3.4 Division

Let's see how to take the differential of a reciprocal.

Law.

The differential law for reciprocals states

\[ \begin{align} \diff \biggl(\frac{1}{v}\biggr) &= \frac{- 1}{v^2} \mult \diff v \end{align} \]

where \( v : \Number \) is a non-zero number, \( v \neq 0 \).

Why?

We can make sense of the reciprocal using the equation

\[ \begin{align} \frac{1}{v} \mult v &= 1. \end{align} \]

Let's take the differential of both sides of this equation.

\[ \begin{align} &\diff \biggl(\frac{1}{v} \mult v\biggr) = \diff (1) \\ &v \mult \diff \biggl(\frac{1}{v}\biggr) + \frac{1}{v} \mult \diff v = 0 \\ &v \mult \diff \biggl(\frac{1}{v}\biggr) = \frac{- 1}{v} \mult \diff v \\ &\diff \biggl(\frac{1}{v}\biggr) = \frac{- 1}{v^2} \mult \diff v \end{align} \]

Voila!

We'll use the differential law for the reciprocal to explain the differential law for division.

Law.

The differential law for division states

\[ \begin{align} \diff \biggl(\frac{u}{v}\biggr) &= \frac{1}{v} \mult \diff u - \frac{u}{v^2} \mult \diff v \end{align} \]

where \( u : \Number \) is a number and \( v : \Number \) is non-zero, \( v \neq 0 \).

Why?

We can rewrite any division as a multiplication with a reciprocal.

\[ \begin{align} \diff \biggl(\frac{u}{v}\biggr) &= \diff \biggl(u \mult \frac{1}{v}\biggr) \\ &= \frac{1}{v} \mult \diff u + u \mult \diff \biggl(\frac{1}{v}\biggr) \\ &= \frac{1}{v} \mult \diff u + u \mult \frac{-1}{v^2} \mult \diff v \\ &= \frac{1}{v} \mult \diff u - \frac{u}{v^2} \mult \diff v \end{align} \]

Let's have an example that uses the differential law for division.

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \frac{x^2}{x + 5}. \end{align} \]

We disallow the input \( x = -5 \) to avoid dividing by zero. Let's compute the differential.

\[ \begin{align} \diff z &= \diff \biggl(\frac{x^2}{x + 5}\biggr) \\ &= \frac{1}{x + 5} \mult \diff \bigl(x^2\bigr) - \frac{x^2}{(x + 5)^2} \mult \diff (x + 5) \\ &= \frac{1}{x + 5} \mult 2 x \mult \diff x - \frac{x^2}{(x + 5)^2} \mult \diff x \\ &= \biggl(\frac{2 x}{x + 5} - \frac{x^2}{(x + 5)^2}\biggr) \mult \diff x \end{align} \]

We've written our answer in standard form.