Answer
$$\frac{{dy}}{{dx}} = 2{x^2}\sin {x^6} + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt $$
Work Step by Step
$$\eqalign{
& y = x\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\frac{d}{{dx}}\left[ x \right] \cr
& {\text{use the fundamental theorem of calculus part 1 with the chain rule }} \cr
& {\text{to }}\frac{d}{{dx}}\left[ {\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] \cr
& \cr
& \frac{{dy}}{{dx}} = x\left( {\sin {{\left( {{x^2}} \right)}^3}} \right)\frac{d}{{dx}}\left( {{x^2}} \right) + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\frac{d}{{dx}}\left[ x \right] \cr
& {\text{compute the derivatives and simplify}} \cr
& \frac{{dy}}{{dx}} = x\left( {\sin {{\left( {{x^2}} \right)}^3}} \right)\left( {2x} \right) + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\left( 1 \right) \cr
& \frac{{dy}}{{dx}} = 2{x^2}\sin {x^6} + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt \cr} $$
