6.5 Simple Differential Laws

The natural exponential and the natural logarithmic functions have simple differential laws.

Law.

The differential law for the natural exponential function states

\[ \begin{align} \diff (\exp u) &= \exp u \mult \diff u \end{align} \]

where \( u : \Number \) is a number.

If you'd like to see why this differential rule holds, you can read Appendix B. The Natural Exponential which explores the natural exponential in much more detail. This appendix also helps explain our curious definition for the natural base, \( \naturalbase \).

Law.

The differential law for the natural logarithm states

\[ \begin{align} \diff (\log v) &= \frac{1}{v} \mult \diff v \end{align} \]

where \( v : \Number \) is any positive number, \( v \gt 0 \).

Why?

We'll use the following variables to solve for our differential law.

\[ \begin{align} u &\depends v & & & v &\depends u \\ u &= \log v & &\andSpaced & v &= \exp u \end{align} \]

We find the logarithmic law by rewriting the corresponding law for the natural exponential.

\[ \begin{align} &\diff (\exp u) = \exp u \mult \diff u \\ &\diff v = v \mult \diff (\log v) \\ &\frac{1}{v} \mult \diff v = \diff (\log v) \end{align} \]

We've found it!

Let's use our differential law for the natural exponential to make a calculation.

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \exp (3 x - 1). \end{align} \]

We'll take the \( x \)-derivative.

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

Let's also see an example that uses a natural logarithm.

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \log \bigl(x^2 + 1\bigr). \end{align} \]

Let's compute the derivative.

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