Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& y = \left( {2{x^7} - {x^2}} \right)\left( {\frac{{x - 1}}{{x + 1}}} \right) \cr
& {\text{multiply}} \cr
& y = \frac{{\left( {2{x^7} - {x^2}} \right)\left( {x - 1} \right)}}{{x + 1}} \cr
& y = \frac{{2{x^7}\left( {x - 1} \right) - {x^2}\left( {x - 1} \right)}}{{x + 1}} \cr
& y = \frac{{2{x^8} - 2{x^7} - {x^3} + {x^2}}}{{x + 1}} \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& = \frac{{\left( {x + 1} \right)\left( {2{x^8} - 2{x^7} - {x^3} + {x^2}} \right)' - \left( {2{x^8} - 2{x^7} - {x^3} + {x^2}} \right)\left( {x + 1} \right)'}}{{{{\left( {x + 1} \right)}^2}}} \cr
& = \frac{{\left( {x + 1} \right)\left( {16{x^7} - 14{x^6} - 3{x^2} + 2x} \right) - \left( {2{x^8} - 2{x^7} - {x^3} + {x^2}} \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr
& {\text{evaluate }}{\left. {dy/dx} \right|_{x = 1}} \cr
& = \frac{{\left( {1 + 1} \right)\left( {16{{\left( 1 \right)}^7} - 14{{\left( 1 \right)}^6} - 3{{\left( 1 \right)}^2} + 2\left( 1 \right)} \right) - \left( {2{{\left( 1 \right)}^8} - 2{{\left( 1 \right)}^7} - {{\left( 1 \right)}^3} + {{\left( 1 \right)}^2}} \right)}}{{{{\left( {1 + 1} \right)}^2}}} \cr
& {\text{simplify}} \cr
& = \frac{{\left( 2 \right)\left( {16 - 14 - 3 + 2} \right) - \left( {2 - 2 - 1 + 1} \right)}}{{{{\left( 2 \right)}^2}}} \cr
& = \frac{{\left( 2 \right)\left( 1 \right) - \left( 0 \right)}}{{{{\left( 2 \right)}^2}}} \cr
& = \frac{1}{2} \cr} $$
