Derivatives of exponential and logarithmic functions

Exponential functions describe growth and decay: populations, money with compound interest, and radioactive material. In this section, we find their derivatives. We also find the derivatives of logarithms, which are their inverses.

For a review, see exponential functions and logarithmic functions.

The derivative of $a^x$

Let $f(x)=a^x$, where $a>0$. Use the definition of the derivative:

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{a^{x+h}-a^x}{h}\end{align*}$

By the rules of exponents, $a^{x+h}=a^x\cdot a^h$:

$\begin{align*}&=\lim_{h\to 0}\dfrac{a^x\cdot a^h-a^x}{h}\end{align*}$

Factor out $a^x$:

$\begin{align*}&=\lim_{h\to 0}a^x\cdot\dfrac{a^h-1}{h}\end{align*}$

The factor $a^x$ does not depend on $h$, so it moves in front of the limit:

$\begin{align*}&f'(x)=a^x\cdot\lim_{h\to 0}\dfrac{a^h-1}{h}\end{align*}$

This says something surprising. The derivative of $a^x$ is $a^x$ times a constant. The constant depends only on $a$. Here are its values for ${a=2}$ and ${a=3}$:

For $a=2$, the constant is about $0.693$. For $a=3$, it is about $1.099$. So for some base between $2$ and $3$, the constant is exactly $1$.

The number $e$

The number $e$ is the base for which

$\displaystyle\lim_{h\to 0}\dfrac{e^h-1}{h}=1$.

Its value is $e\approx 2.71828$. Like $\pi$, it is an irrational number: its decimals never end or repeat. Here is the check:

The constant is $1$, so the derivative of $e^x$ is $e^x$ itself:

$\dfrac{d}{dx}(e^x)=e^x$.

In words: $e^x$ is its own derivative. At every point, the slope of the graph equals its height. For example, at $(0,1)$ the slope is $1$.

The function $e^x$ is called the natural exponential function. It is the most important exponential function in calculus.

With the chain rule, if $u$ is a function of $x$, then

$\dfrac{d}{dx}(e^u)=e^u\cdot u'$.

The natural logarithm

The natural logarithm is the logarithm with base $e$. We write it $\ln x$:

$\ln x=\log_e x$.

On a calculator, the "ln" button gives the natural logarithm. The "log" button gives the logarithm with base $10$.

The functions $\ln x$ and $e^x$ are inverses. Each one undoes the other:

$e^{\ln x}=x$ for $x>0$,   and   $\ln(e^x)=x$.

The domain of $\ln x$ is $x>0$, because $e^y$ is always positive.

The derivative of $\ln x$

$\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$,   for $x>0$.

Why: Let $y=\ln x$. Then $e^y=x$. Differentiate both sides with respect to $x$, using implicit differentiation:

$e^y\cdot y'=1$

Divide both sides by $e^y$:

$\begin{align*}&y'=\dfrac{1}{e^y}\end{align*}$

Since $e^y=x$,

$\begin{align*}&\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\end{align*}$

With the chain rule, if $u$ is a positive function of $x$, then

$\dfrac{d}{dx}(\ln u)=\dfrac{1}{u}\cdot u'=\dfrac{u'}{u}$.

Negative $x$: The function $\ln|x|$ is defined for every $x\ne 0$. Its derivative is also $\dfrac{1}{x}$. For $x<0$, we have $|x|=-x$, so by the chain rule

$\dfrac{d}{dx}\ln(-x)=\dfrac{-1}{-x}=\dfrac{1}{x}$.

Simplify with the properties of logarithms

The properties of logarithms (see logarithmic functions) can turn a hard derivative into an easy one. For $a,b>0$:

• $\ln(ab)=\ln a+\ln b$
• $\ln\left(\dfrac{a}{b}\right)=\ln a-\ln b$
• $\ln(a^p)=p\ln a$

Other bases

Any positive base $a$ can be written as a power of $e$: ${a=e^{\ln a}}$. So

$a^x=\left(e^{\ln a}\right)^x=e^{x\ln a}$.

$\begin{align*}&\dfrac{d}{dx}(a^x)\\&=\dfrac{d}{dx}\left(e^{x\ln a}\right)\end{align*}$

Use the chain rule. The inner function is $x\ln a$. Its derivative is the constant $\ln a$:

$\begin{align*}&=e^{x\ln a}\cdot\ln a\end{align*}$

Since $e^{x\ln a}=a^x$,

$\begin{align*}&\dfrac{d}{dx}(a^x)=a^x\ln a\end{align*}$

This explains the table at the top of the page. The constant for $a=2$ was about $0.693$, and $\ln 2\approx 0.693$. The constant for $a=3$ was about $1.099$, and $\ln 3\approx 1.099$.

For logarithms with base $a$, use the change-of-base formula:

$\log_a x=\dfrac{\ln x}{\ln a}$

$\begin{align*}&\dfrac{d}{dx}(\log_a x)\\&=\dfrac{d}{dx}\left(\dfrac{\ln x}{\ln a}\right)\end{align*}$

The number $\ln a$ is a constant, so it stays in front:

$\begin{align*}&=\dfrac{1}{\ln a}\cdot\dfrac{1}{x}\end{align*}$

Multiply:

$\begin{align*}&\dfrac{d}{dx}(\log_a x)=\dfrac{1}{x\ln a}\end{align*}$

Growth

Logarithmic differentiation

Some functions have $x$ in both the base and the exponent, such as $x^x$. Neither the power rule nor the rule for $a^x$ applies. The power rule needs a constant exponent. The rule for $a^x$ needs a constant base.

The trick is to take the natural logarithm of both sides first. This is called logarithmic differentiation:

1. Take $\ln$ of both sides.
2. Simplify with the properties of logarithms.
3. Differentiate both sides implicitly. The left side, $\ln y$, gives $\dfrac{y'}{y}$.
4. Multiply both sides by $y$, and replace $y$ by its formula.

The power rule for every real exponent

In basic differentiation rules, we said the power rule works for any real number $n$. We proved it for integers and for fractions. Logarithmic differentiation proves it for every real $n$, for example $n=\sqrt{2}$ or $n=\pi$.

Let $y=x^n$, with $x>0$. Then

$\ln y=n\ln x$.

Differentiate both sides:

$\dfrac{y'}{y}=\dfrac{n}{x}$

Multiply both sides by $y=x^n$:

$\begin{align*}&y'=\dfrac{n}{x}\cdot x^n\end{align*}$

Since $\dfrac{x^n}{x}=x^{n-1}$,

$\begin{align*}&\dfrac{d}{dx}(x^n)=nx^{n-1}\end{align*}$

For example, $\dfrac{d}{dx}\left(x^{\sqrt{2}}\right)=\sqrt{2}\,x^{\sqrt{2}-1}$.

Summary

Here $u$ is a differentiable function of $x$ (positive for $\ln u$), and $a>0$, $a\ne 1$:

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