Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 84

Prove Quotient Rule by using Chain Rule and Product Rule

Recall that $\displaystyle \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2}$ for Quotient Rule

We can rewrite $\displaystyle \frac{f(x)}{g(x)}$ as $f(x) [g(x)]^{-1}$ as $f(x) [g(x)]^{-1}$ and by that

We can use Product and Chain Rule. So,


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (f(x) [g(x)]^{-1}) =& \frac{d}{dx} f(x) \cdot [g(x)]^{-1} + f(x) \cdot \frac{d}{dx} [g(x)]^{-1}
\\
\\
\frac{d}{dx} (f(x) [g(x)]^{-1}) =& f'9x) \cdot [g(x)]^{-1} + f(x) \cdot (-1)[g(x)]^{-2} [g'(x)]
\\
\\
\frac{d}{dx} (f(x) [g(x)]^{-1}) =& \frac{f'(x)}{g(x)} + \frac{f(x) (-g'(x))}{[g(x)]^2}

\end{aligned}
\end{equation}
$


By getting the LCD we will be able to prove Quotient Rule as

$\displaystyle \frac{d}{dx} (f(x) [g(x)]^{-1}) = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}$