Calculus I › Differentiation rules › full formula sheet
The derivative of loga x
Any base’s log differentiates to 1/(x ln a) — the change-of-base formula turns it into the ln rule you already know.
Notation: a > 0, a ≠ 1 is the fixed base (2, 10, …). loga x is “the power a needs to make x.”
Before this lesson: Derivative of ln x · Log of a power
Where it comes from
Every logarithm is the natural log in disguise — the change-of-base formula:
Before reading on: loga x = ln x / ln a, and ln a is just a number. If d/dx [ln x] = 1/x, what must d/dx [loga x] be — and why is the ln a in the denominator, not the numerator?
Differentiate: d/dx [ln x / ln a] = (1/ln a)·d/dx [ln x] = (1/ln a)·(1/x) = 1/(x ln a). The ln a is a toll every non-e base pays — the price of not being natural.
The wrong guess to kill: d/dx [log2 x] = 1/x (forgetting the toll)?? At x = 4: the guess gives 0.25; the truth is 1/(4 ln 2) ≈ 0.36. Dead on arrival — base 2’s log grows faster than ln (smaller base, steeper log), so its slope must be bigger than 1/x, not equal.
Derivation
Change of base, then the ln rule — two lines:
Collapse check: for a = e, ln a = ln e = 1, giving 1/x — exactly the ln rule. The general formula contains the natural one.
How to use it
The procedure:
- Plain loga x → 1/(x ln a). Copy 1/x, append the toll ÷ln a.
- Something inside? Chain: d/dx [loga(u)] = u′/(u ln a). Example: d/dx [log2(3x)] = 3/(3x ln 2) = 1/(x ln 2).
- Simplify with log laws first: log5(x²) = 2 log5 x → 2/(x ln 5).
- Base between 0 and 1? ln a < 0 flips the sign — log1/2 x decreases, so negative slopes are correct.
Judgment calls
Memorize via ln: if you blank on the formula, write loga x = ln x/ln a and differentiate — 15 seconds, zero memorization. Common in computer science: log2 appears in algorithm analysis; its derivative 1/(x ln 2) shows up in optimization proofs.
Worked examples
Four problems, easiest first. The toll ln a is the main character.
Example 1 — d/dx [log2 x]
- Copy 1/x, append the toll: = 1/(x ln 2) = 1/(x ln 2).
- Check vs ln: ln 2 ≈ 0.693 < 1, so 1/(x ln 2) > 1/x — base-2 log is steeper than ln, as it should be. ✓
Your turn: d/dx [log3 x]
Answer: 1/(x ln 3)
Copy 1/x, append the toll: 1/(x ln 3). Check vs ln: ln 3 ≈ 1.099 > 1, so 1/(x ln 3) < 1/x — base-3 log is shallower than ln, as it should be.
Example 2 — d/dx [log10(3x)]
- Chain: = d/dx [3x] / (3x ln 10) = 3/(3x ln 10).
- Simplify: = 1/(x ln 10). (Why does the 3 vanish? Same scaling-cancels story as ln(5x).)
Your turn: d/dx [log2(7x)]
Answer: 1/(x ln 2)
Chain: d/dx [7x]/(7x ln 2) = 7/(7x ln 2) = 1/(x ln 2). The 7 cancels — same scaling-cancels story as ln(5x).
Example 3 — d/dx [x·log2 x] (product)
- Structure: multiplied — product rule: f = x, g = log2 x.
- Derivatives: f′ = 1, g′ = 1/(x ln 2).
- Assemble: = log2 x + x·[1/(x ln 2)] = log2 x + 1/ln 2.
Your turn: d/dx [x·log3 x] (product)
Answer: log3 x + 1/ln 3
Product rule: f = x, g = log3 x. f′ = 1, g′ = 1/(x ln 3). Assemble: log3 x + x·[1/(x ln 3)] = log3 x + 1/ln 3.
Example 4 — d/dx [log5(x²)] two ways
Path A — log laws first: log5(x²) = 2 log5 x, so d/dx = 2/(x ln 5).
Path B — chain: = 2x/(x² ln 5) = 2/(x ln 5). Same ✓
- Answer: 2/(x ln 5).
Your turn: d/dx [log4(x³)] two ways
Answer: 3/(x ln 4)
Path A — log laws first: log4(x³) = 3 log4 x, so d/dx = 3/(x ln 4). Path B — chain: 3x²/(x³ ln 4) = 3/(x ln 4). Same.
Memorization tips
- Change-of-base is the whole rule: loga x = ln x/ln a. Blank on the formula? Derive it in 15 seconds.
- The toll is ln(base): 1/(x ln a) — ln of the base, never of x. Mis-aiming the log is the #1 error.
- Base e collapses: ln e = 1 gives back 1/x. If your general formula doesn’t, it’s wrong.
- Smaller base, steeper log: ln 2 < 1 makes 1/(x ln 2) > 1/x. Sanity-check the size, not just the shape.
- Chains: u′/(u ln a): “inside over inside, over ln(base).” Three pieces, one breath.
- Log laws first: loga(xn) = n loga x dodges the chain. Simplify, then differentiate.
Final challenge
Five mixed questions — the ln a toll, chains, and base-e collapse. 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 log_a x?
d/dx[log_a x] = 1/(x·ln a). Proof: change of base gives log_a x = ln x/ln a; the constant 1/ln a pulls out, and d/dx[ln x] = 1/x.
Why does ln a appear?
It’s the change-of-base constant: log_a x = (1/ln a)·ln x. Differentiating, the 1/ln a rides along via the constant multiple rule. Every non-e base pays this “toll.”
What’s the most common mistake?
Writing 1/(x·ln x) — putting x inside the log instead of the base a. The toll is always ln(base): ln 2 for log_2, ln 10 for log_10.
What is d/dx[log_2(3x)]?
1/(x·ln 2). Chain: 3/(3x·ln 2) = 1/(x·ln 2) — the 3 cancels, exactly like ln(5x) → 1/x.
Does the formula work for a = e?
Yes: ln e = 1, so 1/(x·ln e) = 1/x — the ln rule. The general formula contains the natural one as a special case.
More from the codex
Support the codex
This page is free, with no account and no ads. If it helped you learn, consider supporting the indie dev behind it.
Questions or a bug to report? Email [email protected].