Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 63

Given that $f(5) = 1$, $f'(5) = 6$, $g(5) = -3$ and $g'(5) = 2$. Find the following values

$
\begin{equation}
\begin{aligned}
& \text{a.) } (fg)'(5) && \text{b.) } \left( \frac{f}{g} \right)'(5)\\
& \text{c.) } \left( \frac{g}{f} \right)'(5)\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{a.) } (fg)'(5) & = f(5) g'(5) + g(5) f'(5)\\
\\
& = 1(2) + -3(6)\\
\\
& = -16
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{b.) } \left( \frac{f}{g} \right)'(5) &= \frac{g(5) f'(5) - f(5) g'(5)}{[g(5)]^2}\\
\\
&= \frac{-3(6)-1(2)}{(-3)^2}
\\
&= \frac{-20}{9}
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{c.) } \left( \frac{g}{f} \right)'(5) &= \frac{f(5)g'(5)-g(5)f'(5)}{[f(5)]^2}\\
\\
&= \frac{1(2)-(-3)(6)}{(1)^2}\\
\\
&= 20
\end{aligned}
\end{equation}
$