Gilbert Strang’s Calculus: The Exponential Function

Calculus

Exponential Function

Derivatives

Author

Chao Ma

Published

February 13, 2026

Properties

\(\frac{d}{dx}e^x = e^x\)
\(e^a\cdot e^b = e^{a+b}\)
If \(y=e^{cx}\), then \(\frac{dy}{dx}=cy=c e^{cx}\)

Infer the value of \(e\)

Start from the condition that defines the exponential function:

\(y(0)=1\)
\(\frac{dy}{dx}=y\)

If we start with \(y(x)=1\), then

\(y(x)=1\)
\(y'(x)=0\)

This cannot satisfy \(y'=y\), so we need to add an \(x\) term.

Step 1:

\(y(x)=1+x\)
\(y'(x)=1\)

Now \(y'\) is still missing the \(x\) term, so we add a quadratic term.

Step 2:

\(y(x)=1+x+\frac{1}{2}x^2\)
\(y'(x)=1+x\)

Now \(y'\) is missing \(\frac12 x^2\), so we add a cubic term.

Step 3:

\(y(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3\)
\(y'(x)=1+x+\frac{1}{2}x^2\)

Key observation:

This is an infinite correction loop.
Each time we add a term \(\frac{1}{n!}x^n\), its derivative becomes \(\frac{1}{(n-1)!}x^{n-1}\), exactly matching the previous missing term.

So eventually, \[ y(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots+\frac{1}{n!}x^n+\cdots \]

When \(x=1\), we get \[ e=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots+\frac{1}{n!}+\cdots \approx 2.718281828459045. \]

Power-series construction of the exponential function

We are not guessing \(e^x\). We are forcing a function to equal its own derivative, and factorial coefficients are exactly what make this possible.

Graph of \(e^x\)

Graph of the exponential function

At \(x=0\), \(y=1\).
At \(x=1\), \(y=e\).
For \(x<0\), \(0<e^x<1\).
\(e^{-x}=\frac{1}{e^x}\).
The slope is always positive and increases with \(x\).

Example: Computing Compound Interest

Suppose you have $$1 in the bank, with 100% annual interest.

If interest is compounded yearly:

End of year 1: \(\$2\)
End of year 2: \(\$4\)
End of year 3: \(\$8\)

So after \(n\) years, the amount is \(2^n\).

If the same annual rate is compounded monthly, then after one year: \[ \left(1+\frac{1}{12}\right)^{12} \approx 2.613035290224676. \]

If it is compounded daily: \[ \left(1+\frac{1}{365}\right)^{365} \approx 2.7145674820219727. \]

As compounding becomes more frequent, the value approaches \(e\): \[ \left(1+\frac{1}{N}\right)^N \to e. \]

That is why \(e\) naturally appears in continuous growth.

---
title: "Gilbert Strang's Calculus: The Exponential Function"
author: "Chao Ma"
date: "2026-02-13"
categories: [Calculus, Exponential Function, Derivatives]
---

{{< video https://www.youtube.com/watch?v=oo1ZZlvT2LQ&list=PLFW_V3qDH5jRyfpD9uiq6aKVTWIxpbIm3&index=4 >}}

# Properties

- $\frac{d}{dx}e^x = e^x$
- $e^a\cdot e^b = e^{a+b}$
- If $y=e^{cx}$, then $\frac{dy}{dx}=cy=c e^{cx}$

# Infer the value of $e$

Start from the condition that defines the exponential function:

- $y(0)=1$
- $\frac{dy}{dx}=y$

If we start with $y(x)=1$, then

- $y(x)=1$
- $y'(x)=0$

This cannot satisfy $y'=y$, so we need to add an $x$ term.

Step 1:

- $y(x)=1+x$
- $y'(x)=1$

Now $y'$ is still missing the $x$ term, so we add a quadratic term.

Step 2:

- $y(x)=1+x+\frac{1}{2}x^2$
- $y'(x)=1+x$

Now $y'$ is missing $\frac12 x^2$, so we add a cubic term.

Step 3:

- $y(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3$
- $y'(x)=1+x+\frac{1}{2}x^2$

Key observation:

- This is an infinite correction loop.
- Each time we add a term $\frac{1}{n!}x^n$, its derivative becomes $\frac{1}{(n-1)!}x^{n-1}$, exactly matching the previous missing term.

So eventually,
$$
y(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots+\frac{1}{n!}x^n+\cdots
$$

When $x=1$, we get
$$
e=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots+\frac{1}{n!}+\cdots
\approx 2.718281828459045.
$$

![Power-series construction of the exponential function](exponential-function-1.png)

We are not guessing $e^x$. We are forcing a function to equal its own derivative, and factorial coefficients are exactly what make this possible.

# Graph of $e^x$

![Graph of the exponential function](exponential-function-2.png)

- At $x=0$, $y=1$.
- At $x=1$, $y=e$.
- For $x<0$, $0<e^x<1$.
- $e^{-x}=\frac{1}{e^x}$.
- The slope is always positive and increases with $x$.

# Example: Computing Compound Interest

Suppose you have $\$1 in the bank, with 100% annual interest.

If interest is compounded yearly:

- End of year 1: $\$2$
- End of year 2: $\$4$
- End of year 3: $\$8$

So after $n$ years, the amount is $2^n$.

If the same annual rate is compounded monthly, then after one year:
$$
\left(1+\frac{1}{12}\right)^{12} \approx 2.613035290224676.
$$

If it is compounded daily:
$$
\left(1+\frac{1}{365}\right)^{365} \approx 2.7145674820219727.
$$

As compounding becomes more frequent, the value approaches $e$:
$$
\left(1+\frac{1}{N}\right)^N \to e.
$$

That is why $e$ naturally appears in continuous growth.