6.2 Base Functions

Now that we are familiar with negative exponents and fractional exponents, let's take another look at some familiar functions.

function	power	exponent
\( p = b^2 \)	\( p = b^2 \)	\( 2 \)
\( p = b^3 \)	\( p = b^3 \)	\( 3 \)
\( p = \frac{1}{b} \)	\( p = b^{-1} \)	\( -1 \)
\( p = \sqrt[2]{b} \)	\( p = b^{1/2} \)	\( 1/2 \)
\( p = \sqrt[3]{b} \)	\( p = b^{1/3} \)	\( 1/3 \)

Each of these functions is an example of a base function!

Definition.

A base function is a power

\[ \begin{align} p &\depends b \\ p &= b^e \end{align} \]

where the base, \( b \), is allowed to vary and the exponent, \( e \), is constant.

Let's take a look at the differential law for base functions.

Law.

The differential law for base functions states

\[ \begin{align} \diff \bigl(v^c\bigr) &= c \mult v^{c - 1} \mult \diff v \end{align} \]

where \( c : \Number \) is a constant number and the base \( v : \Number \) is a positive number, \( v \gt 0 \).

We won't check the differential base law, but we have already seen significant evidence for it! The differential laws for squaring, cubing, reciprocals, square roots, and cube roots are all special cases of the differential law for base functions.

differential law	differential base law
\( {\displaystyle \diff \bigl(v^2\bigr) = 2 v \mult \diff v } \)	\( {\displaystyle \diff \bigl(v^2\bigr) = 2 v^1 \mult \diff v } \)
\( {\displaystyle \diff \bigl(v^3\bigr) = 3 v^2 \mult \diff v } \)	\( {\displaystyle \diff \bigl(v^3\bigr) = 3 v^2 \mult \diff v } \)
\( {\displaystyle \diff \biggl(\frac{1}{v}\biggr) = \frac{-1}{v^2} \mult \diff v } \)	\( {\displaystyle \diff \bigl(v^{-1}\bigr) = - v^{-2} \mult \diff v } \)
\( {\displaystyle \diff \bigl(\sqrt[2]{v}\bigr) = \frac{1}{2 \mult \sqrt[2]{v}} \mult \diff v } \)	\( {\displaystyle \diff \bigl(v^{1/2}\bigr) = \frac{1}{2} \mult v^{-1/2} \mult \diff v } \)
\( {\displaystyle \diff \bigl(\sqrt[3]{v}\bigr) = \frac{1}{3 \mult \bigl(\sqrt[3]{v}\bigr)^2} \mult \diff v } \)	\( {\displaystyle \diff \bigl(v^{1/3}\bigr) = \frac{1}{3} \mult v^{-2/3} \mult \diff v } \)

The differential law for base functions packs quite a punch!

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \frac{x^4}{2} + \sqrt[2]{3 x} \end{align} \]

defined for positive inputs, \( x \gt 0 \). Let's take the derivative.

\[ \begin{align} \overset{x}{\diff} z &= \overset{x}{\diff} \biggl(\frac{x^4}{2} + \sqrt[2]{3 x}\biggr) \\ &= \overset{x}{\diff} \biggl(\frac{x^4}{2}\biggr) + \overset{x}{\diff} \bigl(\sqrt[2]{3 x}\bigr) \\ &= \frac{1}{2} \mult \overset{x}{\diff} \bigl(x^4\bigr) + \overset{x}{\diff} \bigl((3 x)^{1/2}\bigr) \\ &= \frac{1}{2} \mult 4 x^3 \mult \overset{x}{\diff} x + \frac{1}{2} \mult (3 x)^{-1/2} \mult \overset{x}{\diff} (3 x) \\ &= 2 x^3 + \frac{1}{2} \mult \frac{1}{\sqrt[2]{3 x}} \mult 3 \mult \overset{x}{\diff} x \\ &= 2 x^3 + \frac{3}{2 \mult \sqrt[2]{3 x}} \end{align} \]

Let's use the differential law for base functions to calculate two derivatives of the cube root.

Example.

We can write the cube root as a base function.

\[ \begin{align} z &\depends x \\ z &= x^{1/3} \end{align} \]

Here's the first derivative

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

and here's the second.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} z &= \overset{x}{\diff} \biggl(\frac{1}{3} \mult x^{-2/3}\biggr) \\ &= \frac{1}{3} \mult \overset{x}{\diff} \bigl(x^{-2/3}\bigr) \\ &= \frac{1}{3} \mult \frac{-2}{3} \mult x^{-5/3} \mult \overset{x}{\diff} x \\ &= \frac{-2}{9} \mult x^{-5/3} \end{align} \]

Using the differential law for base functions made this calculation much easier.

As wonderful as the differential base law is, we still have a long way to go to understand powers! We would like to be able to take the differential of any power

\[ \begin{align} &\diff \bigl(v^u\bigr) \end{align} \]

where both the exponent \( u : \Number \) and the base \( v : \Number \) are allowed to vary. In order to get there, we'll need to study exponential functions.