Chapter 11 - Section 11.6 - Implicit Differentiation - Exercises - Page 853: 42

$${\text{ }}{\bf{a}}){\text{ }}m = - \frac{5}{3}{\text{ }}{\bf{b}})y = - \frac{5}{3}x - \frac{5}{3} + \ln 3$$

$$\eqalign{ & {e^{ - xy}} + 2x = 1,{\text{ }}x = - 1 \cr & {\text{We have that }}x = - 1,{\text{ substituting}} \cr & {e^y} + 2\left( { - 1} \right) = 1 \cr & {e^y} - 2 = 1 \cr & {e^y} = 3 \cr & y = \ln 3 \cr & {\text{For }}x = 0{\text{ we obtain the point }}\left( { - 1,\ln 3} \right) \cr & {e^{ - xy}} + 2x = 1 \cr & {\text{Differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left[ {{e^{ - xy}}} \right] + \frac{d}{{dx}}\left[ {2x} \right] = \frac{d}{{dx}}\left[ 1 \right] \cr & {e^{ - xy}}\underbrace {\frac{d}{{dx}}\left[ { - xy} \right]}_{{\text{Product rule}}} + 2 = 0 \cr & {e^{ - xy}}\left( { - x\frac{{dy}}{{dx}} - y} \right) = - 2 \cr & - x{e^{ - xy}}\frac{{dy}}{{dx}} - y{e^{ - xy}} = - 2 \cr & {\text{Solve for }}\frac{{dy}}{{dx}} \cr & x{e^{ - xy}}\frac{{dy}}{{dx}} + y{e^{ - xy}} = 2 \cr & \frac{{dy}}{{dx}} = \frac{{2 - y{e^{ - xy}}}}{{x{e^{ - xy}}}} \cr & \cr & \left( {\bf{a}} \right){\text{The slope at }}\left( { - 1,\ln 3} \right){\text{ is}} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 1,\ln 3} \right)}} = \frac{{2 - \left( { - 1} \right){e^{\ln 3}}}}{{ - {e^{\ln 3}}}} \cr & m = \frac{{2 + 3}}{{ - 3}} \cr & m = - \frac{5}{3} \cr & \left( {\bf{b}} \right){\text{ The equation of the tangent line at }}\left( { - 1,\ln 3} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \ln 3 = - \frac{5}{3}\left( {x + 1} \right) \cr & {\text{Simplify}} \cr & y - \ln 3 = - \frac{5}{3}x - \frac{5}{3} \cr & y = - \frac{5}{3}x - \frac{5}{3} + \ln 3 \cr} $$