Answer
$$\,\,\,\,\,dy = 1.9$$
Work Step by Step
$$\eqalign{
& y = 2{x^3} - 5x;\,\,\,\,\,\,x = - 2{\text{ and }}\Delta x = 0.1 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {2{x^3} - 5x} \right)' \cr
& f'\left( x \right) = 6{x^2} - 5 \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \left( {6{x^2} - 5} \right)dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \left( {6{x^2} - 5} \right)\Delta x \cr
& {\text{Substituting }}x = - 2{\text{ and }}\Delta x = dx = 0.1 \cr
& \,\,\,\,\,dy = \left( {6{{\left( { - 2} \right)}^2} - 5} \right)\left( {0.1} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = 1.9 \cr} $$
