Answer
$$\frac{1}{{\sqrt {x + h} + \sqrt x }}$$
Work Step by Step
$$\eqalign{
& \frac{{\sqrt {x + h} - \sqrt x }}{h},{\text{ for }}h \ne 0 \cr
& {\text{rationalizing}} \cr
& = \frac{{\sqrt {x + h} - \sqrt x }}{h} \times \frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }} \cr
& = \frac{{\left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} + \sqrt x } \right)}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& {\text{use the special product }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} \cr
& = \frac{{{{\left( {\sqrt {x + h} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& {\text{simplifying}} \cr
& = \frac{{x + h - x}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& = \frac{h}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& {\text{cancel h}} \cr
& = \frac{1}{{\sqrt {x + h} + \sqrt x }} \cr} $$
