4.1 Derivatives

In Chapter 3. Differential Laws, we studied the differential, \( \diff \), as an abstract operator. We'll be using these differential laws to define several operators, the first of which is the derivative. The derivative is defined for functions that have a single input variable \( x \). We'll return to studying functions with two input variables in Part II and Part III of this text.

Definition.

The \( x \)-derivative is an operator \( \overset{x}{\diff} \) that satisfies all abstract differential laws. Further, the \( x \)-derivative of the variable \( x \) is one.

\[ \begin{align} \overset{x}{\diff} x &= 1 \end{align} \]

This is known as the unit law for the derivative.

We'll pronounce the notation \( \overset{x}{\diff} z \) as the "\( x \)-derivative of \( z \)."

Example.

Let's take the derivative of the function

\[ \begin{align} z &\depends x \\ z &= x^3 - 2 x. \end{align} \]

We can use all of the differential laws we've studied when calculating the derivative.

\[ \begin{align} \overset{x}{\diff} z &= \overset{x}{\diff} \bigl(x^3 - 2 x\bigr) \\ &= \overset{x}{\diff} \bigl(x^3\bigr) - \overset{x}{\diff} (2 x) \\ &= 3 x^2 \mult \overset{x}{\diff} x - 2 \mult \overset{x}{\diff} x \\ &= 3 x^2 \mult 1 - 2 \mult 1 \\ &= 3 x^2 - 2 \end{align} \]

We've found the derivative!

The unit law allows us to continue our calculation slightly further with the derivative than we could with the differential. As a consequence, our answer does not include the differential variable \( \diff x \).

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \sqrt[2]{x^2 + 1}. \end{align} \]

Let's compute the \( x \)-derivative of \( z \).

\[ \begin{align} \overset{x}{\diff} z &= \overset{x}{\diff} \Bigl(\sqrt[2]{x^2 + 1}\Bigr) \\ &= \frac{1}{2 \mult \sqrt[2]{x^2 + 1}} \mult \overset{x}{\diff} \bigl(x^2 + 1\bigr) \\ &= \frac{1}{2 \mult \sqrt[2]{x^2 + 1}} \mult \overset{x}{\diff} \bigl(x^2\bigr) \\ &= \frac{1}{2 \mult \sqrt[2]{x^2 + 1}} \mult 2 x \mult \overset{x}{\diff} x \\ &= \frac{x}{\sqrt[2]{x^2 + 1}} \end{align} \]

We applied the unit law in the last step of this calculation.

The Derivative and the Differential

It is important that we maintain a distinction between the derivative and the differential: there is no "unit law for the differential."

\[ \begin{align} \diff x &= 1 & &\text{No!} \\ \overset{x}{\diff} x &= 1 & &\text{Yes!} \end{align} \]

That being said, it is easy to convert between the derivative and the differential.

Important.

A function \( z \depends x \) has \( x \)-derivative

\[ \begin{align} \overset{x}{\diff} z &= s \end{align} \]

if and only if the function has differential

\[ \begin{align} \diff z &= s \mult \diff x. \end{align} \]

If we know a function's derivative, we can recover its differential.

Example.

If a function \( z \depends x \) has \( x \)-derivative

\[ \begin{align} \overset{x}{\diff} z &= x^3 - \frac{4}{x} \end{align} \]

then the differential of \( z \) must be

\[ \begin{align} \diff z &= \biggl(x^3 - \frac{4}{x}\biggr) \mult \diff x. \end{align} \]

Converting a differential to a derivative is just as easy.