Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{2{x^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt x }} \cr
& {\text{Using the formula }}\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{1}{{\sqrt {x + \Delta x} }} - \frac{1}{{\sqrt x }}}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{\sqrt x - \sqrt {x + \Delta x} }}{{\sqrt x \sqrt {x + \Delta x} }}}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sqrt x - \sqrt {x + \Delta x} }}{{\Delta x\sqrt x \sqrt {x + \Delta x} }} \times \frac{{\sqrt x + \sqrt {x + \Delta x} }}{{\sqrt x + \sqrt {x + \Delta x} }} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{{\left( {\sqrt x } \right)}^2} - {{\left( {\sqrt {x + \Delta x} } \right)}^2}}}{{\Delta x\left( {\sqrt x \sqrt {x + \Delta x} } \right)\left( {\sqrt x + \sqrt {x + \Delta x} } \right)}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{x - x - \Delta x}}{{\Delta x\left( {\sqrt x \sqrt {x + \Delta x} } \right)\left( {\sqrt x + \sqrt {x + \Delta x} } \right)}} \cr
& \frac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\left( {\sqrt x \sqrt {x + \Delta x} } \right)\left( {\sqrt x + \sqrt {x + \Delta x} } \right)}} \cr
& \cr
& \Delta x \to 0 \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left( {\sqrt x \sqrt {x + 0} } \right)\left( {\sqrt x + \sqrt {x + 0} } \right)}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left( x \right)\left( {2\sqrt x } \right)}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{2{x^{3/2}}}} \cr} $$