Chapter 11 - Section 11.6 - Implicit Differentiation - Exercises - Page 853: 40

$${\text{ }}{\bf{a}}){\text{ }}m = 1,{\text{ }}{\bf{b}})y = x - \frac{1}{e}$$

$$\eqalign{ & \ln \left( {x - y} \right) + 1 = 3{x^2},{\text{ }}x = 0 \cr & {\text{We have that }}x = 0,{\text{ substituting}} \cr & \ln \left( {0 - y} \right) + 1 = 3{\left( 0 \right)^2} \cr & \ln \left( { - y} \right) + 1 = 0 \cr & \ln \left( { - y} \right) = - 1 \cr & {e^{\ln \left( { - y} \right)}} = {e^{ - 1}} \cr & - y = {e^{ - 1}} \cr & y = - \frac{1}{e} \cr & {\text{For }}x = 0{\text{ we obtain the point }}\left( {0, - \frac{1}{e}} \right) \cr & \cr & \ln \left( {x - y} \right) + 1 = 3{x^2} \cr & {\text{Differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left[ {\ln \left( {x - y} \right)} \right] + \frac{d}{{dx}}\left[ 1 \right] = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr & \frac{1}{{x - y}}\frac{d}{{dx}}\left[ {x - y} \right] + 0 = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr & \frac{1}{{x - y}}\left( {1 - \frac{{dy}}{{dx}}} \right) = 6x \cr & \frac{1}{{x - y}} - \frac{1}{{x - y}}\frac{{dy}}{{dx}} = 6x \cr & {\text{Solve for }}\frac{{dy}}{{dx}} \cr & \frac{1}{{x - y}}\frac{{dy}}{{dx}} = \frac{1}{{x - y}} - 6x \cr & \frac{{dy}}{{dx}} = \left( {x - y} \right)\left( {\frac{1}{{x - y}} - 6x} \right) \cr & \frac{{dy}}{{dx}} = 1 - 6x\left( {x - y} \right) \cr & \cr & \left( {\bf{a}} \right){\text{ Calculating the slope at the point}}\left( {0, - \frac{1}{e}} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {0, - \frac{1}{e}} \right)}} = 1 - 6\left( 0 \right)\left( {0 + \frac{1}{e}} \right) \cr & m = 1 \cr & \cr & \left( {\bf{b}} \right){\text{ The equation of the tangent line at }}\left( {0, - \frac{1}{e}} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + \frac{1}{e} = 1\left( {x - 0} \right) \cr & y = x - \frac{1}{e} \cr} $$