Answer
$$dy = 0.03{\text{ and }}\Delta y = 0.03005$$
Work Step by Step
$$\eqalign{
& {\text{Let the function }}f\left( x \right) = 0.5{x^2},{\text{ }} \cr
& x{\text{ - value: }}x = 3,{\text{ }} \cr
& {\text{Differential of }}x:{\text{ }}\Delta x = dx = 0.01 \cr
& \cr
& {\text{Differentiating:}} \cr
& f\left( x \right) = 0.5{x^2} \cr
& f'\left( x \right) = x \cr
& {\text{The differential is }} \cr
& dy = f'\left( x \right)dx \cr
& dy = xdx \cr
& {\text{Substituting }}x = 3{\text{ and }}dx = 0.01 \cr
& dy = \left( 3 \right)\left( {0.01} \right) \cr
& dy = 0.03 \cr
& \cr
& {\text{Now}},{\text{ using }}\Delta x = 0.01,{\text{ the change in }}y{\text{ is}} \cr
& \Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) \cr
& \Delta y = f\left( {3 + 0.01} \right) - f\left( 3 \right) \cr
& \Delta y = f\left( {3.01} \right) - f\left( 3 \right) \cr
& \Delta y = 0.5{\left( {3.01} \right)^2} - 0.5{\left( 3 \right)^2} \cr
& \Delta y = 0.03005 \cr
& \cr
& dy = 0.03{\text{ and }}\Delta y = 0.03005 \cr} $$
