#### Answer

$$\,\,\,\,\,dy = - 0.023$$

#### Work Step by Step

$$\eqalign{
& y = \frac{{2x - 5}}{{x + 1}};\,\,\,\,\,\,x = 2{\text{ and }}\Delta x = - 0.03 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {\frac{{2x - 5}}{{x + 1}}} \right)' \cr
& {\text{by using the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( {2x - 5} \right)' - \left( {2x - 5} \right)\left( {x + 1} \right)'}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( 2 \right) - \left( {2x - 5} \right)\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2x + 2 - 2x + 5}}{{{{\left( {x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{7}{{{{\left( {x + 1} \right)}^2}}} \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \frac{7}{{{{\left( {x + 1} \right)}^2}}}dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \frac{7}{{{{\left( {x + 1} \right)}^2}}}\Delta x \cr
& {\text{Substituting }}x = 2{\text{ and }}\Delta x = dx = - 0.03 \cr
& \,\,\,\,\,dy = \frac{7}{{{{\left( {2 + 1} \right)}^2}}}\left( { - 0.03} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = - 0.023 \cr} $$