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 ) ∗ f ( 1 ) − ( − 3 ) ∗ f ( 1 − 3 x ) {\displaystyle {\frac {d}{dx}}(g(x))={\frac {d}{dx}}\left(\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du\right)=(0)*f(1)-(-3)*f(1-3x)}
which is equal to ( 3 ) ∗ f ( 1 − 3 x ) {\displaystyle (3)*f(1-3x)}
which is= 3 ∗ ( 1 − 3 x ) 3 ( 1 + ( 1 − 3 x ) 2 ) {\displaystyle 3*{\frac {(1-3x)^{3}}{(1+(1-3x)^{2})}}}
or simplified to 3 ∗ ( 1 − 3 x ) 3 ( 1 + ( 1 − 3 x ) 2 ) {\displaystyle {\frac {3*(1-3x)^{3}}{(1+(1-3x)^{2})}}}
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