Answer
$$\frac{{dy}}{{dx}} = 14x\sqrt {2 + {{\cos }^3}\left( {7{x^2}} \right)} $$
Work Step by Step
$$\eqalign{
& y = \int_2^{7{x^2}} {\sqrt {2 + {{\cos }^3}t} } dt \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\int_2^{7{x^2}} {\sqrt {2 + {{\cos }^3}t} } dt} \right] \cr
& {\text{Use the fundamental theorem of calculus part 1}}\cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& {\text{and the chain rule}} \cr
& \frac{{dy}}{{dx}} = \sqrt {2 + {{\cos }^3}\left( {7{x^2}} \right)} \frac{d}{{dx}}\left[ {7{x^2}} \right] \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \sqrt {2 + {{\cos }^3}\left( {7{x^2}} \right)} \left( {14x} \right) \cr
& \frac{{dy}}{{dx}} = 14x\sqrt {2 + {{\cos }^3}\left( {7{x^2}} \right)} \cr} $$
