Answer
\[
\left(f^{\prime} g-f g^{\prime}\right) / g g=F^{\prime}
\]
Work Step by Step
\[
\begin{array}{c}
F g =f\\
g F^{\prime}+F g^{\prime}=f^{\prime}
\end{array}
\]
we used product rule \(-(\mathrm{AB})^{\prime}=\mathrm{AB}^{\prime}+\mathrm{BA}^{\prime}\)
\[
g F^{\prime}=f^{\prime}-F g^{\prime}
\]
subtract both sides by \(\mathrm{F} \mathrm{g}\) '
\[
f^{\prime}-F g^{\prime}=g F^{\prime}
\]
\[
F^{\prime}=\left(f^{\prime}-F g^{\prime}\right) / g
\]
divide both sides by g
\[
F^{\prime}=\left(f^{\prime}-(f / g) g^{\prime}\right) / g
\]
\(\mathrm{f}=\mathrm{Fg}\) dividing both sides by g we get \(\mathrm{F}=\mathrm{f} / \mathrm{g}\)
\[
\begin{array}{l}
F^{\prime}=\left(f^{\prime}-\left(f g^{\prime} / g\right)\right) / g \\
F^{\prime}=\left(\left(f^{\prime} g-f g^{\prime}\right) / g\right) / g
\end{array}
\]
I have trouble writing g square but gg means g square
\[
\left(f^{\prime} g-f g^{\prime}\right) / g g=F^{\prime}
\]
