Answer
$$\left( {\text{a}} \right)F'\left( x \right) = \frac{1}{{\sqrt x }}{\text{ and }}\left( {\text{b}} \right)F'\left( x \right) = \frac{1}{{\sqrt x }}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( x \right) = \int_4^x {\frac{1}{{\sqrt t }}dt} \cr
& \cr
& \left( {\text{a}} \right){\text{Use Part 2 of the Fundamental Theorem of Calculus to calculate }}F'\left( x \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& {\text{Then}} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_4^x {\frac{1}{{\sqrt t }}dt} } \right],\,\,\,\,\left\{ {{\text{With }}f\left( t \right) = \frac{1}{{\sqrt t }}} \right\} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = f\left( x \right) = \frac{1}{{\sqrt x }} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{1}{{\sqrt x }} \cr
& \cr
& \left( {\text{b}} \right){\text{Integrating}} \cr
& \,\,\,\,\,\,\,F\left( x \right) = \int_4^x {\frac{1}{{\sqrt t }}dt} \cr
& \,\,\,\,\,\,\,F\left( x \right) = \left[ {2\sqrt t } \right]_4^x \cr
& \,\,\,\,\,\,\,F\left( x \right) = 2\sqrt x - 2\sqrt 4 \cr
& \,\,\,\,\,\,\,F\left( x \right) = 2\sqrt x - 4 \cr
& {\text{Differentiating}} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{d}{{dx}}\left[ {2\sqrt x - 4} \right] \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = 2\left( {\frac{1}{{2\sqrt x }}} \right) - 0 \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{1}{{\sqrt x }} \cr} $$
