Answer
$$\,\,\,\,\,dy = - 4.8$$
Work Step by Step
$$\eqalign{
& y = 2{x^3} + {x^2} - 4x;\,\,\,\,\,\,x = 2{\text{ and }}\Delta x = - 0.2 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {2{x^3} + {x^2} - 4x} \right)' \cr
& f'\left( x \right) = 6{x^2} + 2x - 4 \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \left( {6{x^2} + 2x - 4} \right)dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \left( {6{x^2} + 2x - 4} \right)\Delta x \cr
& {\text{Substituting }}x = 2{\text{ and }}\Delta x = dx = - 0.2 \cr
& \,\,\,\,\,dy = \left( {6{{\left( 2 \right)}^2} + 2\left( 2 \right) - 4} \right)\left( { - 0.2} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = \left( {24} \right)\left( { - 0.2} \right) \cr
& \,\,\,\,\,dy = - 4.8 \cr} $$
