Answer
$$-\frac{{13}}{8}$$
Work Step by Step
$$\eqalign{
& y = \frac{{4x + 1}}{{{x^2} - 5}} \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& = \frac{{\left( {{x^2} - 5} \right)\left( {4x + 1} \right)' - \left( {4x + 1} \right)\left( {{x^2} - 5} \right)'}}{{{{\left( {{x^2} - 5} \right)}^2}}} \cr
& = \frac{{\left( {{x^2} - 5} \right)\left( 4 \right) - \left( {4x + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 5} \right)}^2}}} \cr
& = \frac{{4{x^2} - 20 - 8{x^2} - 2x}}{{{{\left( {{x^2} - 5} \right)}^2}}} \cr
& = \frac{{ - 4{x^2} - 2x - 20}}{{{{\left( {{x^2} - 5} \right)}^2}}} \cr
& = - \frac{{4{x^2} + 2x + 20}}{{{{\left( {{x^2} - 5} \right)}^2}}} \cr
& {\text{evaluate }}{\left. {dy/dx} \right|_{x = 1}} \cr
& = - \frac{{4{{\left( 1 \right)}^2} + 2\left( 1 \right) + 20}}{{{{\left( {{{\left( 1 \right)}^2} - 5} \right)}^2}}} \cr
& {\text{simplify}} \cr
& = - \frac{{4 + 2 + 20}}{{{{\left( { - 4} \right)}^2}}} \cr
& = - \frac{{26}}{{16}} \cr
& = -\frac{{13}}{8} \cr} $$
