## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = \frac{{4{x^3} + 7{x^2} - 20x}}{{{{\left( {x + 2} \right)}^2}}}$$
\eqalign{ & y = \frac{{2{x^3} - 5{x^2}}}{{x + 2}} \cr & {\text{differentiate both sides}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{2{x^3} - 5{x^2}}}{{x + 2}}} \right] \cr & {\text{use quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {x + 2} \right)\frac{d}{{dx}}\left[ {2{x^3} - 5{x^2}} \right] - \left( {2{x^3} - 5{x^2}} \right)\left[ {x + 2} \right]}}{{{{\left( {x + 2} \right)}^2}}} \cr & {\text{find derivatives}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {x + 2} \right)\left( {6{x^2} - 10x} \right) - \left( {2{x^3} - 5{x^2}} \right)\left( 1 \right)}}{{{{\left( {x + 2} \right)}^2}}} \cr & {\text{simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{6{x^3} - 10{x^2} + 12{x^2} - 20x - 2{x^3} + 5{x^2}}}{{{{\left( {x + 2} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{6{x^3} - 2{x^3} + \left( { - 10{x^2} + 12{x^2} + 5{x^2}} \right) - 20x}}{{{{\left( {x + 2} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{4{x^3} + 7{x^2} - 20x}}{{{{\left( {x + 2} \right)}^2}}} \cr}