Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 97

Differentiate $f=Fg$ using the Product Rule and solve the resulting equation for $F$
and complete the results of $F'$ using Quotient Rule for $\displaystyle F= \frac{f}{g}$.

Using Product Rule,
$f = Fg$
$f' = F'g+Fg''$

$\begin{equation} \begin{aligned} \frac{F'\cancel{g}}{\cancel{g}} &= \frac{f'-Fg'}{g}\\ F' &= \frac{f'-Fg'}{g} ; \text{ but } F = \frac{f}{g}\\ F' &= \frac{f' - \left(\frac{f}{g}\right) (g')}{g}\\ F' &= \frac{f'g-fg'}{g^2} \end{aligned} \end{equation}$


Using Quotient Rule,

$\begin{equation} \begin{aligned} F &= \frac{f}{g}\\ F' &= \frac{f'g-fg'}{g^2} \end{aligned} \end{equation}$


It shows that the results agree by using either of the two rules.