Derivative of eˣ Explained — Why It’s Its Own Derivative & Quizzes

Where it comes from

Apply the definition to f(x) = e^x and factor:

Before reading on: e^x crosses (0,1) at exactly 45° — slope 1. Of all possible bases a, only one makes a^x hit slope 1 at x = 0. What does that say about why e is special?

[e x+h - e x] / h = e x \cdot (e h -1)/h Say it: the difference quotient of e to the x equals e to the x times the difference quotient of e to the x at 0

So d/dx [e^x] = e^x · lim_h→0(e^h−1)/h. The whole question is what that limit equals — and e is defined to make it 1. That’s not a coincidence discovered later; it’s the reason e ≈ 2.71828 is special among all possible bases.

The wrong guess to kill: the power rule, d/dx [e^x] = x·e^x−1?? Test at x = 1: the guess gives 1·e⁰ = 1; the true slope is e ≈ 2.72. Dead on arrival — and structurally wrong, since the power rule needs a constant exponent.

slope of e^x at 0

=

e⁰ = 1

The curve crosses (0,1) at exactly 45° — the signature of e.

slope of 2^x at 0

=

ln 2 ≈ 0.69

Shallower — 2 isn’t the “natural” base. Only e hits slope 1.

Derivation

We show d/dx [e^x] = e^x · L with L = lim_h→0(e^h−1)/h, then prove L = 1 from the definition of e:

d/dx [e^x]

=

lim_h→0 [e^x+h − e^x]/h

Step 1 — the definition.

=

e^x · lim_h→0 (e^h−1)/h

Step 2 — factor e^x. e^x+h = e^x·e^h; e^x doesn’t depend on h, so it leaves the limit.

L

=

lim_h→0 (e^h−1)/h = 1

Step 3 — key step: the definition of e. Set t = e^h−1, so h = ln(1+t) and t → 0 as h → 0: L = lim_t→0 t/ln(1+t) = lim 1/ln((1+t)^1/t) = 1/ln e = 1, since e = lim_n→∞(1+1/n)ⁿ.

=

e^x · 1 = e^x

Step 4 — assemble. ∎

The miracle is manufactured: e is chosen as the base where L = 1. For 2^x the same proof gives 2^x·ln 2 — the a^x rule, next page.

How to use it

The procedure:

Plain e^x → e^x. It’s its own derivative — the easiest rule in the chapter.
Something in the exponent? Chain: d/dx [e^u] = e^u·u′. Example: d/dx [e^2x] = 2e^2x.
Constants ride along: d/dx [3e^x] = 3e^x.
Products need the product rule: x·e^x → e^x + x·e^x = e^x(1+x).

The three lookalikes

e^x (this page: itself) vs. e^x² (chain: 2x·e^x²) vs. x^e (power rule — constant exponent!: e·x^e−1). Sort by asking: where’s the x — exponent, inside the exponent, or base?

Common mistake: writing d/dx [e^2x] = e^2x — treating it like plain e^x. The exponent 2x runs twice as fast; the ×2 is mandatory.

Worked examples

Four problems, easiest first. The chain rule does most of the work here.

Example 1 — d/dx [3e^x]

Pull out the 3: = 3·d/dx [e^x].
e^x rule: = 3e^x = 3e^x. (Why unchanged? e^x reproduces itself under differentiation.)

Common mistake: 3x·e^x−1 — power rule on an exponential. Power rule needs a constant exponent; x is not constant.

Your turn: d/dx [−2e^x]

Answer: −2e^x

Pull out the −2: −2·d/dx [e^x] = −2e^x. e^x reproduces itself — the constant just rides along.

Example 2 — d/dx [e^2x] (chain)

Layers: outer e^u, inner u = 2x.
Outside: e^2x. Inside’s derivative: 2.
Multiply: = 2e^2x.

Common mistake: e^2x alone — the forgotten ×2. The 2x inside doubles every rate.

Your turn: d/dx [e^−3x] (chain)

Answer: −3e^−3x

Layers: outer e^u, inner u = −3x. Outside: e^−3x. Inside’s derivative: −3. Multiply: −3e^−3x.

Example 3 — d/dx [x·e^x] (product)

Structure: multiplied — product rule: f = x, g = e^x.
Derivatives: f′ = 1, g′ = e^x.
Assemble: = 1·e^x + x·e^x = e^x(1 + x). (Why factor? e^x is common — factored form differentiates cleaner next time.)

Common mistake: e^x + e^x = 2e^x — differentiating x as if it were e^x’s chain factor. Product rule: differentiate each factor in turn.

Your turn: d/dx [x²·e^x] (product)

Answer: x·e^x(2 + x)

Product rule: f = x², g = e^x. f′ = 2x, g′ = e^x. Assemble: 2x·e^x + x²·e^x = x·e^x(2 + x).

Example 4 — d/dx [e^x/(x+1)] (quotient)

Structure: divided — quotient rule: f = e^x, g = x+1.
Derivatives: f′ = e^x, g′ = 1.
Chant: = [e^x(x+1) − e^x(1)]/(x+1)² = x·e^x/(x+1)². (Why so clean? e^x(x+1−1) = x·e^x — factor e^x first.)

Common mistake: leaving it as [e^x(x+1) − e^x]/(x+1)². Always factor e^x — it’s never zero, so it’s always safe to pull out.

Your turn: d/dx [e^x/(x−1)] (quotient)

Answer: (x−2)·e^x/(x−1)²

Quotient rule: f = e^x, g = x−1. Chant: [e^x(x−1) − e^x(1)]/(x−1)². Factor e^x: e^x(x−1−1)/(x−1)² = (x−2)·e^x/(x−1)².

Memorization tips

Its own derivative: e^x → e^x. The one rule you can’t get wrong — unless there’s something in the exponent.
e is defined for this: e is the base where lim(e^h−1)/h = 1. The rule isn’t luck; it’s the definition of e paying off.
Exponent? Chain: e^2x → 2e^2x, e^x² → 2x·e^x². The ×u′ is never optional.
Sort the lookalikes: e^x (itself), e^x² (chain), x^e (power rule!). Ask where the x lives.
Factor e^x always: it’s never zero, so factoring it out of sums is always safe — and answers collapse beautifully.
Slope 1 at 0: e^x crosses (0,1) at 45°. If your chain version doesn’t give slope k at 0 for e^kx, recheck the ×k.

Final challenge

Five mixed questions — the definition of e, chains, and the x^e lookalike. Score 5/5 and it’s yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the derivative of eˣ?

d/dx[eˣ] = eˣ — it’s its own derivative. From the definition: [e^(x+h)−eˣ]/h = eˣ·(e^h−1)/h, and lim(h→0)(e^h−1)/h = 1 by the very definition of e.

Why is e special?

e ≈ 2.71828 is defined as the base making lim(h→0)(b^h−1)/h = 1 — equivalently e = lim(n→∞)(1+1/n)ⁿ. For any other base b, d/dx[bˣ] = bˣ·ln b; only e gives the clean self-reproducing rule.

What is d/dx[e^(2x)]?

2e^(2x). Chain rule: e^(2x) for the outside, times the inside’s derivative 2. Forgetting the ×2 is the classic error.

What’s the difference between eˣ, e^(x²), and x^e?

eˣ differentiates to itself; e^(x²) needs the chain rule (2x·e^(x²)); x^e has a constant exponent, so the power rule gives e·x^(e−1). Three structures, three rules.

Is d/dx[e²] = e²?

No — e² ≈ 7.389 is a constant (no x), so its derivative is 0 by the constant rule. The eˣ rule needs an x in the exponent.

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The derivative of ex

Where it comes from

Derivation

How to use it

The three lookalikes

Worked examples

Example 1 — d/dx [3ex]

Example 2 — d/dx [e2x] (chain)

Example 3 — d/dx [x·ex] (product)

Example 4 — d/dx [ex/(x+1)] (quotient)