Chapter 2 - The Derivative - 2.2 The Derivative Function - Exercises Set 2.2 - Page 131: 14

$$\,\,\,f'\left( x \right) = \frac{1}{{\sqrt {2x + 1} }},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{3}x + \frac{5}{3}$$

$$\eqalign{ & f\left( x \right) = \sqrt {2x + 1} ;\,\,\,\,a = 4 \cr & {\text{Evaluate at }}a = 8 \cr & f\left( 4 \right) = \sqrt {2\left( 4 \right) + 1} \cr & f\left( 4 \right) = 3 \cr & f\left( 4 \right) = 3 \Rightarrow \left( {4,3} \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{{\sqrt {2\left( {x + h} \right) + 1} - \sqrt {2x + 1} }}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {2\left( {x + h} \right) + 1} - \sqrt {2x + 1} }}{h} \times \frac{{\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} }}{{\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} }} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {2\left( {x + h} \right) + 1} } \right)}^2} - {{\left( {\sqrt {2x + 1} } \right)}^2}}}{{h\left( {\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {x + h} \right) + 1 - 2x - 1}}{{h\left( {\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2x + 2h + 1 - 2x - 1}}{{h\left( {\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{{h\left( {\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{2}{{\sqrt {2\left( {x + h} \right) + 1} + \sqrt {2x + 1} }} \cr & h \to 0 \cr & \,\,\,f'\left( x \right) = \frac{2}{{\sqrt {2\left( {x + 0} \right) + 1} + \sqrt {2x + 1} }} \cr & \,\,\,f'\left( x \right) = \frac{2}{{2\sqrt {2x + 1} }} \cr & \,\,\,f'\left( x \right) = \frac{1}{{\sqrt {2x + 1} }} \cr & \cr & {\text{Calculate the tangent line }}y = f\left( x \right){\text{ at }}x = a \cr & \,\,\,m = f'\left( 4 \right) = \frac{1}{{\sqrt {2\left( 4 \right) + 1} }} \cr & \,\,\,m = \frac{1}{3} \cr & {\text{The tangent line is given by }}y - {y_1} = m\left( {x - {x_1}} \right){\text{ at }}\left( {4,3} \right) \cr & \,\,y - 3 = \frac{1}{3}\left( {x - 4} \right) \cr & \,\,y - 3 = \frac{1}{3}x - \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\,y = \frac{1}{3}x + \frac{5}{3} \cr & \cr & \,\,\,f'\left( x \right) = \frac{1}{{\sqrt {2x + 1} }},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{3}x + \frac{5}{3} \cr} $$