## Calculus with Applications (10th Edition)

$$y = - \frac{{37}}{{11}}x + \frac{{59}}{{11}}$$
\eqalign{ & \ln \left( {x + y} \right) = {x^3}{y^2} + \ln \left( {{x^2} + 2} \right) - 4;\,\,\,\,\,\,\,\left( {1,2} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {\ln \left( {x + y} \right)} \right) = \frac{d}{{dx}}\left( {{x^3}{y^2}} \right) + \frac{d}{{dx}}\left( {\ln \left( {{x^2} + 2} \right)} \right) \cr & {\text{use product rule as follows}} \cr & \frac{d}{{dx}}\left( {\ln \left( {x + y} \right)} \right) = {x^3}\frac{d}{{dx}}\left( {{y^2}} \right) + {y^2}\frac{d}{{dx}}\left( {{x^3}} \right) + \frac{d}{{dx}}\left( {\ln \left( {{x^2} + 2} \right)} \right) \cr & {\text{solve the derivatives}} \cr & \frac{1}{{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right) = 2y{x^3}\frac{{dy}}{{dx}} + 3{x^2}{y^2} + \frac{{2x}}{{{x^2} + 2}} \cr & \frac{1}{{x + y}} + \frac{1}{{x + y}}\frac{{dy}}{{dx}} = 2y{x^3}\frac{{dy}}{{dx}} + 3{x^2}{y^2} + \frac{{2x}}{{{x^2} + 2}} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & \frac{1}{{x + y}}\frac{{dy}}{{dx}} - 2y{x^3}\frac{{dy}}{{dx}} = 3{x^2}{y^2} + \frac{{2x}}{{{x^2} + 2}} - \frac{1}{{x + y}} \cr & \left( {\frac{1}{{x + y}} - 2y{x^3}} \right)\frac{{dy}}{{dx}} = 3{x^2}{y^2} + \frac{{2x}}{{{x^2} + 2}} - \frac{1}{{x + y}} \cr & \frac{{dy}}{{dx}} = \frac{{3{x^2}{y^2} + \frac{{2x}}{{{x^2} + 2}} - \frac{1}{{x + y}}}}{{\frac{1}{{x + y}} - 2y{x^3}}} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,2} \right)}} = \frac{{3{{\left( 1 \right)}^2}{{\left( 2 \right)}^2} + \frac{{2\left( 1 \right)}}{{{{\left( 1 \right)}^2} + 2}} - \frac{1}{{1 + 2}}}}{{\frac{1}{{1 + 2}} - 2\left( 2 \right){{\left( 1 \right)}^3}}} \cr & m = \frac{{12 + \frac{2}{3} - \frac{1}{3}}}{{\frac{1}{3} - 4}} = \frac{{37/3}}{{ - 11/3}} = - \frac{{37}}{{11}} \cr & {\text{find the equation of the tangent line at the point }}\left( {1,2} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = - \frac{{37}}{{11}}\left( {x - 1} \right) \cr & y - 2 = - \frac{{37}}{{11}}x + \frac{{37}}{{11}} \cr & y = - \frac{{37}}{{11}}x + \frac{{37}}{{11}} + 2 \cr & y = - \frac{{37}}{{11}}x + \frac{{59}}{{11}} \cr}