# Quotient rule in calculus

## 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.

$$f'(x) = \frac{g'(x)}{j(x)} - \frac{g(x) * j'(x)}{j(x)^{2}}$$

$$f'(x) = \frac{g'(x) * j(x)}{j(x)^2} - \frac{g(x) * j'(x)}{j(x)^{2}}$$

$$f'(x) = \frac{g'(x) * j(x) - g(x) * j'(x)}{j(x)^{2}}$$