## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = 3{\left( {{x^3} + 1} \right)^{10}}{\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)^2}$$
\eqalign{ & y = {\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)^3} \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)}^3}} \right] \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dx}} = 3{\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)^2}\frac{d}{{dx}}\left[ {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}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}} = 3{\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)^2}{\left( {{x^3} + 1} \right)^{10}} \cr & \frac{{dy}}{{dx}} = 3{\left( {{x^3} + 1} \right)^{10}}{\left( {\int_0^x {{{\left( {{t^3} + 1} \right)}^{10}}dt} } \right)^2} \cr}