Answer
$$\frac{{dy}}{{dx}} = \sqrt {2 + {{\cos }^3}x} $$
Work Step by Step
$$\eqalign{
& y = \int_2^x {\sqrt {2 + {{\cos }^3}t} } dt \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\int_2^x {\sqrt {2 + {{\cos }^3}t} } dt} \right] \cr
& {\text{Use the fundamental theorem of calculus part 1 }}\cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& \frac{{dy}}{{dx}} = \sqrt {2 + {{\cos }^3}\left( x \right)} \cr
& \frac{{dy}}{{dx}} = \sqrt {2 + {{\cos }^3}x} \cr} $$
