Answer
$$\,\,\,\,\,dy = - 0.0097$$
Work Step by Step
$$\eqalign{
& y = \frac{{6x - 3}}{{2x + 1}};\,\,\,\,\,\,x = 3{\text{ and }}\Delta x = - 0.04 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {\frac{{6x - 3}}{{2x + 1}}} \right)' \cr
& {\text{by using the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( {6x - 3} \right)' - \left( {6x - 3} \right)\left( {2x + 1} \right)'}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( 6 \right) - \left( {6x - 3} \right)\left( 2 \right)}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{12x + 6 - 12x + 6}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{12}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \frac{{12}}{{{{\left( {2x + 1} \right)}^2}}}dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \frac{{12}}{{{{\left( {2x + 1} \right)}^2}}}\Delta x \cr
& {\text{Substituting }}x = 3{\text{ and }}\Delta x = dx = - 0.04 \cr
& \,\,\,\,\,dy = \frac{{12}}{{{{\left( {2\left( 3 \right) + 1} \right)}^2}}}\left( { - 0.04} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = - 0.0097 \cr} $$
