Answer
$$g'\left( x \right) = 3{x^8} - 3{x^5}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \int_1^{{x^3}} {\left( {{t^2} - t} \right)} dt \cr
& {\text{Differentiate }} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\int_1^{{x^3}} {\left( {{t^2} - t} \right)} dt} \right] \cr
& {\text{Using the formula }}\left( {18} \right){\text{ }}\frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)} dt} \right] = f\left( {g\left( x \right)} \right)g'\left( x \right) \cr
& g'\left( x \right) = \left( {{{\left( {{x^3}} \right)}^2} - \left( {{x^3}} \right)} \right)\left( {3{x^2}} \right) \cr
& g'\left( x \right) = \left( {{x^6} - {x^3}} \right)\left( {3{x^2}} \right) \cr
& g'\left( x \right) = 3{x^8} - 3{x^5} \cr
& \cr
& {\text{Evaluating the integral}} \cr
& g\left( x \right) = \int_1^{{x^3}} {\left( {{t^2} - t} \right)} dt = \left[ {\frac{{{t^3}}}{3} - \frac{{{t^2}}}{2}} \right]_1^{{x^3}} \cr
& g\left( x \right) = \left( {\frac{{{{\left( {{x^3}} \right)}^3}}}{3} - \frac{{{{\left( {{x^3}} \right)}^2}}}{2}} \right) - \left( {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^2}}}{2}} \right) \cr
& g\left( x \right) = \frac{{{x^9}}}{3} - \frac{{{x^6}}}{2} + \frac{1}{6} \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = \frac{{9{x^8}}}{3} - \frac{{6{x^5}}}{2} + 0 \cr
& g'\left( x \right) = 3{x^8} - 3{x^5} \cr} $$
