Chapter 11 - Section 11.6 - Implicit Differentiation - Exercises - Page 853: 41

$${\text{ }}{\bf{a}}){\text{ }}m = - 0.1897,{\text{ }}{\bf{b}})y = - 0.1897x + 1.4717$$

$$\eqalign{ & {e^{xy}} - x = 4x \cr & {\text{We have that }}x = 3,{\text{ substituting}} \cr & {e^{3y}} - 3 = 4\left( 3 \right) \cr & {e^{3y}} = 15 \cr & y = \frac{1}{3}\ln 15 \cr & {\text{For }}x = 3{\text{ we obtain the point}}\left( {3,\frac{1}{3}\ln 15} \right) \cr & {e^{xy}} = 5x \cr & {\text{Differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left[ {{e^{xy}}} \right] = \frac{d}{{dx}}\left[ {5x} \right] \cr & {\text{Therefore,}} \cr & {e^{xy}}\underbrace {\frac{d}{{dx}}\left[ {xy} \right]}_{{\text{Product rule}}} = \frac{d}{{dx}}\left[ {5x} \right] \cr & {e^{xy}}\left( {x\frac{{dy}}{{dx}} + y} \right) = 5 \cr & x{e^{xy}}\frac{{dy}}{{dx}} + y{e^{xy}} = 5 \cr & \frac{{dy}}{{dx}} = \frac{{5 - y{e^{xy}}}}{{x{e^{xy}}}} \cr & \cr & \left( {\bf{a}} \right){\text{ Find the slope at }}\left( {3,\frac{1}{3}\ln 15} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {3,\frac{1}{3}\ln 15} \right)}} = \frac{{5 - \frac{1}{3}\ln 15{e^{3\ln 15}}}}{{3{e^{3\ln 15}}}} \cr & m = - 0.1897 \cr & \cr & \left( {\bf{b}} \right){\text{The equation of the tangent line at }}\left( {10, - 1} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{1}{3}\ln 15 = - 0.1897\left( {x - 3} \right) \cr & {\text{Simplifying}} \cr & y - 0.9026 = - 0.1897x + 0.5691 \cr & y = - 0.1897x + 1.4717 \cr} $$