Answer
$$\frac{7}{{16}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{2x - 1}}{{x + 3}} \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& = \frac{{\left( {x + 3} \right)\left( {2x - 1} \right)' - \left( {2x - 1} \right)\left( {x + 3} \right)'}}{{{{\left( {x + 3} \right)}^2}}} \cr
& = \frac{{\left( {x + 3} \right)\left( 2 \right) - \left( {2x - 1} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}} \cr
& = \frac{{2x + 6 - 2x + 1}}{{{{\left( {x + 3} \right)}^2}}} \cr
& = \frac{7}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{evaluate }}{\left. {dy/dx} \right|_{x = 1}} \cr
& = \frac{7}{{{{\left( {1 + 3} \right)}^2}}} \cr
& {\text{simplify}} \cr
& = \frac{7}{{16}} \cr} $$
