Exponential Functions, Logarithms & Their Derivatives

Today we’re going deep into the two most powerful function families in calculus: exponentials and logarithms. You’ll see how the infinite series for \(e^x\) proves that it is its own derivative (how cool is that?), learn how to take derivatives of any exponential like \(2^x\) or \(10^x\), and meet the natural logarithm \(\ln(x)\). By the end, you’ll even get a sneak peek at Euler’s formula, which ties together exponentials, sine, cosine, and imaginary numbers in one jaw-dropping equation.

TipWhy Exponentials and Logarithms Matter

Exponential functions show up everywhere something grows or decays:

• Population growth: bacteria double every 20 minutes — that’s exponential!
• Compound interest: your savings account grows exponentially over time
• Radioactive decay: scientists use exponential decay to date ancient fossils
• Sound volume: decibels use logarithms to measure how loud something is
• Earthquakes: the Richter scale is logarithmic — a magnitude 7 quake is 10x stronger than a magnitude 6

The number \(e \approx 2.718\) is the “natural” base for all of this. Today we’ll discover where \(e\) comes from and why it makes calculus beautifully simple!

Topics Covered

• Binomial expansion of \((1 + r/n)^n\) term by term
• Taking the limit as \(n \to \infty\) to discover the infinite series for \(e^r\)
• The number \(e \approx 2.718\ldots\) and why it’s special
• Maclaurin series (power series) — representing functions as infinite polynomials
• Proving that \(\frac{d}{dx}(e^x) = e^x\) using the power series
• Derivative of \(a^x\): rewrite as \(e^{x \ln a}\), get \(a^x \cdot \ln(a)\)
• The natural logarithm \(\ln(x)\) as the inverse of \(e^x\)
• Proving \(\frac{d}{dx}\ln(x) = \frac{1}{x}\)
• Preview of Euler’s formula: \(e^{i\theta} = \cos\theta + i\sin\theta\)

Lecture Video

Key Frames from the Lecture

What You Need to Know First

NoteWhat is the Binomial Theorem?

The Binomial Theorem tells us how to expand \((a + b)^n\):

\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient (“\(n\) choose \(k\)”).

For example:

\[(1 + x)^3 = 1 + 3x + 3x^2 + x^3\]

The coefficients \(1, 3, 3, 1\) come from Pascal’s triangle. In this lesson, we’ll use the binomial theorem on \((1 + r/n)^n\) and then let \(n\) get really, really big!

NoteWhat is a factorial?

A factorial is a product of all positive integers up to \(n\):

\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\]

Examples: \(3! = 6\), \(4! = 24\), \(5! = 120\).

By convention, \(0! = 1\).

Factorials grow incredibly fast — \(10! = 3{,}628{,}800\) and \(20!\) is already in the quintillions!

NoteWhat is compound interest?

If you invest money at interest rate \(r\) (as a decimal), compounded \(n\) times per year, then after 1 year your money is multiplied by:

\[\left(1 + \frac{r}{n}\right)^n\]

For example, with \(r = 1\) (100% interest) compounded monthly (\(n = 12\)):

\[\left(1 + \frac{1}{12}\right)^{12} \approx 2.613\]

What happens if you compound more and more often — every second, every microsecond? The answer approaches \(e \approx 2.718\). That’s exactly what we’ll prove today!

NoteWhat is the chain rule?

The chain rule tells us how to differentiate a function inside another function:

\[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]

Think of it as peeling layers: differentiate the outer function, then multiply by the derivative of the inner function.

For example: \(\frac{d}{dx}(3x+1)^5 = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4\).

Key Concepts

From Compound Interest to the Number \(e\)

We start with the compound interest formula and expand it using the binomial theorem:

\[\left(1 + \frac{r}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \left(\frac{r}{n}\right)^k\]

Writing out the binomial coefficient:

\[\binom{n}{k} \frac{r^k}{n^k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k! \cdot n^k} \cdot r^k\]

Now here’s the magic. Look at the coefficient of \(r^k\):

\[\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}\]

This is a product of \(k\) fractions: \(\frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots\)

Each fraction looks like \(1 - \frac{\text{something}}{n}\). As \(n \to \infty\), every one of these fractions approaches 1. So the whole coefficient approaches \(\frac{1}{k!}\), and we get:

\[\lim_{n \to \infty}\left(1 + \frac{r}{n}\right)^n = \sum_{k=0}^{\infty} \frac{r^k}{k!} = 1 + r + \frac{r^2}{2!} + \frac{r^3}{3!} + \frac{r^4}{4!} + \cdots\]

Explore — watch the compound interest value approach \(e\) as \(n\) increases:

Set \(r = 1\) and watch the blue curve approach \(e \approx 2.718\) (red dashed line) as \(x\) grows. Try other values of \(r\)!

The Definition of \(e\)

Plugging \(r = 1\) into our series:

\[e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots \approx 2.71828\ldots\]

The number \(e\) is irrational — its decimal digits never repeat and never end. It’s one of the most important constants in all of mathematics, right alongside \(\pi\).

Power Series (Maclaurin Series)

The expression we found is called a power series — we’ve written \(e^r\) as an infinite polynomial:

\[e^r = 1 + r + \frac{r^2}{2!} + \frac{r^3}{3!} + \frac{r^4}{4!} + \cdots\]

This is also called the Maclaurin series for \(e^r\). The idea of representing functions as infinite sums of powers of \(x\) is one of the most powerful tools in all of mathematics. It lets us turn complicated functions into polynomials that we can differentiate, integrate, and compute with easily.

Proving \(\frac{d}{dx}(e^x) = e^x\)

This is one of the most elegant results in calculus. We differentiate the power series term by term:

\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots\]

Take the derivative of each term:

\[\frac{d}{dx}(e^x) = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \frac{5x^4}{5!} + \cdots\]

Now simplify each fraction. For instance, \(\frac{3x^2}{3!} = \frac{3x^2}{3 \cdot 2!} = \frac{x^2}{2!}\). In general, \(\frac{kx^{k-1}}{k!} = \frac{x^{k-1}}{(k-1)!}\).

So we get:

\[\frac{d}{dx}(e^x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = e^x\]

The series reproduces itself! That’s why \(e^x\) is special: it is its own derivative.

Explore — see that \(e^x\) and its derivative are the same curve:

Drag the slider for \(a\). Notice that the slope of the tangent line always equals the height of the point — because \(\frac{d}{dx}(e^x) = e^x\)!

Derivative of \(a^x\) (Any Exponential Base)

What about \(\frac{d}{dx}(2^x)\) or \(\frac{d}{dx}(10^x)\)? The trick is to rewrite any exponential using \(e\):

\[a^x = e^{x \ln a}\]

Why? Because \(e^{\ln a} = a\), so \(e^{x \ln a} = (e^{\ln a})^x = a^x\).

Now apply the chain rule. Let \(u = x \ln a\):

\[\frac{d}{dx}(a^x) = \frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx} = e^{x \ln a} \cdot \ln a = a^x \cdot \ln a\]

ImportantKey Idea: Derivative of Any Exponential

To differentiate an exponential with any base, rewrite it using \(e\) and apply the chain rule. The result picks up a factor of \(\ln a\) – this is why \(e\) is the “cleanest” base (since \(\ln e = 1\)).

\[\boxed{\frac{d}{dx}(a^x) = a^x \cdot \ln a}\]

Notice: when \(a = e\), we get \(\ln e = 1\), so \(\frac{d}{dx}(e^x) = e^x \cdot 1 = e^x\). It all fits together!

Explore — compare \(a^x\) with its derivative \(a^x \ln a\) for different bases:

Set \(a = e \approx 2.718\) and the two curves overlap perfectly — that’s the magic of \(e\)!

The Natural Logarithm

The natural logarithm \(\ln(x)\) is the inverse of \(e^x\). It answers the question:

\[\ln(a) = \text{"what power of } e \text{ gives } a\text{?"}\]

So \(\ln(e) = 1\) (because \(e^1 = e\)), \(\ln(1) = 0\) (because \(e^0 = 1\)), and \(\ln(e^3) = 3\).

Key properties:

• \(e^{\ln x} = x\) for all \(x > 0\)
• \(\ln(e^x) = x\) for all \(x\)
• \(\ln(ab) = \ln a + \ln b\)
• \(\ln(a^n) = n \ln a\)

Derivative of \(\ln(x) = \frac{1}{x}\)

Start with the identity:

\[e^{\ln x} = x\]

Differentiate both sides using the chain rule:

\[e^{\ln x} \cdot \frac{d}{dx}(\ln x) = 1\]

But \(e^{\ln x} = x\), so:

\[x \cdot \frac{d}{dx}(\ln x) = 1\]

ImportantKey Idea: Derivative of the Natural Logarithm

The derivative of \(\ln(x)\) turns a “transcendental” function into a simple algebraic one. This result comes straight from differentiating the identity \(e^{\ln x} = x\) and solving.

\[\boxed{\frac{d}{dx}\ln(x) = \frac{1}{x}}\]

This is remarkable: the derivative of \(\ln(x)\) is \(\frac{1}{x}\) — a simple algebraic function! And later we’ll see the flip side: the integral of \(\frac{1}{x}\) is \(\ln|x| + C\).

Explore — the slope of \(\ln(x)\) at any point equals \(\frac{1}{x}\):

Drag the slider and compare: the tangent line’s slope at \(x = a\) always equals \(\frac{1}{a}\) (green dashed curve).

Preview: Euler’s Formula

Here’s a mind-blowing teaser. If we plug an imaginary number \(i\theta\) into the power series for \(e^x\):

\[e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots\]

Using \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and separating real and imaginary parts, we get Euler’s formula:

ImportantKey Idea: Euler’s Formula

This single equation unifies exponentials, trigonometry, and imaginary numbers. It comes from plugging \(i\theta\) into the power series for \(e^x\) and separating real and imaginary parts.

\[\boxed{e^{i\theta} = \cos\theta + i\sin\theta}\]

This connects exponentials, trigonometry, and complex numbers in one equation. Setting \(\theta = \pi\) gives Euler’s identity: \(e^{i\pi} + 1 = 0\) — often called the most beautiful equation in mathematics.

TipHomework Challenge

Use Euler’s formula and the power series for \(e^{i\theta}\) to derive:

• The Maclaurin series for \(\cos\theta\) (collect the real parts)
• The Maclaurin series for \(\sin\theta\) (collect the imaginary parts)

Then differentiate those series term by term to show that \(\frac{d}{d\theta}\sin\theta = \cos\theta\) and \(\frac{d}{d\theta}\cos\theta = -\sin\theta\).

Cheat Sheet

What you want	Formula
Definition of \(e\)	\(e = \displaystyle\sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots \approx 2.718\)
Power series for \(e^r\)	\(e^r = \displaystyle\sum_{k=0}^{\infty} \frac{r^k}{k!} = 1 + r + \frac{r^2}{2!} + \frac{r^3}{3!} + \cdots\)
Derivative of \(e^x\)	\(\frac{d}{dx}(e^x) = e^x\)
Derivative of \(a^x\)	\(\frac{d}{dx}(a^x) = a^x \cdot \ln a\)
Natural log definition	$(a) = $ “what power of \(e\) gives \(a\)?”
Derivative of \(\ln(x)\)	\(\frac{d}{dx}\ln(x) = \frac{1}{x}\)
Euler’s formula	\(e^{i\theta} = \cos\theta + i\sin\theta\)

The Big Chain of Ideas

\[\text{Compound interest} \;\longrightarrow\; \text{Binomial expansion} \;\longrightarrow\; \text{Let } n \to \infty \;\longrightarrow\; \text{Power series for } e^r \;\longrightarrow\; \frac{d}{dx}e^x = e^x\]