Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 29

$y = \frac{1}{{\sqrt 3 }}x + 4$

$$\eqalign{ & {x^{2/3}} + {y^{2/3}} = 4,{\text{ }}\left( { - 3\sqrt 3 ,1} \right) \cr & {\text{Taking the derivative of both sides of the equation with }} \cr & {\text{respect to }}x \cr & \frac{d}{{dx}}\left[ {{x^{2/3}}} \right] + \frac{d}{{dx}}\left[ {{y^{2/3}}} \right] = \frac{d}{{dx}}\left[ 4 \right] \cr & {\text{Using the general power rule}} \cr & \frac{2}{3}{x^{2/3 - 1}} + \frac{2}{3}{y^{2/3 - 1}}\frac{{dy}}{{dx}} = 0 \cr & \frac{2}{3}{x^{ - 1/3}} + \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = 0 \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = - \frac{2}{3}{x^{ - 1/3}} \cr & {y^{ - 1/3}}\frac{{dy}}{{dx}} = - {x^{ - 1/3}} \cr & \frac{{dy}}{{dx}} = \frac{{ - {x^{ - 1/3}}}}{{{y^{ - 1/3}}}} \cr & \frac{{dy}}{{dx}} = - \frac{{\root 3 \of y }}{{\root 3 \of x }} \cr & \cr & {\text{Find the slope }}m{\text{ at the point }}\left( { - 3\sqrt 3 ,1} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 3\sqrt 3 ,1} \right)}} = - \frac{{\root 3 \of y }}{{\root 3 \of x }} \cr & m = - \frac{{\root 3 \of 1 }}{{\root 3 \of { - 3\sqrt 3 } }} \cr & m = \frac{1}{{{{\left( {{3^{3/2}}} \right)}^{1/3}}}} \cr & m = \frac{1}{{\sqrt 3 }} \cr & {\text{Find the equation of the tangent line at the given point}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & \underbrace {\left( { - 3\sqrt 3 ,1} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = - 3\sqrt 3 {\text{ and }}{y_1} = 1 \cr & {\text{Therefore}} \cr & y - 1 = \frac{1}{{\sqrt 3 }}\left( {x + 3\sqrt 3 } \right) \cr & {\text{Simplify}} \cr & y = \frac{1}{{\sqrt 3 }}x + 3 + 1 \cr & y = \frac{1}{{\sqrt 3 }}x + 4 \cr} $$