# Discrete Calculus: Derivative Rules

June 17, 2020 · 18:42 pm · Discrete CalculusMath

Now that we know what discrete derivates are, let’s see if we can derive (no pun intended) similar rules found in regular calculus.

## Product Rule

\begin{aligned} \Delta_h [fg](n) &= \frac{[fg](n+h) - [fg](n)}{h} \\ &= \frac{f(n+h)g(n+h) - f(n)g(n)}{h} \\ &= \frac{f(n+h)g(n+h) - f(n+h)g(n) + f(n+h)g(n) - f(n)g(n)}{h} \\ &= \frac{f(n+h)(g(n+h) - g(n)) + g(n)(f(n+h)-f(n))}{h} \\ &= f(n+h) \frac{g(n+h) - g(n)}{h} + g(n) \frac{f(n+h) - f(n)}{h} \\ &= f(n+h) \Delta_h g(n) + g(n) \Delta_h f(n) \\ \end{aligned}

## Quotient Rule

\begin{aligned} \Delta_h \left[\frac{f}{g}\right](n) &= \frac{\left[\frac{f}{g}\right](n+h) - \left[\frac{f}{g}\right](n)}{h} \\ &= \frac{\frac{f(n+h)}{g(n+h)} - \frac{f(n)}{g(n)}}{h} \\ &= \frac{1}{h} \frac{f(n+h)g(n) - f(n)g(n+h)}{g(n+h)g(n)} \\ &= \frac{1}{g(n+h)g(n)} \frac{f(n+h)g(n) - f(n)g(n) + f(n)g(n) - f(n)g(n+h)}{h} \\ &= \frac{1}{g(n+h)g(n)} \frac{g(n)(f(n+h) - f(n)) + f(n)(g(n) - g(n+h))}{h} \\ &= \frac{1}{g(n+h)g(n)} \left(g(n) \frac{f(n+h) - f(n)}{h} - f(n) \frac{g(n+h) - g(n)}{h}\right) \\ &= \frac{g(n) \Delta_h f(n) - f(n) \Delta_h g(n)}{g(n+h)g(n)} \\ \end{aligned}

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