Answer
$$\frac{{dy}}{{dx}} = - \csc x $$
Work Step by Step
$$\eqalign{
& y = \int_{\sec x}^2 {\frac{1}{{{t^2} + 1}}} dt \cr
& {\text{Use the integral property }}\cr
&\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr
& y = - \int_2^{\sec x} {\frac{1}{{{t^2} + 1}}} dt \cr
& {\text{Use the fundamental theorem of calculus part 1 }}\cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& {\text{and the chain rule}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {\sec x} \right)}^2} + 1}}\frac{d}{{dx}}\left[ {\sec x} \right] \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\sec }^2}x + 1}}\left( {\sec x\tan x} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\tan }^2}x}}\left( {\sec x\tan x} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{\sec x}}{{\tan x}} \cr
& \frac{{dy}}{{dx}} = - \left( {\frac{1}{{\cos x}}} \right)\left( {\frac{{\cos x}}{{\sin x}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sin x}} \cr
& \frac{{dy}}{{dx}} = - \csc x \cr} $$
