Answer
$$\Delta y \approx 0.0048$$
Work Step by Step
$$\eqalign{
& y = \frac{x}{{{x^2} + 1}};{\text{ from }}x = 2{\text{ to }}x = 1.96 \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}dx \cr
& f\left( x \right) = \frac{x}{{{x^2} + 1}} \cr
& f'\left( x \right) = \frac{{{x^2} + 1 - x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{{x^2} + 1 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& dx = \Delta x \cr
& dx = 1.96 - 2 \cr
& dx = - 0.04 \cr
& \cr
& \Delta y \approx f'\left( x \right)dx = dy \cr
& \Delta y \approx \frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx \cr
& {\text{Substitute }}dx = - 0.04{\text{ and }}x = 2 \cr
& \Delta y \approx \frac{{1 - {{\left( 2 \right)}^2}}}{{{{\left( {{{\left( 2 \right)}^2} + 1} \right)}^2}}}\left( { - 0.04} \right) \cr
& \Delta y \approx 0.0048 \cr} $$
