Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 25

$$\left. {\bf{a}} \right)y = \frac{4}{7}x + \frac{{48}}{7},\,\,\,\,\,\,\,\left. {\bf{b}} \right)y = \frac{2}{3}x + 6$$

$$\eqalign{ & f\left( x \right) = 4\sqrt x ;\,\,\,\,\,\,\,\,\,{x_1} = 9,\,\,\,\,\,\,{x_2} = 16 \cr & \cr & \left. {\bf{a}} \right) \cr & {\text{Evaluate }}f\left( {{x_1}} \right){\text{ and }}f\left( {{x_2}} \right) \cr & {y_1} = f\left( {{x_1}} \right) = 4\sqrt 9 = 12 \cr & {y_1} = 12 \cr & \left( {{x_1},{y_1}} \right) = \left( {9,12} \right) \cr & and \cr & {y_2} = f\left( {{x_2}} \right) = f\left( {16} \right) = 4\sqrt {16} = 16 \cr & {y_2} = 16 \cr & \left( {{x_2},{y_2}} \right) = \left( {16,16} \right) \cr & \cr & {\text{The slope of the secant line is }} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{16 - 12}}{{16 - 9}} = \frac{4}{7} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 12 = \frac{4}{7}\left( {x - 9} \right) \cr & y - 12 = \frac{4}{7}x - \frac{{36}}{7} \cr & y = \frac{4}{7}x + \frac{{48}}{7} \cr & \cr & \left. {\bf{b}} \right) \cr & {\text{Using the definition of the derivative}}: \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4\sqrt {x + h} - 4\sqrt x }}{h} \cr & f'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h}\left( {\frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }}} \right) \cr & f'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {x + h} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr & f'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \frac{{x + h - x}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr & f'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {x + h} + \sqrt x }} \cr & f'\left( x \right) = \frac{2}{{\sqrt x }} \cr & \cr & {\text{The slope of the tangent line at the point }}\left( {{x_1},{y_1}} \right) = \left( {9,12} \right){\text{ is}} \cr & f'\left( 9 \right) = \frac{2}{{\sqrt 9 }} \cr & f'\left( 9 \right) = \frac{2}{3} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 12 = \frac{2}{3}\left( {x - 9} \right) \cr & y - 12 = \frac{2}{3}x - 6 \cr & y = \frac{2}{3}x + 6 \cr} $$