Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 34

$y = - \frac{{\sqrt 3 }}{5}x + \frac{{11}}{5}$

$$\eqalign{ & {x^2}{y^2} = {\left( {y + 1} \right)^2}\left( {4 - {y^2}} \right),{\text{ }}\left( {2\sqrt 3 ,1} \right) \cr & {\text{Taking the derivative of both sides of the equation with }} \cr & {\text{respect to }}x \cr & \frac{d}{{dx}}\left[ {{x^2}{y^2}} \right] = \frac{d}{{dx}}\left[ {{{\left( {y + 1} \right)}^2}\left( {4 - {y^2}} \right)} \right] \cr & {\text{Use the product rule on both sides}} \cr & {x^2}\frac{d}{{dx}}\left[ {{y^2}} \right] + {y^2}\frac{d}{{dx}}\left[ {{x^2}} \right] = {\left( {y + 1} \right)^2}\frac{d}{{dx}}\left[ {\left( {4 - {y^2}} \right)} \right] \cr & + \left( {4 - {y^2}} \right)\frac{d}{{dx}}\left[ {{{\left( {y + 1} \right)}^2}} \right] \cr & {\text{Using the general power rule}} \cr & {x^2}\left( {2y} \right)\frac{{dy}}{{dx}} + {y^2}\left( {2x} \right) = \left( { - 2y} \right){\left( {y + 1} \right)^2}\frac{{dy}}{{dx}} + 2\left( {4 - {y^2}} \right)\left( {y + 1} \right)\frac{{dy}}{{dx}} \cr & {\text{Collecting the terms that contains }}\frac{{dy}}{{dx}} \cr & 2{x^2}y\frac{{dy}}{{dx}} + 2y{\left( {y + 1} \right)^2}\frac{{dy}}{{dx}} - 2\left( {4 - {y^2}} \right)\left( {y + 1} \right)\frac{{dy}}{{dx}} = - 2x{y^2} \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \left[ {2{x^2}y + 2y{{\left( {y + 1} \right)}^2} - 2\left( {4 - {y^2}} \right)\left( {y + 1} \right)} \right]\frac{{dy}}{{dx}} = - 2x{y^2} \cr & \frac{{dy}}{{dx}} = \frac{{ - 2x{y^2}}}{{2{x^2}y + 2y{{\left( {y + 1} \right)}^2} - 2\left( {4 - {y^2}} \right)\left( {y + 1} \right)}} \cr & \cr & {\text{Find the slope }}m{\text{ at the point }}\left( {2\sqrt 3 ,1} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {2\sqrt 3 ,1} \right)}} = \frac{{ - 2\left( {2\sqrt 3 } \right){{\left( 1 \right)}^2}}}{{2{{\left( {2\sqrt 3 } \right)}^2} + 2{{\left( {1 + 1} \right)}^2} - 2\left( {4 - 1} \right)\left( {1 + 1} \right)}} \cr & m = \frac{{ - 4\sqrt 3 }}{{24 + 8 - 12}} \cr & m = - \frac{{\sqrt 3 }}{5} \cr & \cr & {\text{Find the equation of the tangent line at the given point}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & \underbrace {\left( {2\sqrt 3 ,1} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 2\sqrt 3 {\text{ and }}{y_1} = 1 \cr & {\text{Therefore}} \cr & y - 1 = - \frac{{\sqrt 3 }}{5}\left( {x - 2\sqrt 3 } \right) \cr & {\text{Simplify}} \cr & y - 1 = - \frac{{\sqrt 3 }}{5}x + \frac{6}{5} \cr & y = - \frac{{\sqrt 3 }}{5}x + \frac{{11}}{5} \cr} $$