Statement
The derivative for quotients or fractions is the following.
$$ \frac{d}{dx}\left(\frac{g(x)}{j(x)}\right) = \frac{g'(x) * j(x) - g(x) * j'(x)}{j(x)^2} $$
Proof
Define function $ f $ and write it as a product.
$$ f(x) = \frac{g(x)}{j(x)} $$
$$ f(x) = g(x) * j(x)^{-1} $$
Differentiate $ f $ using the product rule.
$$ f'(x) = \tfrac{d}{dx}\bigg(g(x)\bigg) * j(x)^{-1} + g(x) * \tfrac{d}{dx}\bigg(j(x)^{-1}\bigg) $$
$$ f'(x) = g'(x) * j(x)^{-1} + g(x) * \tfrac{d}{dx}\bigg(j(x)^{-1}\bigg) $$
Differentiate $ \tfrac{d}{dx}\bigg(j(x)^{-1}\bigg) $ using the power rule and the chain rule.
$$ f'(x) = g'(x) * j(x)^{-1} + g(x) * -1 * j(x)^{-2} * \tfrac{d}{dx}\bigg(j(x)\bigg) $$
$$ f'(x) = g'(x) * j(x)^{-1} - g(x) * j(x)^{-2} * j'(x) $$
Write negative exponents as fractions and make a common denominator.
$$ f'(x) = \frac{g'(x)}{j(x)} - \frac{g(x) * j'(x)}{j(x)^{2}} $$
$$ f'(x) = \frac{g'(x) * j(x)}{j(x)^2} - \frac{g(x) * j'(x)}{j(x)^{2}} $$
$$ f'(x) = \frac{g'(x) * j(x) - g(x) * j'(x)}{j(x)^{2}} $$