} + 3} \right)\frac{d}{{dx}}\left[ {\left( {2x + 5} \right)} \right] + \left( {2x + 5} \right)\frac{d}{{dx}}\left[ {\left( {4{x^2} + 3} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \left( {4{x^2} + 3} \right)\left( 2 \right) + \left( {2x + 5} \right)\left( {8x} \right) \cr
& {\text{Use the distributive property}} \cr
& \frac{{dy}}{{dx}} = 8{x^2} + 6 + 16{x^2} + 40x \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = 24{x^2} + 40x + 6 \cr} $$
