## Calculus with Applications (10th Edition)

$$y = 7x - 5$$
\eqalign{ & \ln \left( {{x^2} + {y^2}} \right) = \ln 5x + \frac{y}{x} - 2;\,\,\,\,\,\,\,\left( {1,2} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {\ln \left( {{x^2} + {y^2}} \right)} \right) = \frac{d}{{dx}}\left( {\ln 5x} \right) + \frac{d}{{dx}}\left( {{x^{ - 1}}y} \right) - \frac{d}{{dx}}\left( 2 \right) \cr & {\text{solve the derivatives}} \cr & \frac{2}{{{x^2} + {y^2}}}\left( {x + y\frac{{dy}}{{dx}}} \right) = \frac{1}{x} + {x^{ - 1}}\frac{{dy}}{{dx}} - \frac{y}{{{x^2}}} \cr & \frac{{2x}}{{{x^2} + {y^2}}} + \frac{{2y}}{{{x^2} + {y^2}}}\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{1}{x}\frac{{dy}}{{dx}} - \frac{y}{{{x^2}}} \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & \frac{{2y}}{{{x^2} + {y^2}}}\frac{{dy}}{{dx}} - \frac{1}{x}\frac{{dy}}{{dx}} = \frac{1}{x} - \frac{y}{{{x^2}}} - \frac{{2x}}{{{x^2} + {y^2}}} \cr & \left( {\frac{{2y}}{{{x^2} + {y^2}}} - \frac{1}{x}} \right)\frac{{dy}}{{dx}} = \frac{1}{x} - \frac{y}{{{x^2}}} - \frac{{2x}}{{{x^2} + {y^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{\frac{1}{x} - \frac{y}{{{x^2}}} - \frac{{2x}}{{{x^2} + {y^2}}}}}{{\frac{{2y}}{{{x^2} + {y^2}}} - \frac{1}{x}}} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,2} \right)}} = \frac{{\frac{1}{1} - \frac{2}{{{1^2}}} - \frac{{2\left( 1 \right)}}{{{1^2} + {2^2}}}}}{{\frac{{2\left( 2 \right)}}{{{1^2} + {2^2}}} - \frac{1}{1}}} \cr & m = \frac{{1 - 2 - 2/5}}{{4/5 - 1}} = \frac{{ - 7/5}}{{ - 1/5}} = 7 \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 = 7\left( {x - 1} \right) \cr & y - 2 = 7x - 7 \cr & y = 7x - 5 \cr}