#### Answer

$$\,\,\,\,\,dy = 0.1$$

#### Work Step by Step

$$\eqalign{
& y = {x^3} - 2{x^2} + 3;\,\,\,\,\,\,x = 1{\text{ and }}\Delta x = - 0.1 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {{x^3} - 2{x^2} + 3} \right)' \cr
& f'\left( x \right) = 3{x^2} - 4x \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \left( {3{x^2} - 4x} \right)dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \left( {3{x^2} - 4x} \right)\Delta x \cr
& {\text{Substituting }}x = 1{\text{ and }}\Delta x = dx = - 0.1 \cr
& \,\,\,\,\,dy = \left( {3{{\left( 1 \right)}^2} - 4\left( 1 \right)} \right)\left( { - 0.1} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = 0.1 \cr} $$