Calculus I › Differentiation rules › full formula sheet
The derivative of ln x
The natural log’s derivative is 1/x — proved by flipping the exponential inside out with implicit differentiation.
Notation: ln x is the natural log (base e), defined for x > 0. d/dx[ln(u)] = u′/u when there’s something inside.
Before this lesson: Implicit differentiation · Derivative of ex
Where it comes from
We need the derivative of the inverse of ex. The trick: write y = ln x as x = ey, then differentiate implicitly — treating y as a function of x:
Before reading on: y = ln x means x = ey. Differentiate both sides with respect to x — where does dy/dx come from, and what do you get when you solve for it?
The wrong guess to kill: d/dx [ln x] = 1/ln x?? At x = e: the guess gives 1/ln e = 1; the truth is 1/e ≈ 0.368. Dead on arrival — and dimensionally absurd (the derivative of a log shouldn’t contain a log).
Derivation
The implicit proof above, stated cleanly. (A limit-definition proof exists too — see the hint.)
Via the definition: [ln(x+h)−ln x]/h = (1/h)·ln(1+h/x) = (1/x)·ln((1+h/x)x/h) → (1/x)·ln e = 1/x, using e = limn→∞(1+1/n)n. Same answer, longer road.
How to use it
The procedure:
- Plain ln x → 1/x.
- Something inside? Chain: d/dx [ln(u)] = u′/u — “the derivative of the inside, over the inside.” Example: d/dx [ln(5x)] = 5/(5x) = 1/x.
- For x < 0 use ln|x|: d/dx [ln|x|] = 1/x too (chain: (1/|x|)·(|x|/x)… — the signs cancel). So the formula 1/x works for all x ≠ 0.
- Simplify with log laws first: ln(x²) = 2 ln x differentiates to 2/x — same as the chain gives (2x/x²), with less pain.
Judgment calls
ln(5x) → 1/x, not 1/(5x). The chain’s 5 cancels: 5/(5x) = 1/x. Shifting or scaling inside a log never changes the derivative’s shape — a favorite exam trap.
Worked examples
Four problems, easiest first. “Inside over inside” — then simplify.
Example 1 — d/dx [ln(5x)] (the trap)
- Chain: = d/dx [5x] / (5x) = 5/(5x).
- Simplify: = 1/x. (Why does the 5 vanish? Scaling x before logging just shifts the graph — slopes don’t care.)
Your turn: d/dx [ln(8x)] (the trap)
Answer: 1/x
Chain: d/dx [8x]/(8x) = 8/(8x) = 1/x. The 8 cancels — scaling before logging never changes the derivative.
Example 2 — d/dx [x² ln x] (product)
- Structure: multiplied — product rule: f = x², g = ln x.
- Derivatives: f′ = 2x, g′ = 1/x.
- Assemble: = 2x ln x + x²·(1/x) = 2x ln x + x = x(2 ln x + 1).
Your turn: d/dx [x³ ln x] (product)
Answer: x²(3 ln x + 1)
Product rule: f = x³, g = ln x. f′ = 3x², g′ = 1/x. Assemble: 3x² ln x + x³·(1/x) = 3x² ln x + x² = x²(3 ln x + 1).
Example 3 — d/dx [ln(x²+1)]
- Chain: = d/dx [x²+1] / (x²+1) = 2x/(x²+1).
- Answer: 2x/(x²+1). (No simplification available — and none needed.)
Your turn: d/dx [ln(4x²)]
Answer: 2/x
Chain: d/dx [4x²]/(4x²) = 8x/(4x²) = 2/x. (Or log laws: ln 4 + 2 ln x → 0 + 2/x. Same.)
Example 4 — d/dx [(ln x)/x] (quotient)
- Structure: divided — quotient rule: f = ln x, g = x.
- Derivatives: f′ = 1/x, g′ = 1.
- Chant: = [(1/x)(x) − (ln x)(1)]/x² = (1 − ln x)/x².
Your turn: d/dx [x/ln x] (quotient, flipped)
Answer: (ln x − 1)/(ln x)²
Quotient rule: f = x, g = ln x. f′ = 1, g′ = 1/x. Chant: [(1)(ln x) − x(1/x)]/(ln x)² = (ln x − 1)/(ln x)².
Memorization tips
- “Inside over inside”: d/dx [ln(u)] = u′/u. Say it as you write — numerator is the inside’s derivative.
- ln(kx) → 1/x: scaling inside a log always cancels. If your answer still has the k, the chain factor died.
- Inverse-pair check: ey = x differentiated implicitly gives 1/x. If the formula slips, re-derive from the inverse in 30 seconds.
- Log laws first: ln(x²) = 2 ln x → 2/x beats chaining through 2x/x². Simplify, then differentiate.
- ln|x| covers negatives: d/dx [ln|x|] = 1/x for all x ≠ 0. The absolute value’s chain signs cancel out.
- Never 1/ln x: the derivative of a log is algebraic (1/x), not logarithmic. If a log survives in your answer, recheck.
Final challenge
Five mixed questions — the 1/x shape, chains, and the ln(5x) trap. Score 5/5 and it’s 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 derivative of ln x?
d/dx[ln x] = 1/x. Proof: write y = ln x as x = e^y, differentiate implicitly to get 1 = e^y·dy/dx, so dy/dx = 1/e^y = 1/x.
What is d/dx[ln(5x)]?
1/x — not 1/(5x). Chain rule: 5/(5x) = 1/x. Scaling inside a logarithm never survives; it only shifts the graph.
How does the chain rule work with ln?
d/dx[ln(u)] = u′/u: “the derivative of the inside, over the inside.” Example: d/dx[ln(x²+1)] = 2x/(x²+1).
What about ln|x| for negative x?
d/dx[ln|x|] = 1/x as well, for all x ≠ 0. The absolute value’s extra chain signs cancel, so one formula covers both sides.
Can I prove it from the limit definition?
Yes: [ln(x+h)−ln x]/h = (1/x)·ln((1+h/x)^(x/h)) → (1/x)·ln e = 1/x, using e = lim(n→∞)(1+1/n)ⁿ. Longer than the implicit proof, same answer.
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