##### Theorem2.10.1

\begin{align*} \diff{}{x} \log x &= \frac{1}{x} \end{align*}where \(\log x\) is the logarithm base \(e\text{.}\)

The chain rule opens the way to understanding derivatives of more complicated function. Not only compositions of known functions as we have seen the examples of the previous section, but also functions which are defined implicitly.

Consider the logarithm base \(e\) — \(\log_e(x)\) is the power that \(e\) must be raised to to give \(x\text{.}\) That is, \(\log_e(x)\) is defined by

\begin{align*} e^{\log_e x} &= x \end{align*}i.e. — it is the inverse of the exponential function with base \(e\text{.}\) Since this choice of base works so cleanly and easily with respect to differentiation, this base turns out to be (arguably) the most natural choice for the base of the logarithm. And as we saw in our whirlwind review of logarithms in Section 2.7, it is easy to use logarithms of one base to compute logarithms with another base:

\begin{align*} \log_q x &= \frac{\log_e x}{\log_e q} \end{align*}So we are (relatively) free to choose a base which is convenient for our purposes.

The logarithm with base \(e\text{,}\) is called the “natural logarithm”. The “naturalness” of logarithms base \(e\) is exactly that this choice of base works very nicely in calculus (and so wider mathematics) in ways that other bases do not ^{ 1 } . There are several different “standard” notations for the logarithm base \(e\text{;}\)

We recommend that you be able to recognise all of these.

In this text we will write the natural logarithm as “\(\log\)” with no base. The reason for this choice is that base \(e\) is the standard choice of base for logarithms in mathematics ^{ 2 }

The natural logarithm inherits many properties of general logarithms ^{ 3 } . So, for all \(x,y \gt 0\) the following hold:

- \(e^{\log x}=x\text{,}\)
- for any real number \(X\text{,}\) \(\log \big(e^X\big)=X\text{,}\)
- for any \(a \gt 1\text{,}\) \(\log_a x=\tfrac{\log x}{\log a}\) and \(\log x=\tfrac{\log_a x}{\log_a e}\)
- \(\log 1=0\text{,}\) \(\log e=1\)
- \(\log(xy)=\log x+\log y\)
- \(\log\big(\tfrac{x}{y}\big)=\log x-\log y\text{,}\) \(\log\big(\tfrac{1}{y}\big)=-\log y\)
- \(\log(x^X)=X\log x\)
- \(\lim\limits_{x\rightarrow\infty}\log x=\infty\text{,}\) \(\lim\limits_{x\rightarrow0}\log x=-\infty\)

And finally we should remember that \(\log x\) has domain (i.e. is defined for) \(x \gt 0\) and range (i.e. takes all values in) \(-\infty \lt x \lt \infty\text{.}\)

To compute the derivative of \(\log x\) we could attempt to start with the limit definition of the derivative

\begin{align*} \diff{}{x}\log x &= \lim_{h \to 0} \frac{\log(x+h) - \log(x)}{h}\\ &= \lim_{h\to 0} \frac{\log( (x+h)/x )}{h} \\ &= \text{um\dots} \end{align*}This doesn't look good. But all is not lost — we have the chain rule, and we know that the logarithm satisfies the equation:

\begin{align*} x &= e^{\log x} \end{align*}Since both sides of the equation are the same function, both sides of the equation have the same derivative. i.e. we are using ^{ 4 }

So now differentiate both sides:

\begin{align*} \diff{}{x} x &= \diff{}{x} e^{\log x} \\ \end{align*} The left-hand side is easy, and the right-hand side we can process using the chain rule with \(f(u)=e^u\) and \(u=\log x\text{.}\) \begin{align*} 1 &= \diff{f}{u} \cdot \diff{u}{x} \\ &= e^u \cdot \underbrace{\diff{}{x} \log x }_\text{what we want to compute}\\ \\ \end{align*} Recall that \(e^u = e^{\log x} = x\text{,}\) so \begin{align*} 1 &= x \cdot \underbrace{\diff{}{x} \log x }_\text{now what?}\\ \end{align*} We can now just rearrange this equation to make the thing we want the subject: \begin{align*} \diff{}{x} \log x &= \frac{1}{x} \end{align*}Thus we have proved:

where \(\log x\) is the logarithm base \(e\text{.}\)

We can extend Theorem 2.10.1 to compute the derivative of logarithms of other bases in a straightforward way. Since for any positive \(a \neq 1\text{:}\)

\begin{align*} \log_a x &= \frac{\log x}{\log a} = \frac{1}{\log a} \cdot \log x & \text{since $a$ is a constant}\\ \diff{}{x} \log_a x &= \frac{1}{\log a} \cdot \frac{1}{x} \end{align*}