5.3 The Fundamental Theorem of Calculus/17

$y=\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}du$

${\frac {d}{dx}}(y)={\frac {d}{dx}}\left(\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}\,du\right)=(0)\cdot {\frac {(1)^{3}}{1+(1)^{2}}}-(-3)\cdot {\frac {(1-3x)^{3}}{1+(1-3x)^{2}}}$

${\text{Therefore, }}y'={\frac {3(1-3x)^{3}}{1+(1-3x)^{2}}}$