\(\bf g(x)=x^2\int\limits_{3}^{x}[4f(s)-9]ds\\\\
-----------------------------\\\\
\textit{let us use the product rule to get g'(x)}
\\\\
\cfrac{dg}{dx}\implies 2x\left[ \int\limits_{3}^{x}[4f(s)-9]ds \right]\quad +\quad x^2[4f(x)-9]
\\\\
g'(8)\implies
\begin{array}{llll}
2(8)\left[ \int\limits_{3}^{8}[4f(s)-9]ds \right]\quad +\quad (8)^2[4f(8)-9]
\\\\
2(8)\left[ 4\underline{\int\limits_{3}^{8}f(s)ds} - \int\limits_{3}^{8} 9\cdot ds \right]\quad +\quad (8)^2[4\underline{f(8)}-9]
\end{array}\)
\(\bf \textit{now, we know that}\implies
\begin{cases}
f(8)=5
\\\\
\int\limits_{3}^{8}f(s)ds=3
\end{cases}\\\\
-----------------------------\\\\
g'(8)\implies 2(8)\left[ 4\cdot \underline{3} - \left[\cfrac{}{} 9s \right]_3^8\right]\quad +\quad (8)^2[4\cdot \underline{5}-9]\)
now, apply the bounds to \(\bf \left[\cfrac{}{} 9s \right]_3^8\) to get the definite integral value, and simplify away
