Answer
$f'\left( 6 \right) = \frac{2}{5}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {4x + 1} ,{\text{ }}a = 6 \cr
& {\text{Using the definition of the derivative}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {4\left( {a + h} \right) + 1} - \sqrt {4a + 1} }}{h} \cr
& {\text{Rationalizing the numerator}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {4\left( {a + h} \right) + 1} - \sqrt {4a + 1} }}{h} \times \frac{{\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1} }}{{\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1} }} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {4\left( {a + h} \right) + 1} } \right)}^2} - {{\left( {\sqrt {4a + 1} } \right)}^2}}}{{h(\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1}) }} \cr
& {\text{Simplifying}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4\left( {a + h} \right) + 1 - \left( {4a + 1} \right)}}{{h(\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1}) }} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4a + 4h + 1 - 4a - 1}}{{h(\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1}) }} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4h}}{{h(\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1}) }} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{4}{{\sqrt {4\left( {a + h} \right) + 1} + \sqrt {4a + 1} }} \cr
& {\text{Evaluate the limit when }}h \to 0 \cr
& f'\left( a \right) = \frac{4}{{\sqrt {4a + 1} + \sqrt {4a + 1} }} \cr
& f'\left( a \right) = \frac{2}{{\sqrt {4a + 1} }} \cr
& {\text{Find }}f'\left( a \right){\text{ at }}a = 6 \cr
& f'\left( 6 \right) = \frac{2}{{\sqrt {4\left( 6 \right) + 1} }} \cr
& f'\left( 6 \right) = \frac{2}{5} \cr} $$
