## Calculus with Applications (10th Edition)

$$y = - \frac{1}{4}x + 3$$
\eqalign{ & y + \frac{{\sqrt x }}{y} = 3;{\text{ }}\left( {4,2} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {y + \frac{{\sqrt x }}{y}} \right) = \frac{d}{{dx}}\left( 3 \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( y \right) + \frac{d}{{dx}}\left( {\frac{{\sqrt x }}{y}} \right) = \frac{d}{{dx}}\left( 3 \right) \cr & {\text{use quotient rule}} \cr & \frac{d}{{dx}}\left( y \right) + \frac{{y\frac{d}{{dx}}\left( {\sqrt x } \right) - \sqrt x \frac{d}{{dx}}\left( y \right)}}{{{y^2}}} = \frac{d}{{dx}}\left( 3 \right) \cr & {\text{solve the derivatives}} \cr & \frac{{dy}}{{dx}} + \frac{1}{{{y^2}}}\left( {y\left( {\frac{1}{{2\sqrt x }}} \right) - \sqrt x \frac{{dy}}{{dx}}} \right) = 0 \cr & \frac{{dy}}{{dx}} + \frac{1}{{2y\sqrt x }} - \frac{{\sqrt x }}{{{y^2}}}\frac{{dy}}{{dx}} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} - \frac{{\sqrt x }}{{{y^2}}}\frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x y}} \cr & \left( {1 - \frac{{\sqrt x }}{{{y^2}}}} \right)\frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x y}} \cr & \left( {\frac{{{y^2} - \sqrt x }}{{{y^2}}}} \right)\frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x y}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x y}}\left( {\frac{{{y^2}}}{{{y^2} - \sqrt x }}} \right) \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {4,2} \right)}} = - \frac{1}{{2\sqrt 4 \left( 2 \right)}}\left( {\frac{{{{\left( 2 \right)}^2}}}{{{{\left( 2 \right)}^2} - \sqrt 4 }}} \right) \cr & m = - \frac{1}{4} \cr & {\text{find the equation of the tangent line at the point }}\left( {4,2} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = - \frac{1}{4}\left( {x - 4} \right) \cr & y - 2 = - \frac{1}{4}x + 1 \cr & y = - \frac{1}{4}x + 3 \cr}