Answer
$$\Delta y = 2x\Delta x + \Delta {x^2} - 2\Delta x\,\,\,\,{\text{ and}}\,\,\,{\text{ }}dy = \left( {2x - 2} \right)dx$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {x^2} - 2x + 1 \cr
& \cr
& {\text{Calculating }}\Delta y.\,\,\,f\left( x \right) = {x^2} - 2x + 1 \cr
& \Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) \cr
& {\text{Then evaluating }} \cr
& \Delta y = {\left( {x + \Delta x} \right)^2} - 2\left( {x + \Delta x} \right) + 1 - \left( {{x^2} - 2x + 1} \right) \cr
& \Delta y = {x^2} + 2x\Delta x + \Delta {x^2} - 2x - 2\Delta x + 1 - {x^2} + 2x - 1 \cr
& {\text{Simplifying}} \cr
& \Delta y = 2x\Delta x + \Delta {x^2} - 2\Delta x \cr
& \cr
& {\text{and}} \cr
& y = {x^2} - 2x + 1 \cr
& \frac{{dy}}{{dx}} = \left( {2x - 2} \right),{\text{ so }}dy = \left( {2x - 2} \right)dx \cr
& \cr
& \Delta y = 2x\Delta x + \Delta {x^2} - 2\Delta x\,\,\,\,{\text{ and}}\,\,\,{\text{ }}dy = \left( {2x - 2} \right)dx \cr} $$
