Chapter 2 - The Derivative - 2.2 The Derivative Function - Exercises Set 2.2 - Page 131: 9

$$\,\,\,f'\left( x \right) = 4x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 4x - 2$$

$$\eqalign{ & f\left( x \right) = 2{x^2};\,\,\,\,a = 1 \cr & {\text{Evaluate at }}a = 1 \cr & f\left( 1 \right) = 2{\left( 1 \right)^2} \cr & f\left( 1 \right) = 2 \Rightarrow \left( {1,2} \right) \cr & {\text{The function }}f'\left( x \right){\text{ is defined by the formula}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & {\text{Then}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( {x + h} \right)}^2} - 2{x^2}}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {{x^2} + 2xh + {h^2}} \right) - 2{x^2}}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^2} + 4xh + 2{h^2} - 2{x^2}}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4xh + 2{h^2}}}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {4x + 2h} \right) \cr & h \to 0 \cr & \,\,\,f'\left( x \right) = 4x + 2\left( 0 \right) \cr & \,\,\,f'\left( x \right) = 4x \cr & \cr & {\text{Calculate the tangent line }}y = f\left( x \right){\text{ at }}x = a \cr & \,\,\,m = f'\left( 1 \right) = 4\left( 1 \right) \cr & \,\,\,m = 4 \cr & {\text{The tangent line is given by }}y - {y_1} = m\left( {x - {x_1}} \right){\text{ at }}\left( {1,2} \right) \cr & \,\,y - 2 = 4\left( {x - 1} \right) \cr & \,\,y - 2 = 4x - 4 \cr & \,\,\,\,\,\,\,\,\,\,y = 4x - 2 \cr & \cr & \,\,\,f'\left( x \right) = 4x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 4x - 2 \cr} $$