Answer
$$\frac{{dy}}{{dx}} = 0$$
Work Step by Step
$$\eqalign{
& y = \int_{ - 1}^x {\frac{{{t^2}}}{{{t^2} + 4}}} dt - \int_3^x {\frac{{{t^2}}}{{{t^2} + 4}}} dt \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\int_{ - 1}^x {\frac{{{t^2}}}{{{t^2} + 4}}} dt} \right) - \frac{d}{{dx}}\left( {\int_3^x {\frac{{{t^2}}}{{{t^2} + 4}}} dt} \right) \cr
& {\text{use the fundamental theorem of calculus part}} 1 \cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& \frac{{dy}}{{dx}} = \frac{{{{\left( x \right)}^2}}}{{{{\left( x \right)}^2} + 4}} - \frac{{{{\left( x \right)}^2}}}{{{{\left( x \right)}^2} + 4}} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = 0 \cr} $$
