Answer
$$\Delta y \approx - 0.01$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{x^2} + 8} ;{\text{ from }}x = 1{\text{ to }}x = 0.97 \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}dx \cr
& f\left( x \right) = \sqrt {{x^2} + 8} \cr
& f'\left( x \right) = \frac{{2x}}{{2\sqrt {{x^2} + 8} }} \cr
& f'\left( x \right) = \frac{x}{{\sqrt {{x^2} + 8} }} \cr
& dx = \Delta x \cr
& dx = 0.97 - 1 \cr
& dx = - 0.03 \cr
& \cr
& \Delta y \approx f'\left( x \right)dx = dy \cr
& \Delta y \approx \frac{x}{{\sqrt {{x^2} + 8} }}dx \cr
& {\text{Substitute }}dx = - 0.03{\text{ and }}x = 1 \cr
& \Delta y \approx \frac{1}{{\sqrt {{{\left( 1 \right)}^2} + 8} }}\left( { - 0.03} \right) \cr
& \Delta y \approx - 0.01 \cr} $$
