Answer
$g'\left( w \right) = \sin \left( {1 + {w^3}} \right)$
Work Step by Step
$$\eqalign{
& g\left( w \right) = \int_0^w {\sin \left( {1 + {t^3}} \right)} dt \cr
& {\text{Differentiate both sides with respect to }}w \cr
& g'\left( w \right) = \frac{d}{{dw}}\left[ {\int_a^w {\sin \left( {1 + {t^3}} \right)} dt} \right] \cr
& {\text{Use Part 1 of the Fundamental Theorem of calculus}} \cr
& g\left( w \right) = \int_0^w {f\left( t \right)} dt,{\text{ }}a \leqslant w \leqslant b,{\text{ }} \Rightarrow {\text{ }}g'\left( w \right) = f\left( w \right) \cr
& {\text{Let }}f\left( t \right) = \sin \left( {1 + {t^3}} \right) \cr
& f\left( w \right) = \sin \left( {1 + {{\left( w \right)}^3}} \right) \cr
& f\left( w \right) = \sin \left( {1 + {w^3}} \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& g'\left( w \right) = \sin \left( {1 + {w^3}} \right) \cr} $$
