Answer
$$\left( {\text{a}} \right) - \frac{1}{{2{{\left( {{x_0}} \right)}^{3/2}}}},\,\,\,\,\left( {\text{b}} \right) - \frac{1}{{16}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt x }},\,\,\,\,{x_0} = 4 \cr
& \left( {\text{a}} \right){\text{Calculating a formula for the slope of the tangent line at }}{x_0} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{\sqrt {{x_0} + h} }} - \frac{1}{{\sqrt {{x_0}} }}}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {{x_0}} - \sqrt {{x_0} + h} }}{{h\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)}} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {{x_0}} - \sqrt {{x_0} + h} }}{{h\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)}} \times \frac{{\sqrt {{x_0}} + \sqrt {{x_0} + h} }}{{\sqrt {{x_0}} + \sqrt {{x_0} + h} }} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {{x_0}} } \right)}^2} - {{\left( {\sqrt {{x_0} + h} } \right)}^2}}}{{h\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)\left( {\sqrt {{x_0}} + \sqrt {{x_0} + h} } \right)}} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{x_0} - {x_0} - h}}{{h\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)\left( {\sqrt {{x_0}} + \sqrt {{x_0} + h} } \right)}} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)\left( {\sqrt {{x_0}} + \sqrt {{x_0} + h} } \right)}} \cr
& {m_{\tan }} = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0} + h} } \right)\left( {\sqrt {{x_0}} + \sqrt {{x_0} + h} } \right)}} \cr
& h \to 0 \cr
& {m_{\tan }} = - \frac{1}{{\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0}} } \right)\left( {\sqrt {{x_0}} + \sqrt {{x_0}} } \right)}} \cr
& {m_{\tan }} = - \frac{1}{{{x_0}\left( {2\sqrt {{x_0}} } \right)}} \cr
& {m_{\tan }} = - \frac{1}{{2{{\left( {{x_0}} \right)}^{3/2}}}} \cr
& \cr
& \left( {\text{b}} \right){\text{Using the formula obtained in part }}\left( {\text{a}} \right){\text{ to find the slope}} \cr
& {\text{at }}{x_0} = 4 \cr
& {m_{\tan }} = - \frac{1}{{2{{\left( 4 \right)}^{3/2}}}} \cr
& {m_{\tan }} = - \frac{1}{{16}} \cr} $$
