Derivative of aˣ Explained — aˣ·ln a Proof & Quizzes

Q: What is the derivative of aˣ?

d/dx[aˣ] = aˣ·ln a. Rewrite aˣ = e^(x·ln a), then the chain rule gives e^(x·ln a)·ln a = aˣ·ln a. The ln a is the chain factor from the base change.

Q: Where does the ln a come from?

From writing aˣ as e^(x·ln a): differentiating e^(x·ln a) by the chain rule multiplies by d/dx[x·ln a] = ln a. It’s not an extra rule — it’s the eˣ rule plus the chain rule.

Q: Why can’t I use the power rule on 2ˣ?

The power rule needs a variable base with a constant exponent (xⁿ). In 2ˣ the base is constant and the exponent varies — the mirror image. Test at x = 0: power rule gives 0, but the true slope is ln 2 ≈ 0.693.

Q: What if the base is less than 1?

Then ln a is negative, so the derivative is negative — matching exponential decay. Example: d/dx[(1/2)ˣ] = (1/2)ˣ·ln(1/2) = −(1/2)ˣ·ln 2.

Q: What is d/dx[2^(3x)]?

3·2^(3x)·ln 2. Two chain factors: ln 2 from the base-a rule, and 3 from the 3x inside the exponent.

Where it comes from

We know d/dx [e^x] = e^x because e was defined for that. But 2^x? Run the definition’s factorization: d/dx [2^x] = 2^x · lim_h→0(2^h−1)/h. That limit is some number — numerically about 0.693. And 0.693 ≈ ln 2. Coincidence? No: it’s the chain rule waiting to happen.

The power-rule trap must die first: d/dx [2^x] = x·2^x−1??

Before reading on: a^x = e^x ln a. Chain-ruling e^u with u = x ln a — what extra factor must appear next to a^x, and why doesn’t e^x need one?

the guess at x = 0

=

0·2⁻¹ = 0

The power rule applied to 2^x as if x were a constant exponent.

the true slope

=

2⁰·ln 2 = ln 2 ≈ 0.693

0 ≠ 0.693 — dead on arrival. The power rule needs a constant exponent; here the base is constant and the exponent moves.

The key insight: a^x = e^x ln a (since e^ln a = a). Every exponential is e^x wearing a linear disguise — and the disguise’s slope, ln a, becomes the chain factor.

Derivation

Rewrite the base-a exponential as a base-e exponential, then chain:

a^x

=

e^{ln(a^x)} = e^x ln a

Step 1 — the rewrite. a = e^ln a (for a > 0), so a^x = (e^ln a)^x = e^x ln a. Log power law.

d/dx [a^x]

=

d/dx [e^x ln a]

Step 2 — differentiate the rewrite. Now it’s e^u with u = x ln a.

=

e^x ln a · ln a

Step 3 — chain rule. d/du [e^u] = e^u; d/dx [x ln a] = ln a (it’s just a constant slope).

=

a^x · ln a

Step 4 — rewrite back. e^x ln a = a^x. ∎

Consistency check: for a = e, the rule gives e^x·ln e = e^x·1 = e^x — exactly the e^x rule. The general rule contains the special one.

How to use it

The procedure:

Plain a^x → a^x ln a. Copy the exponential, append ×ln(base).
Something in the exponent? Chain again: d/dx [a^u] = a^u ln a·u′. Example: d/dx [2^3x] = 2^3x ln 2·3.
Base between 0 and 1? ln a is negative — the derivative flips sign, matching exponential decay: d/dx [(1/2)^x] = (1/2)^x ln(1/2) = −(1/2)^x ln 2.
Base 1? ln 1 = 0 — and indeed 1^x = 1 is constant. The rule agrees with the constant rule.

The three exponential lookalikes

2^x (this rule: 2^x ln 2) vs. x² (power rule: 2x) vs. x^x (neither — needs logarithmic differentiation, a later topic). Ask: is the x in the exponent, the base, or both?

Common mistake: writing d/dx [2^x] = x·2^x−1 (power rule). Power rule = variable base, constant exponent. a^x = constant base, variable exponent — the mirror image needs its own rule.

Worked examples

Four problems, easiest first. Copy the exponential, append ×ln(base).

Example 1 — d/dx [2^x]

Copy the exponential: 2^x.
Append ×ln(base): = 2^x ln 2. (Why ln 2? From the e^x ln 2 rewrite’s chain factor.)
Check at 0: ln 2 ≈ 0.693 — matches the numerical limit from the definition. ✓

Common mistake: x·2^x−1. At x = 0 that gives 0; the true slope is 0.693. Power rule on an exponential is always wrong.

Your turn: d/dx [3^x]

Answer: 3^x ln 3

Copy the exponential, append ×ln(base): 3^x ln 3. Check at 0: ln 3 ≈ 1.099 — the true slope of 3^x at the origin.

Example 2 — d/dx [10^3x] (double chain)

Outer (a^u rule): 10^3x ln 10.
Inner’s derivative: d/dx [3x] = 3.
Multiply: = 3·10^3x ln 10. (Why two factors? The ln 10 comes from the base rewrite; the 3 from the 3x inside.)

Common mistake: 10^3x ln 10 without the ×3, or 3·10^3x without the ln 10. Both chain factors are mandatory.

Your turn: d/dx [5^−2x] (double chain)

Answer: −2·5^−2x ln 5

Outer (a^u rule): 5^−2x ln 5. Inner’s derivative: −2. Multiply: −2·5^−2x ln 5.

Example 3 — d/dx [(1/2)^x] (decay)

Copy + ln(base): = (1/2)^x ln(1/2).
Simplify the log: ln(1/2) = −ln 2, so = −(1/2)^x ln 2. (Why negative? The function decays — falling means negative slope.)

Common mistake: dropping the sign and writing +(1/2)^x ln 2. A decaying exponential can’t have positive slope — the graph convicts it.

Your turn: d/dx [(1/3)^x] (decay)

Answer: −(1/3)^x ln 3

Copy + ln(base): (1/3)^x ln(1/3). Simplify: ln(1/3) = −ln 3, so −(1/3)^x ln 3. Negative — the function decays.

Example 4 — d/dx [x·2^x] (product)

Structure: multiplied — product rule: f = x, g = 2^x.
Derivatives: f′ = 1, g′ = 2^x ln 2.
Assemble: = 1·2^x + x·2^x ln 2 = 2^x(1 + x ln 2). (Why factor? 2^x is never zero — always safe.)

Common mistake: 2^x + x·2^x — using the e^x rule (no ln 2) on a base-2 exponential. Only base e reproduces cleanly.

Your turn: d/dx [x·3^x] (product)

Answer: 3^x(1 + x ln 3)

Product rule: f = x, g = 3^x. f′ = 1, g′ = 3^x ln 3. Assemble: 3^x + x·3^x ln 3 = 3^x(1 + x ln 3).

Memorization tips

Copy, then ×ln(base): 2^x → 2^x ln 2. Two pen strokes — the ln a is the whole rule.
The ln a is a chain factor: from a^x = e^x ln a. It’s not decoration; it’s the disguise’s slope.
Mirror of the power rule: power = variable base, constant exponent; a^x = constant base, variable exponent. Say which mirror you’re in before writing.
Decay means negative: base < 1 → ln a < 0 → negative slopes. If your decaying exponential has positive slope, the sign died.
a = e collapses: e^x ln e = e^x. If your general rule doesn’t reproduce the e^x rule, it’s wrong.
a = 1 collapses: ln 1 = 0, and 1^x = 1 is constant. Both rules agree — a free consistency check.

Final challenge

Five mixed questions — the ln a factor, chains, and the power-rule trap. 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 aˣ?

d/dx[aˣ] = aˣ·ln a. Rewrite aˣ = e^(x·ln a), then the chain rule gives e^(x·ln a)·ln a = aˣ·ln a. The ln a is the chain factor from the base change.

Where does the ln a come from?

From writing aˣ as e^(x·ln a): differentiating e^(x·ln a) by the chain rule multiplies by d/dx[x·ln a] = ln a. It’s not an extra rule — it’s the eˣ rule plus the chain rule.

Why can’t I use the power rule on 2ˣ?

The power rule needs a variable base with a constant exponent (xⁿ). In 2ˣ the base is constant and the exponent varies — the mirror image. Test at x = 0: power rule gives 0, but the true slope is ln 2 ≈ 0.693.

What if the base is less than 1?

Then ln a is negative, so the derivative is negative — matching exponential decay. Example: d/dx[(1/2)ˣ] = (1/2)ˣ·ln(1/2) = −(1/2)ˣ·ln 2.

What is d/dx[2^(3x)]?

3·2^(3x)·ln 2. Two chain factors: ln 2 from the base-a rule, and 3 from the 3x inside the exponent.

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The derivative of ax

Where it comes from

Derivation

How to use it

The three exponential lookalikes

Worked examples

Example 1 — d/dx [2x]

Example 2 — d/dx [103x] (double chain)

Example 3 — d/dx [(1/2)x] (decay)

Example 4 — d/dx [x·2x] (product)