Chapter 2 - The Derivative - 2.2 The Derivative Function - Exercises Set 2.2 - Page 131: 13

$$\,\,\,f'\left( x \right) = \frac{1}{{2\sqrt {x + 1} }},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{6}x + \frac{5}{3}$$

$$\eqalign{ & f\left( x \right) = \sqrt {x + 1} ;\,\,\,\,a = 8 \cr & {\text{Evaluate at }}a = 8 \cr & f\left( 8 \right) = \sqrt {8 + 1} \cr & f\left( 8 \right) = 3 \cr & f\left( 8 \right) = 3 \Rightarrow \left( {8,3} \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 {x + h + 1} - \sqrt {x + 1} }}{h} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h + 1} - \sqrt {x + 1} }}{h} \times \frac{{\sqrt {x + h + 1} + \sqrt {x + 1} }}{{\sqrt {x + h + 1} + \sqrt {x + 1} }} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {x + h + 1} } \right)}^2} - {{\left( {\sqrt {x + 1} } \right)}^2}}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{x + h + 1 - x - 1}}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{h}{{h\left( {\sqrt {x + h + 1} + \sqrt {x + 1} } \right)}} \cr & \,\,\,f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {x + h + 1} + \sqrt {x + 1} }} \cr & h \to 0 \cr & \,\,\,f'\left( x \right) = \frac{1}{{\sqrt {x + 0 + 1} + \sqrt {x + 1} }} \cr & \,\,\,f'\left( x \right) = \frac{1}{{2\sqrt {x + 1} }} \cr & \cr & {\text{Calculate the tangent line }}y = f\left( x \right){\text{ at }}x = a \cr & \,\,\,m = f'\left( 8 \right) = \frac{1}{{2\sqrt {\left( 8 \right) + 1} }} \cr & \,\,\,m = \frac{1}{6} \cr & {\text{The tangent line is given by }}y - {y_1} = m\left( {x - {x_1}} \right){\text{ at }}\left( {8,3} \right) \cr & \,\,y - 3 = \frac{1}{6}\left( {x - 8} \right) \cr & \,\,y - 3 = \frac{1}{6}x - \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\,y = \frac{1}{6}x + \frac{5}{3} \cr & \cr & \,\,\,f'\left( x \right) = \frac{1}{{2\sqrt {x + 1} }},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{6}x + \frac{5}{3} \cr} $$