Introduction
As you study math, a number like this suddenly shows up.
$$e \approx 2.71828182845904\ldots$$Why this particular number? Unlike \(\pi = 3.14159\ldots\), there is no obvious geometric reason such as the circumference of a circle, and textbooks usually just introduce it as "Euler's number" and move on.
The one-sentence answer first — \(e\) is the base of the unique exponential function \(e^x\) whose shape does not change at all when you differentiate it.
Let me unpack that a little. "To differentiate" means to find the slope of the graph (how steep the tangent line is) — see Why Is the Derivative the "Instantaneous Slope"? for why this is so. But the exponential function \(a^x\) has a remarkable property — when you differentiate it, it comes back as the original function multiplied by some constant.
$$\frac{d}{dx}\,a^x = (\text{some multiplier}) \times a^x$$This "multiplier" is determined by what the base \(a\) is (about 0.69 for base 2, about 1.10 for base 3). But there is exactly one base for which this multiplier is precisely 1. That base is \(e \approx 2.718\). When the multiplier is 1, multiplying changes nothing, so \(e^x\) becomes \(e^x\) itself when differentiated.
The formal name of this "multiplier" is the natural logarithm \(\ln a\) (the logarithm with base \(e\)). For now it is enough to think of it simply as "some multiplier determined by each base," and we will check directly with numbers in the main text why that value is \(\ln a\).
In this article we will derive this reason from the definition of the derivative without skipping a single step.
Why this matters
Why is "being its own derivative" so special?
The rate at which heat spreads, the rate at which a radioactive element decays, the rate at which a population grows — every natural phenomenon that "changes in proportion to its current amount" satisfies one and the same equation.
$$\frac{dy}{dt} = k \cdot y$$"The rate of change is \(k\) times the current value." The solution of this equation is \(y = Ce^{kt}\). Because the derivative of \(e^x\) is again \(e^x\) itself, the function that solves this equation is exactly \(e^{kt}\). Without \(e\), we could not solve the most fundamental equation describing natural phenomena.
Step 1 — Approaching \(a^x\) through the definition of the derivative
We write the definition of the derivative. For \(f(x) = a^x\):
$$f'(x) = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}$$In the numerator we use \(a^{x+h} = a^x \cdot a^h\) to factor out \(a^x\).
$$f'(x) = \lim_{h \to 0} \frac{a^x \cdot a^h - a^x}{h} = a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}$$Since \(a^x\) does not depend on \(h\), we can pull it outside the limit. The limit that remains is now:
$$\lim_{h \to 0} \frac{a^h - 1}{h}$$What this value is, that is the key.
Step 2 — The identity of the remaining limit
The limit left over after pulling \(a^x\) outside in Step 1,
$$\lim_{h \to 0}\frac{a^h-1}{h}$$the identity of this value determines everything. Since the value is fixed for each base \(a\), let us name it \(L(a)\).
First, let us put into words what this expression means. Since \(a^0 = 1\), the numerator \(a^h - 1\) is the vertical change — "how much the function value changed (from 1) when we move \(x\) by \(h\) starting from 0." Dividing this by the distance moved, \(h\), gives the average slope over that short interval, and sending \(h \to 0\) gives the instantaneous slope at \(x = 0\). That is:
\(L(a)\) is the slope of the curve \(y = a^x\) at the moment it passes through height 1 (\(x = 0\)).
Seen this way, the meaning of the Step 1 result \(f'(x) = a^x \cdot L(a)\) also becomes clear — the slope at any position is "the starting slope \(L(a)\) × the height \(a^x\) at that point." (The value shown as the slope of the tangent line at \(x = 0\) in this article's interactive is exactly this \(L(a)\).)
Now if we plug in numbers directly for a few values of \(a\) and compute \(L(a)\), a pattern emerges.
| \(a\) | \(L(a)\) (the limit above) | Note |
|---|---|---|
| 2 | \(\approx 0.693\) | \(= \ln 2\) |
| 3 | \(\approx 1.099\) | \(= \ln 3\) |
| 4 | \(\approx 1.386\) | \(= \ln 4\) |
| ? | 1.000 | the value we want |
Do you see the pattern? The \(L(a)\) values in the table (0.693, 1.099, 1.386…) are exactly equal to the natural logarithm \(\ln a\) of that \(a\).
But why does a logarithm pop out at all? The table has a clue. \(L(4) = 1.386\) is exactly twice \(L(2) = 0.693\). This is no coincidence — \(4 = 2 \times 2\), and \(L\) has this property.
$$L(a \times b) = L(a) + L(b)$$That is, \(L\) turns multiplication into addition. And the function that turns multiplication into addition is precisely the logarithm (see Why Does the Logarithm Turn Multiplication into Addition? for why). So \(L(a)\) can only be some logarithm. The only remaining question is "what is its base" — and a logarithm equals 1 at its own base (\(\log_b b = 1\)). So we name the base that makes \(L(a) = 1\) the number \(e\), and then \(L(a) = \log_e a = \ln a\).
This is the natural logarithm \(\ln a\) — "the logarithm with base \(e\)," that is, the value indicating to what power \(e\) must be raised to get \(a\) (for example, \(\ln 2 \approx 0.693\) means \(e^{0.693} \approx 2\)). A rigorous proof requires the definition of the natural logarithm, but in fact many textbooks take this very limit as the definition of \(\ln a\) — which amounts to running the direction backward.
In the end:
$$\frac{d}{dx} a^x = a^x \cdot \ln a$$Step 3 — Why e
From the result above, for the derivative to equal the original function:
$$a^x \cdot \ln a = a^x$$Dividing both sides by \(a^x\) (since \(a^x > 0\)):
$$\ln a = 1$$Solving this equation for \(a\):
$$a = e^1 = e$$This is the definition of \(e\). \(e\) is the unique positive number satisfying \(\ln a = 1\) — that is, the number whose natural logarithm is 1.
To summarize:
$$\frac{d}{dx} e^x = e^x \cdot \underbrace{\ln e}_{=\,1} = e^x$$Differentiating \(e^x\) gives \(e^x\). No other base can do this.
Comparison with other bases
What about \(a = 2\)?
$$\frac{d}{dx} 2^x = 2^x \cdot \ln 2 \approx 2^x \cdot 0.693$$The derivative is only about 69.3% of the original function. The slope is always smaller than the value.
For \(a = 3\):
$$\frac{d}{dx} 3^x = 3^x \cdot \ln 3 \approx 3^x \cdot 1.099$$The derivative is 109.9% of the original function — that is, the slope is larger than the value.
Only when \(a = e\) does the ratio become exactly 100%.
Find it yourself
In the interactive below, change the base \(a\) with the slider. The blue curve is \(a^x\), and the orange line is the tangent at \(x = 0\). Find for yourself the moment when the slope of the tangent line matches the function value (always 1).
When you press the preset "a = e ≈ 2.718," the tangent turns green and "slope = function value ✓" appears. When \(a = 2\) the slope is short at 69.3%, and when \(a = 3\) it is long at 109.9%.
Bonus — Finding e through compound interest
There is another path to \(e\). If you split an annual interest rate of 100% into \(n\) compounding periods:
$$\left(1 + \frac{1}{n}\right)^n$$| \(n\) | Result |
|---|---|
| 1 (once a year) | \(2.000\) |
| 2 (semiannual) | \(2.250\) |
| 4 (quarterly) | \(2.441\) |
| 12 (monthly) | \(2.613\) |
| 365 (daily) | \(2.7146\ldots\) |
| \(\infty\) | \(e = 2.71828\ldots\) |
This definition looks like a completely different starting point from differentiation, but it is proved mathematically that the two definitions point to the same number. Whether you start from the economic concept of compound interest or from differentiation, you arrive at the same \(e\).
Key takeaways
| Question | Answer |
|---|---|
| What is the derivative of \(a^x\)? | \(a^x \cdot \ln a\) |
| What makes derivative = original function? | \(\ln a = 1\), i.e. \(a = e\) |
| What is the value of \(e\)? | \(e \approx 2.71828\ldots\) |
| Why does it matter? | The solution of equations of the form \(\frac{dy}{dt} = ky\) is \(e^{kt}\) |
\(e\) is not a number invented by nature but the number you inevitably and logically arrive at when you want a "function that does not change under differentiation."
Studying this with AI
AI explains the definition and properties of \(e\) well. Just be aware that during the derivation it may skip steps it calls "obvious," so ask specifically about the parts where you get stuck.
Examples of useful questions:
- "Show rigorously why the limit of \((a^h-1)/h\) as \(h \to 0\) is \(\ln a\)."
- "Why does differentiating the natural logarithm \(\ln x\) give \(1/x\)?"
- "Expand \(e^x\) as a Taylor series."
Wrapping up
\(e = 2.718\ldots\) is not an arbitrarily chosen number. It is determined logically by the condition that the derivative equals the function itself. And because "change proportional to a quantity" in nature requires exactly this condition, \(e\) shows up everywhere in physics, biology, and economics.
Related reading — Read first: Why Is the Derivative the "Instantaneous Slope"? → Up next: Why Does the Logarithm Turn Multiplication into Addition? · Why Does the Arctangent Show Up? · Draw the graphs yourself: Desmos · Wolfram Alpha