Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 201: 52

$$\eqalign{ & \left( {\bf{a}} \right) \cr & y = x - 1 \cr & y = - \frac{1}{2}x - \frac{1}{2} \cr & y = - \frac{1}{2}x + \frac{3}{2} \cr & \left( {\bf{b}} \right){\text{ See the graph}} \cr} $$

$$\eqalign{ & x + {y^3} - y = 1;\,\,\,x = 1 \cr & {\text{Calculate }}y{\text{ for }}x = 1 \cr & 1 + {y^3} - y = 1 \cr & {\text{ }}{y^3} - y = 0 \cr & y\left( {{y^2} - 1} \right) = 0 \cr & y = 0,{\text{ }}y = - 1,{\text{ }}y = 1 \cr & {\text{We obtain the points}} \cr & \left( {1,0} \right),{\text{ }}\left( {1, - 1} \right),{\text{ }}\left( {1,1} \right) \cr & {\text{Differentiate implicitly with respect to }}x \cr & 1 + 3{y^2}\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = 0 \cr & {\text{Solve for }}\frac{{dy}}{{dx}} \cr & 3{y^2}\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = - 1 \cr & \frac{{dy}}{{dx}}\left( {3{y^2} - 1} \right) = - 1 \cr & \frac{{dy}}{{dx}} = \frac{1}{{1 - 3{y^2}}} \cr & {\text{Calculate the slope }}m{\text{ for the points }}\left( {1,0} \right),{\text{ }}\left( {1, - 1} \right),{\text{ }}\left( {1,1} \right) \cr & \left( {1,0} \right) \Rightarrow m = \frac{1}{{1 - 3{{\left( 0 \right)}^2}}} = 1 \cr & \left( {1, - 1} \right) \Rightarrow m = \frac{1}{{1 - 3{{\left( { - 1} \right)}^2}}} = - \frac{1}{2} \cr & \left( {1,1} \right) \Rightarrow m = \frac{1}{{1 - 3{{\left( { 1} \right)}^2}}} = - \frac{1}{2} \cr & {\text{Find the equations of the tangent lines at the points }} \cr & \left( {1,0} \right),{\text{ }}\left( {1, - 1} \right),{\text{ }}\left( {1,1} \right),{\text{ then}} \cr & {\bf{For }}\left( {{\bf{1,0}}} \right) \cr & y - 0 = 1\left( {x - 1} \right) \cr & y = x - 1 \cr & {\bf{For }}\left( {{\bf{1, - 1}}} \right) \cr & y + 1 = - \frac{1}{2}\left( {x - 1} \right) \cr & {\text{ }}y = - \frac{1}{2}x - \frac{1}{2} \cr & {\bf{For }}\left( {{\bf{1,1}}} \right) \cr & y - 1 = - \frac{1}{2}\left( {x - 1} \right) \cr & {\text{ }}y = - \frac{1}{2}x + \frac{3}{2} \cr} $$