Chapter 2 - The Derivative - 2.5 Derivatives of Trigonometric Functions - Exercises Set 2.5 - Page 151: 25

$$\left( {\text{a}} \right)y = x,\,\,\,\,\,\left( {\text{b}} \right)y = 2x - \frac{\pi }{2} + 1,\,\,\,\,\,\left( {\text{c}} \right)y = 2x + \frac{\pi }{2} - 1$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \tan x \cr & f'\left( x \right) = {\sec ^2}x \cr & \cr & \left( {\text{a}} \right){\text{ At }}x = 0 \cr & \,\,\,\,f\left( 0 \right) = \tan \left( 0 \right) = 0 \cr & {\text{We obtain the point }}\left( {0,0} \right) \cr & {\text{The slope at }}x = 0{\text{ is given by}} \cr & \,\,m = f'\left( 0 \right) = {\sec ^2}\left( 0 \right) \cr & \,\,m = 1 \cr & {\text{The equation of the line tangent to the graph is}} \cr & \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr & \,\,y - 0 = 1\left( {x - 0} \right) \cr & \,\,y = x \cr & \cr & \left( {\text{b}} \right){\text{ At }}x = \frac{\pi }{4} \cr & \,\,\,\,f\left( {\frac{\pi }{4}} \right) = \tan \left( {\frac{\pi }{4}} \right) = 1 \cr & {\text{We obtain the point }}\left( {\frac{\pi }{4},1} \right) \cr & {\text{The slope at }}x = 0{\text{ is given by}} \cr & \,\,m = f'\left( {\frac{\pi }{4}} \right) = {\sec ^2}\left( {\frac{\pi }{4}} \right) \cr & \,\,m = 2 \cr & {\text{The equation of the line tangent to the graph is}} \cr & \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr & \,\,y - 1 = 2\left( {x - \frac{\pi }{4}} \right) \cr & \,\,y - 1 = 2x - \frac{\pi }{2} \cr & \,\,\,\,\,\,\,\,\,y = 2x - \frac{\pi }{2} + 1 \cr & \cr & \left( {\text{c}} \right){\text{ At }}x = - \frac{\pi }{4} \cr & \,\,\,\,f\left( { - \frac{\pi }{4}} \right) = \tan \left( { - \frac{\pi }{4}} \right) = - 1 \cr & {\text{We obtain the point }}\left( { - \frac{\pi }{4}, - 1} \right) \cr & {\text{The slope at }}x = 0{\text{ is given by}} \cr & \,\,m = f'\left( { - \frac{\pi }{4}} \right) = {\sec ^2}\left( { - \frac{\pi }{4}} \right) \cr & \,\,m = 2 \cr & {\text{The equation of the line tangent to the graph is}} \cr & \,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr & \,\,y + 1 = 2\left( {x + \frac{\pi }{4}} \right) \cr & \,\,y + 1 = 2x + \frac{\pi }{2} \cr & \,\,\,\,\,\,\,\,\,y = 2x + \frac{\pi }{2} - 1 \cr & \cr & \left( {\text{a}} \right)y = x,\,\,\,\,\,\left( {\text{b}} \right)y = 2x - \frac{\pi }{2} + 1,\,\,\,\,\,\left( {\text{c}} \right)y = 2x + \frac{\pi }{2} - 1 \cr} $$