Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 28

$y = - x + \frac{\pi }{4}$

$$\eqalign{ & \tan \left( {x + y} \right) + \sec \left( {x - y} \right) = 2,{\text{ }}\left( {\frac{\pi }{8},\frac{\pi }{8}} \right) \cr & \cr & {\text{Taking the derivative of both sides of the equation with }} \cr & {\text{respect to }}x \cr & \frac{d}{{dx}}\left[ {\tan \left( {x + y} \right)} \right] + \frac{d}{{dx}}\left[ {\sec \left( {x - y} \right)} \right] = \frac{d}{{dx}}\left[ 2 \right] \cr & \cr & {\text{Using the formulas }} \cr & \frac{d}{{dx}}\left[ {\tan u} \right] = {\sec ^2}u\frac{{du}}{{dx}},\,{\text{ }}and{\text{ }}\frac{d}{{dx}}\left[ {\sec u} \right] = \sec u\tan u\frac{{du}}{{dx}},{\text{ so}} \cr & {\sec ^2}\left( {x + y} \right)\frac{d}{{dx}}\left[ {x + y} \right] + \sec \left( {x - y} \right)\tan \left( {x - y} \right)\frac{d}{{dx}}\left[ {x - y} \right] = 0 \cr & \cr & {\text{Computing derivatives}} \cr & {\sec ^2}\left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right) + \sec \left( {x - y} \right)\tan \left( {x - y} \right)\left( {1 - \frac{{dy}}{{dx}}} \right) = 0 \cr & {\sec ^2}\left( {x + y} \right) + {\sec ^2}\left( {x + y} \right)\frac{{dy}}{{dx}} + \sec \left( {x - y} \right)\tan \left( {x - y} \right) + \cr & - \sec \left( {x - y} \right)\tan \left( {x - y} \right)\frac{{dy}}{{dx}} = 0 \cr & \cr & {\text{Collecting the terms that contains }}\frac{{dy}}{{dx}} \cr & {\sec ^2}\left( {x + y} \right)\frac{{dy}}{{dx}} - \sec \left( {x - y} \right)\tan \left( {x - y} \right)\frac{{dy}}{{dx}} = - {\sec ^2}\left( {x + y} \right) \cr & - \sec \left( {x - y} \right)\tan \left( {x - y} \right) \cr & \cr & {\text{Factor and solve for }}\frac{{dy}}{{dx}} \cr & \left[ {{{\sec }^2}\left( {x + y} \right) - \sec \left( {x - y} \right)\tan \left( {x - y} \right)} \right]\frac{{dy}}{{dx}} = - {\sec ^2}\left( {x + y} \right) \cr & - \sec \left( {x - y} \right)\tan \left( {x - y} \right) \cr & \frac{{dy}}{{dx}} = \frac{{ - {{\sec }^2}\left( {x + y} \right) - \sec \left( {x - y} \right)\tan \left( {x - y} \right)}}{{{{\sec }^2}\left( {x + y} \right) - \sec \left( {x - y} \right)\tan \left( {x - y} \right)}} \cr & \cr & {\text{Find the slope }}m{\text{ at the point }}\left( {\frac{\pi }{8},\frac{\pi }{8}} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{\pi }{8},\frac{\pi }{8}} \right)}} = \frac{{ - {{\sec }^2}\left( {\frac{\pi }{4}} \right) - \sec \left( 0 \right)\tan \left( 0 \right)}}{{{{\sec }^2}\left( {\frac{\pi }{4}} \right) - \sec \left( 0 \right)\tan \left( 0 \right)}} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{\pi }{8},\frac{\pi }{8}} \right)}} = - 1 \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( {\frac{\pi }{8},\frac{\pi }{8}} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = \frac{\pi }{8}{\text{ and }}{y_1} = \frac{\pi }{8} \cr & {\text{Therefore}} \cr & y - \frac{\pi }{8} = - \left( {x - \frac{\pi }{8}} \right) \cr & {\text{Simplify}} \cr & y - \frac{\pi }{8} = - x + \frac{\pi }{8} \cr & y = - x + \frac{\pi }{4} \cr} $$