# Product rule in calculus

## Statement

The derivative of a product $g(x) * j(x)$ is $g'(x) * j(x) + g(x) * j'(x)$.

$$\tfrac{d}{dx}\bigg(g(x) * j(x)\bigg) = g'(x) * j(x) + g(x) * j'(x)$$

## Proof

Define function $f$.

$$f(x) = g(x) * j(x)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{f(x + h) - f(x)}{h}\right)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{g(x + h) * j(x + h) - g(x) * j(x)}{h}\right)$$

Now subtract and add $g(x) * j(x + h)$.

$$f'(x) = \lim_{h \to 0}\left(\frac{g(x + h) * j(x + h) - \overbrace{ g(x) * j(x + h) }^\text{subtract} + \overbrace{ g(x) * j(x + h) }^\text{add} - g(x) * j(x)}{h}\right)$$

Split the limit.

$$f'(x) = \lim_{h \to 0}\left(\frac{g(x + h) * j(x + h) - g(x) * j(x + h)}{h}\right) + \lim_{h \to 0}\left(\frac{g(x) * j(x + h) - g(x) * j(x)}{h}\right)$$

Factor out $j(x + h)$ in the first limit, and factor out $g(x)$ in the second. Then move those factors out of the fraction.

$$f'(x) = \lim_{h \to 0}\left(\frac{j(x + h) * \bigg(g(x + h) - g(x)\bigg)}{h}\right) + \lim_{h \to 0}\left(\frac{g(x) * \bigg(j(x + h) - j(x)\bigg)}{h}\right)$$

$$f'(x) = \lim_{h \to 0}\left(\frac{g(x + h) - g(x)}{h} * j(x + h) \right) + \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h} * g(x) \right)$$

Now split the limits again.

$$f'(x) = \lim_{h \to 0}\left(\frac{g(x + h) - g(x)}{h}\right) * \lim_{h \to 0}\bigg(j(x + h)\bigg) + \lim_{h \to 0}\left(\frac{j(x + h) - j(x)}{h}\right) * \lim_{h \to 0}\bigg(g(x)\bigg)$$

Notice:

• The first and third limit are the definition of the derivative for $g(x)$ and $j(x)$.
• In the second limit, $h$ goes to $0$, so $j(x + h)$ goes to $j(x + 0) = j(x)$.
• There is no $h$ in the fourth limit, so this limit just becomes $g(x)$.

$$f'(x) = g'(x) * j(x) + j'(x) * g(x)$$

## Proofs building upon this proof

### Quotient rule in calculus

This proof shows that the derivative for the quotient or fraction a/b is (a'b - ab') / b^2.