Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 50

$$\eqalign{ & {\bf{a}}.\,y = 10x - 20 \cr & {\bf{b}}.y = 12x - 33 \cr} $$

$$\eqalign{ & {\text{Let }}g\left( x \right) = {x^2} + f\left( x \right) \cr & {\bf{a}} \cr & g\left( 3 \right) = {\left( 3 \right)^2} + f\left( 3 \right) \cr & g\left( 3 \right) = 9 + 1 \cr & g\left( 3 \right) = 10 \cr & {\text{Point }}\left( {3,10} \right) \cr & \cr & {\text{Calculate }}g'\left( 3 \right) \cr & g'\left( x \right) = 2x + f'\left( x \right) \cr & g'\left( 3 \right) = 2\left( 3 \right) + f'\left( 3 \right) \cr & g'\left( 3 \right) = 6 + 4 \cr & g'\left( 3 \right) = 10 \cr & {\text{The equation of the tangent line is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 10 = 10\left( {x - 3} \right) \cr & y - 10 = 10x - 30 \cr & y = 10x - 20 \cr & \cr & {\bf{b}}. \cr & {\text{Let }}h\left( x \right) = 3f\left( x \right) \cr & h\left( 3 \right) = 3f\left( 3 \right) \cr & h\left( 3 \right) = 3\left( 1 \right) \cr & h\left( 3 \right) = 3 \cr & {\text{Point }}\left( {3,3} \right) \cr & \cr & {\text{Calculate }}h'\left( 3 \right) \cr & h'\left( x \right) = 3f'\left( x \right) \cr & h'\left( 3 \right) = 3f'\left( 3 \right) \cr & h'\left( 3 \right) = 3\left( 4 \right) \cr & h'\left( 3 \right) = 12 \cr & {\text{The equation of the tangent line is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 3 = 12\left( {x - 3} \right) \cr & y - 3 = 12x - 36 \cr & y = 12x - 33 \cr} $$