Calculus · Proof

Why eˣ Differentiates to Itself

The exponential function f(x) = eˣ is the unique (up to a constant multiple) non-zero differentiable function that equals its own derivative. This remarkable property is not a coincidence — it follows directly from how the constant e is defined.

Step 1: The Difference Quotient for aˣ

For any base a > 0, apply the limit definition to f(x) = aˣ:

f′(x) = limh→0 [a^(x+h) − aˣ] / h = limh→0 aˣ · (aʰ − 1) / h

Factor out aˣ (which does not depend on h):

f′(x) = aˣ · limh→0 (aʰ − 1) / h

The remaining limit depends only on the base a. Call it L(a):

L(a) = limh→0 (aʰ − 1) / h

For a = 2, L(2) ≈ 0.693. For a = 3, L(3) ≈ 1.099. There is exactly one base for which L(a) = 1 — and that base is defined to be e.

Step 2: The Definition of e

The number e is defined as the unique positive real number satisfying:

limh→0 (eʰ − 1) / h = 1

Equivalently, e can be characterised as:

e = limn→∞ (1 + 1/n)ⁿ ≈ 2.71828182845...

With L(e) = 1, the derivative of eˣ is simply:

d/dx eˣ = eˣ · 1 = eˣ

Step 3: Confirmation via the Taylor Series

The Taylor series of eˣ centred at 0 can be derived by requiring the series Σ Cₙxⁿ to equal its own term-by-term derivative. This forces C(n+1)·(n+1) = Cₙ, with C₀ = e⁰ = 1. Solving the recurrence gives Cₙ = 1/n!, so:

eˣ = Σn=0 xⁿ/n! = 1 + x + x²/2! + x³/3! + ···

Differentiating term by term (valid since the series converges uniformly on bounded intervals):

d/dx eˣ = 0 + 1 + x/1! + x²/2! + ··· = Σn=0 xⁿ/n! = eˣ

The series is unchanged by differentiation — a third confirmation that eˣ is its own derivative.

Derivative of aˣ for Other Bases

Using logarithmic differentiation, write aˣ = e^(x·ln a). By the chain rule:

d/dx aˣ = d/dx e^(x·ln a) = e^(x·ln a) · ln(a) = aˣ · ln(a)

When a = e, ln(e) = 1 and the formula reduces to d/dx eˣ = eˣ, consistent with everything above.

Worked Example

Differentiate f(x) = e^(3x²).

  • Outer function: eᵘ, derivative eᵘ.
  • Inner function: u = 3x², derivative 6x.
  • Chain rule: f′(x) = e^(3x²) · 6x.
f′(x) = 6x·e^(3x²)

At x = 1: f′(1) = 6e³ ≈ 6 × 20.086 ≈ 120.5. Numerical verification with h = 10⁻⁶ confirms this value.

Frequently Asked Questions

Why is d/dx eˣ = eˣ?

Because e is precisely the base for which the exponential function equals its own derivative. Formally, e is defined so that lim_{h→0} (eʰ−1)/h = 1. This makes the difference quotient for eˣ evaluate to eˣ itself.

What is the Taylor series for eˣ?

eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ··· = Σ_{n=0}^∞ xⁿ/n!. This series converges for all real (and complex) x. Differentiating term by term gives the same series, confirming d/dx eˣ = eˣ.

How is e defined as a limit?

e = lim_{n→∞} (1 + 1/n)ⁿ ≈ 2.71828... Equivalently, e = lim_{h→0} (1+h)^{1/h}. This limit arises naturally when computing continuous compound interest and in the definition of the natural logarithm.

What about d/dx aˣ for other bases?

For any positive base a ≠ 1, d/dx aˣ = aˣ·ln(a). When a = e, ln(e) = 1, so the formula reduces to d/dx eˣ = eˣ·1 = eˣ. The factor ln(a) measures how far a deviates from the 'natural' base e.

Adapted and rewritten from Silas Maths source notes; formulas reviewed for CalxSolver publication.