## Calculus with Applications (10th Edition)

$$y = - \frac{5}{9}x + \frac{{16}}{9}$$
\eqalign{ & y = \frac{x}{{{x^2} - 1}},\,\,\,\,\,x = 2 \cr & {\text{find the derivative of }}y \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{x}{{{x^2} - 1}}} \right] \cr & {\text{by using the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} - 1} \right)\left( 1 \right) - x\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{{x^2} - 1 - 2{x^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{ - 1 - {x^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr & {\text{Find the slope of the tangent line at }}x = 2 \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} \cr & m = \frac{{ - 1 - {{\left( 2 \right)}^2}}}{{{{\left( {{2^2} - 1} \right)}^2}}} \cr & m = \frac{{ - 5}}{9} \cr & \cr & {\text{Evaluate the function at }}x = 2 \cr & y\left( 2 \right) = \frac{2}{{{2^2} - 1}} \cr & y\left( 2 \right) = \frac{2}{3} \cr & {\text{we know the point }}\left( {2,\frac{2}{3}} \right){\text{ and the slope }}m = - \frac{5}{9} \cr & {\text{find the equation of the tangent line using the point - slope form of a line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{2}{3} = - \frac{5}{9}\left( {x - 2} \right) \cr & {\text{simplifying}} \cr & y - \frac{2}{3} = - \frac{5}{9}x + \frac{{10}}{9} \cr & y = - \frac{5}{9}x + \frac{{10}}{9} + \frac{2}{3} \cr & y = - \frac{5}{9}x + \frac{{16}}{9} \cr}