How do you differentiate xˣ?

Logarithmic differentiation: ln y = x ln x (bring the exponent down), then y'/y = ln x + 1, so y' = xˣ(ln x + 1). Neither the power rule nor the exponential rule applies directly.

Calculus I › Exponential & log laws › full formula sheet

ln(a^r) = r · ln a

Say it: “the natural log of a to the r equals r times the natural log of a”

Log of a power

Q: What is the log-of-a-power law?

An exponent trapped inside a log comes down in front: ln(aʳ) = r·ln a (for a > 0). For example, ln(x³) = 3 ln x.

Q: Is ln(x²) equal to 2 ln x?

Only for x > 0. The safe form is ln(x²) = 2 ln|x| (x ≠ 0): at x = −3, ln 9 = 2 ln 3, but ln(−3) is undefined. Forgetting the absolute value is the exam trap.

Exponents trapped inside a log come down in front — the key move for differentiating x^x-type monsters.

Notation on this page: ln is the natural log; a > 0 and r is any real number.

Before this lesson: Log of a product

Where it comes from

Expressions like ln(x²), ln(2^x), ln(√x) trap an exponent inside the log, where the derivative can't reach it cleanly. The reflex is to apply the exponent to the log itself:

ln(a r) = (ln a) r ?? the tempting — and wrong — guess

Kill it with a = e, r = 2:

ln(e²)

The true value: ln undoes the power of e. Answer 2.

(ln e)²

1² = 1

The naive guess gives 1. 2 ≠ 1 — dead on arrival.

Here is the intuition. For an integer exponent, a power is just repeated multiplication — and we already know what ln does to products:

ln(a³)

ln(a · a · a)

a³ is three copies multiplied.

ln a + ln a + ln a = 3 · ln a

The product law, three times: each copy contributes one ln a. The exponent counts the copies — so it becomes a multiplier.

This law is the product law applied r times — all three log laws are one family. And it has a killer application: it is the only way to differentiate functions like x^x, where the variable is both base and exponent. We'll do that in Example 4.

Before reading on: ln(a³) = ln(a·a·a). Apply the product law three times — where does the 3 end up?

Derivation

The integer case is the product law repeated. The general case needs one definition: for a > 0, a^r := e^r·ln a (real powers are defined through e^x).

ln(aⁿ)

ln(a·…·a) = ln a + … + ln a

Step 1 — integer n. aⁿ is n copies multiplied; the product law splits it into n copies of ln a added.

n · ln a

Step 2 — count. n copies of ln a added = n·ln a. Integer case proved.

ln(a^r)

ln(e^r·ln a) = r · ln a

Step 3 — key step: real r. Replace a^r by its definition e^r·ln a, then the inverse pair collapses ln(e^z) to z. One substitution, one cancellation — the whole extension. ∎

The absolute-value trap. The law needs a > 0. So ln(x²) = 2·ln x is wrong for negative x — at x = −3, ln 9 = 2·ln 3 works, but ln(−3) is undefined. The safe form: ln(x²) = 2·ln|x| (x ≠ 0). Examiners love this trap; now it can't catch you.

Before reading on: y = x^x: the exponent is trapped where the power rule can’t reach it. What function could pull it down in front?

How to use it

The procedure, every time:

Confirm the exponent is inside the log. ln(x⁵) ✓. (ln x)⁵ ✗ — the exponent is outside; nothing comes down. ln(5x) ✗ — that is the product law.
Check the base is positive (or wrap in |·|). ln(x²) → 2·ln|x|.
Bring it down: ln(x⁵) = 5·ln x (x > 0). The exponent becomes a plain multiplier.
Use it before differentiating. d/dx [ln(x⁴)] becomes d/dx [4·ln x] = 4/x — the chain rule gives (1/x⁴)·4x³ = 4/x too, but the bring-down is one line.

The killer app: logarithmic differentiation

For y = x^x (x > 0), neither the power rule (exponent isn't constant) nor the exponential rule (base isn't constant) applies. Take ln of both sides and the exponent comes down:

ln y = x \cdot ln x \Rightarrow y'/y = ln x + 1 \Rightarrow y' = x x (ln x + 1) the exponent-down move makes the undifferentiable differentiable Say it: take the log, differentiate implicitly, then solve for y prime

Roots are powers too

ln(√x) = ln(x^1/2) = ½·ln x (x > 0). Any root is a fractional exponent — bring it down the same way. And ln(e^3x) = 3x·ln e = 3x (or straight from the inverse pair).

Common mistake: writing ln(x²) = 2·ln x and then evaluating at x = −3 to get 2·ln(−3) — undefined. The correct move was 2·ln|−3| = 2·ln 3 = ln 9. ✓ Always ask: “can my base be negative?”

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: ln(2³)

Confirm: exponent inside the log, base positive. ✓
Bring it down: ln(2³) = 3·ln 2.
Check. ln 8 ≈ 2.0794; 3 · 0.6931 = 2.0794. ✓

Common mistake: (ln 2)³ ≈ 0.333 ≠ 2.079. The exponent multiplies the log — it never becomes the log's own exponent.

Your turn: Simplify ln(5⁴).

Answer: 4 · ln 5.

Bring the exponent down: 4·ln 5. Check: ln 625 ≈ 6.438 = 4 · 1.609 ✓.

Example 2 — roots: ln(√x), x > 0

Rewrite the root as a power: √x = x^1/2. (Why? Fractional exponents are roots — same object.)
Bring it down: ln(x^1/2) = ½·ln x.
Check at x = 4. ln 2 ≈ 0.6931; ½ · ln 4 = ½ · 1.3863 = 0.6931. ✓

Common mistake: ln(√x) = √(ln x). The ½ comes down as a multiplier, not as a root of the log. At x = 4: √(ln 4) ≈ 1.177 ≠ 0.693.

Your turn: Simplify ln(∛√x), x > 0.

Answer: (1/3) · ln x.

Roots are fractional powers: ∛√x = x^1/3, so ln gives (1/3) ln x ✓.

Example 3 — differentiate smart: d/dx [ln(1/x²)]

Rewrite: 1/x² = x⁻², so ln(1/x²) = ln(x⁻²).
Bring it down: = −2·ln x (x > 0).
Differentiate: d/dx [−2·ln x] = −2/x.
Check via chain rule. (x²)·(−2x⁻³) = −2/x. ✓ Same — but the bring-down path never juggles negative powers inside a derivative.

Common mistake: bringing down −2 and then writing d/dx [−2·ln x] = −2 (treating −2·ln x as a constant times… nothing). The ln x is still a function of x — its derivative is 1/x, so the answer keeps a /x.

Your turn: Differentiate d/dx [ln(1/x³)], x > 0.

Answer: −3/x.

Rewrite: ln(x⁻³) = −3 ln x; differentiate to −3/x. (Chain-rule check: x³ · (−3x⁻⁴) = −3/x ✓.)

Example 4 — logarithmic differentiation: y = x^x, x > 0

Take ln of both sides. ln y = ln(x^x). (Why? The exponent x is trapped inside — ln is the tool that frees it.)
Bring the exponent down. ln y = x·ln x.
Differentiate implicitly. Left: (1/y)·y′ (chain rule). Right: product rule — 1·ln x + x·(1/x) = ln x + 1. So y′/y = ln x + 1.
Solve for y′. y′ = x^x(ln x + 1).
Sanity check at x = 1. y′(1) = 1¹·(0+1) = 1. Numerically: 1.01^1.01 ≈ 1.0101, so the slope near 1 is ≈ (1.0101−1)/0.01 ≈ 1.01 ≈ 1. ✓

Common mistake: applying the power rule: d/dx [x^x] = x·x^x−1 = x^x. At x = 1 that gives 1 — accidentally right! — but at x = 2 it gives 4 while the true derivative is 2²·(ln 2 + 1) ≈ 6.77. The power rule needs a constant exponent.

Your turn: Find y′ for y = x^2x, x > 0.

Answer: y′ = x^2x(2 ln x + 2).

ln y = 2x ln x; implicit differentiation: y′/y = 2 ln x + 2x·(1/x) = 2 ln x + 2; multiply by y.

Memorization tips

Point and drag: put your finger on the exponent inside the log and drag it down in front. The gesture is the law.
The integer anchor: ln(a³) = ln(a·a·a) = ln a + ln a + ln a. If you ever blank, rebuild from three copies.
One family: product law (× → +), quotient law (÷ → −), power law (exponent → ×). The power law is just the product law applied r times.
The |x| trap: ln(x²) = 2·ln|x|, not 2·ln x. Examiners set this trap on purpose — the absolute value is the whole question.
Self-check: ln(e²) = 2 = 2·1. If your rule doesn't give 2 here, it's wrong.
The killer app: see x^x (or anything^{something-with-x}) → take ln, bring the exponent down. That reflex alone is worth marks.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the log of a power is yours.

← Back to the Calculus I formula sheet

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 log-of-a-power law?

An exponent trapped inside a log comes down in front: ln(a^r) = r·ln a (for a > 0). For example, ln(x³) = 3·ln x.

Why isn’t ln(a^r) equal to (ln a)^r?

Test a = e, r = 2: ln(e²) = 2, but (ln e)² = 1² = 1. The exponent comes down as a multiplier; it is not applied to the log.

Why does the exponent come down?

For integers it’s the product law applied r times: ln(a³) = ln(a·a·a) = ln a + ln a + ln a = 3·ln a. The general case follows from writing a^r = e^r·ln a.

Is ln(x²) equal to 2·ln x?

Only for x > 0. The safe form is ln(x²) = 2·ln|x| (x ≠ 0): at x = −3, ln 9 = 2·ln 3, but ln(−3) is undefined. Forgetting the absolute value is the exam trap.

How do you differentiate x^x?

Logarithmic differentiation: ln y = x·ln x (bring the exponent down), then y′/y = ln x + 1, so y′ = x^x(ln x + 1). Neither the power rule nor the exponential rule applies directly.

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Log of a power

Where it comes from

Derivation

How to use it

The killer app: logarithmic differentiation

Roots are powers too

Worked examples

Example 1 — the basic move: ln(2³)

Example 2 — roots: ln(√x), x > 0

Example 3 — differentiate smart: d/dx [ln(1/x²)]

Example 4 — logarithmic differentiation: y = xx, x > 0

Memorization tips

Final challenge

How to learn a formula here

Frequently asked questions

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Example 4 — logarithmic differentiation: y = x^x, x > 0