Chapter 2 - The Derivative - 2.2 The Derivative Function - Exercises Set 2.2 - Page 131: 12

$$\,\,\,f'\left( x \right) = 6{x^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 6x + 5$$

$$\eqalign{ & f\left( x \right) = 2{x^3} + 1;\,\,\,\,a = - 1 \cr & {\text{Evaluate at }}a = - 1 \cr & f\left( { - 1} \right) = 2{\left( { - 1} \right)^3} + 1 \cr & f\left( { - 1} \right) = - 2 + 1 \cr & f\left( { - 1} \right) = - 1 \cr & f\left( { - 1} \right) = - 1 \Rightarrow \left( { - 1, - 1} \right) \cr & {\text{The function }}f'\left( x \right){\text{ is defined by the formula}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & {\text{Then}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( {x + h} \right)}^3} + 1 - \left( {2{x^3} + 1} \right)}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) + 1 - \left( {2{x^3} + 1} \right)}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^3} + 6{x^2}h + 6x{h^2} + 2{h^3} + 1 - 2{x^3} - 1}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{6{x^2}h + 6x{h^2} + 2{h^3}}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {6{x^2} + 6xh + 2{h^2}} \right) \cr & h \to 0 \cr & \,\,\,f'\left( x \right) = 6{x^2} + 6x\left( 0 \right) + 2{\left( 0 \right)^2} \cr & \,\,\,f'\left( x \right) = 6{x^2} \cr & \cr & {\text{Calculate the tangent line }}y = f\left( x \right){\text{ at }}x = a \cr & \,\,\,m = f'\left( { - 1} \right) = 6{\left( { - 1} \right)^2} \cr & \,\,\,m = 6 \cr & {\text{The tangent line is given by }}y - {y_1} = m\left( {x - {x_1}} \right){\text{ at }}\left( { - 1, - 1} \right) \cr & \,\,y + 1 = 6\left( {x + 1} \right) \cr & \,\,y + 1 = 6x + 6 \cr & \,\,\,\,\,\,\,\,\,\,y = 6x + 5 \cr & \cr & \,\,\,f'\left( x \right) = 6{x^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 6x + 5 \cr} $$