Answer
$$\frac{{dy}}{{dx}} = - \frac{6}{{3 + {x^4}}}$$
Work Step by Step
$$\eqalign{
& y = \int_x^1 {\frac{6}{{3 + {t^4}}}} dt \cr
& {\text{Use the integral property }}\cr
&\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr
& y = - \int_1^x {\frac{6}{{3 + {t^4}}}} dt \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{dx}}\left[ {\int_1^x {\frac{6}{{3 + {t^4}}}} dt} \right] \cr
& {\text{Use the fundamental theorem of calculus part 1 }}\cr
&\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& \frac{{dy}}{{dx}} = - \left( {\frac{6}{{3 + {{\left( x \right)}^4}}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{6}{{3 + {x^4}}} \cr} $$
