g ( x ) = ∫ 1 − 3 x 1 u 3 ( 1 + u 2 ) d u {\displaystyle g(x)=\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du}
d d x ( g ( x ) ) = d d x ( ∫ 1 − 3 x 1 u 3 ( 1 + u 2 ) d u ) = ( 0 ) ⋅ ( 1 − 3 x ) 3 ( 1 + ( 1 − 3 x ) 2 ) − ( − 3 ) ⋅ ( − 3 ) 3 ( 1 + ( − 3 ) 2 ) {\displaystyle {\frac {d}{dx}}(g(x))={\frac {d}{dx}}\left(\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du\right)=(0)\cdot {\frac {(1-3x)^{3}}{(1+(1-3x)^{2})}}-(-3)\cdot {\frac {(-3)^{3}}{(1+(-3)^{2})}}}
Therefore, g ′ ( x ) = y 2 sin ( y ) {\displaystyle {\text{Therefore, }}g'(x)=y^{2}\sin {(y)}}
