Answer
$$g'\left( x \right) = \frac{1}{{\sqrt {2x - 3} }}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}g\left( x \right) = \sqrt {2x - 3} \cr
& {\text{Differentiate using the definition of the derivative}} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {2\left( {x + h} \right) - 3} - \sqrt {2x - 3} }}{h} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {2x + 2h - 3} - \sqrt {2x - 3} }}{h} \cr
& {\text{Rationalizing}} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {2x + 2h - 3} - \sqrt {2x - 3} }}{h} \times \frac{{\sqrt {2x + 2h - 3} + \sqrt {2x - 3} }}{{\sqrt {2x + 2h - 3} + \sqrt {2x - 3} }} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {2x + 2h - 3} } \right)}^2} - {{\left( {\sqrt {2x - 3} } \right)}^2}}}{{h\left( {\sqrt {2x + 2h - 3} + \sqrt {2x - 3} } \right)}} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2x + 2h - 3 - 2x + 3}}{{h\left( {\sqrt {2x + 2h - 3} + \sqrt {2x - 3} } \right)}} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{{h\left( {\sqrt {2x + 2h - 3} + \sqrt {2x - 3} } \right)}} \cr
& g'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{2}{{\sqrt {2x + 2h - 3} + \sqrt {2x - 3} }} \cr
& {\text{Evaluate the limit}} \cr
& g'\left( x \right) = \frac{2}{{\sqrt {2x + 2\left( 0 \right) - 3} + \sqrt {2x - 3} }} \cr
& g'\left( x \right) = \frac{2}{{\sqrt {2x - 3} + \sqrt {2x - 3} }} \cr
& g'\left( x \right) = \frac{2}{{2\sqrt {2x - 3} }} \cr
& g'\left( x \right) = \frac{1}{{\sqrt {2x - 3} }} \cr} $$
