Answer
$$y = - 4x + 8$$
Work Step by Step
$$\eqalign{
& {\text{let }}y = \ln \left( {5 - {x^2}} \right) \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {5 - {x^2}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2x}}{{5 - {x^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2} - 5}} \cr
& {\text{Find the slope evaluating }}\frac{{dy}}{{dx}}{\text{ at }}x = 2 \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} = \frac{{2\left( 2 \right)}}{{{{\left( 2 \right)}^2} - 5}} \cr
& m = - 4 \cr
& {\text{Evaluate }}y{\text{ at }}x = 2 \cr
& y\left( 2 \right) = \ln \left( {5 - {{\left( 2 \right)}^2}} \right) \cr
& y\left( 2 \right) = \ln \left( 1 \right) \cr
& y\left( 2 \right) = 0 \cr
& \cr
& {\text{Find the equation of the tangent line to the graph using the }} \cr
& {\text{point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr
& {\text{With }}{x_1} = 2,\,\,{y_1} = 0 \cr
& y - 0 = - 4\left( {x - 2} \right) \cr
& {\text{Simplify}} \cr
& y = - 4x + 8 \cr} $$
