Calculus I › Differentiation rules › full formula sheet
The derivative of arccos x
The inverse cosine’s derivative is negative 1/√(1−x²) — arcsin’s mirror image, with the cofunction minus.
Notation: arccos x is the angle in [0, π] whose cosine is x. Defined for |x| ≤ 1; the derivative for |x| < 1.
Before this lesson: Derivative of arcsin x · Implicit differentiation
Where it comes from
Write y = arccos x as cos y = x and differentiate implicitly:
Before reading on: arcsin x + arccos x = π/2 — a constant. If arcsin’s derivative is 1/√(1−x²), what must arccos’s derivative be — and what in the implicit computation produces the minus?
The graph demands the minus: arccos falls from π to 0 as x goes from −1 to 1 — at x = 0 its slope is −1. The no-minus guess +1/√(1−x²) gives +1 at 0 ✗. Dead on arrival.
Mirror check: arcsin x + arccos x = π/2 (constant!), so their derivatives must be negatives: 1/√(1−x²) + (−1/√(1−x²)) = 0 ✓. If your arccos formula doesn’t negate arcsin’s, it’s wrong.
Derivation
The implicit proof, stated cleanly. Compare each step with arcsin’s — the only difference is cosine’s minus:
How to use it
The procedure:
- Plain arccos x → −1/√(1−x²). Minus first — it’s a cofunction.
- Something inside? Chain: d/dx [arccos(u)] = −u′/√(1−u²). Example: d/dx [arccos(2x)] = −2/√(1−4x²).
- Watch the domain: |u| < 1 — vertical tangents at ±1, same as arcsin.
- Mirror check: your answer should be exactly arcsin’s answer negated. If not, find the dropped minus.
Judgment calls
Prefer arcsin when free: arccos x = π/2 − arcsin x, so any arccos problem converts — but the direct rule is one step. Decreasing means negative: arccos falls left-to-right; if your derivative comes out positive anywhere, the minus died.
Worked examples
Four problems, easiest first. Minus first, mirror of arcsin.
Example 1 — d/dx [arccos(2x)]
- Layers: outer arccos u, inner u = 2x.
- Outside: −1/√(1−(2x)²) = −1/√(1−4x²).
- Inside’s derivative: 2. Multiply: = −2/√(1−4x²), for |x| < 1/2.
Your turn: d/dx [arccos(3x)]
Answer: −3/√(1−9x²), for |x| < 1/3
Layers: outer arccos u, inner u = 3x. Outside: −1/√(1−(3x)²) = −1/√(1−9x²). Inside’s derivative: 3. Multiply: −3/√(1−9x²).
Example 2 — d/dx [x·arccos x] (product)
- Structure: multiplied — product rule: f = x, g = arccos x.
- Derivatives: f′ = 1, g′ = −1/√(1−x²).
- Assemble: = arccos x − x/√(1−x²).
Your turn: d/dx [x²·arccos x] (product)
Answer: 2x arccos x − x²/√(1−x²)
Product rule: f = x², g = arccos x. f′ = 2x, g′ = −1/√(1−x²). Assemble: 2x arccos x + x²(−1/√(1−x²)).
Example 3 — d/dx [arccos(1−x)]
- Layers: outer arccos u, inner u = 1−x.
- Outside: −1/√(1−(1−x)²). Inside’s derivative: −1.
- Multiply: (−1)·(−1)/√(1−(1−2x+x²)) = 1/√(2x−x²). (Why positive? Two minuses — the rule’s and the chain’s — make a plus.)
Your turn: d/dx [arccos(x−1)]
Answer: −1/√(2x−x²)
Layers: outer arccos u, inner u = x−1. Outside: −1/√(1−(x−1)²). Inside’s derivative: 1. Simplify: 1−(x−1)² = 2x−x², so −1/√(2x−x²).
Example 4 — the mirror check: d/dx [arcsin x + arccos x]
- Term by term: 1/√(1−x²) + (−1/√(1−x²)).
- Sum: = 0.
- Why: arcsin x + arccos x = π/2, a constant — and constants differentiate to 0. The formulas confirm each other ✓
Your turn: The mirror check at x = 1/2: d/dx [arcsin x + arccos x]
Answer: 0
Term by term: 1/√(1−1/4) − 1/√(1−1/4) = 2/√3 − 2/√3 = 0. The sum is the constant π/2 — the formulas confirm each other.
Memorization tips
- arcsin’s mirror, negated: arccos′ = −arcsin′. If you know one, you know both — just flip the sign.
- The π/2 detector: arcsin + arccos = π/2, so their derivatives sum to 0. A two-second consistency check.
- Cofunction → minus: arccos starts with “arc-co” — write the minus first, like cos, cot, csc.
- Flip and chain: cos y = x → −sin y·y′ = 1. Re-derive live in 30 seconds if you blank.
- Range picks the root’s sign: y ∈ [0, π] → sin y ≥ 0 → positive root; the minus out front survives.
- Decreasing → negative: arccos falls left to right. Positive derivative anywhere = dropped minus.
Final challenge
Five mixed questions — the mirror minus, chains, and the π/2 sum. 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 arccos x?
d/dx[arccos x] = −1/√(1−x²), for |x| < 1. Proof: y = arccos x means cos y = x; implicit differentiation gives −sin y·dy/dx = 1, so dy/dx = −1/sin y = −1/√(1−x²).
Why the minus sign?
Two reasons: the graph (arccos falls, so slopes are negative) and the proof (cosine’s derivative −sin y seeds it). Also arcsin x + arccos x = π/2 is constant, so their derivatives must be negatives of each other.
How is it related to arcsin’s derivative?
It’s exactly the negation: arcsin′ = +1/√(1−x²), arccos′ = −1/√(1−x²). Same root, opposite signs — learn them as a signed pair.
What is d/dx[arccos(2x)]?
−2/√(1−4x²), for |x| < 1/2. Chain: −1/√(1−(2x)²) times the inside’s derivative 2.
What happens at x = ±1?
The derivative blows up to −∞: vertical tangents at (±1, 0/π). Same as arcsin — the formula predicts infinite steepness, with the negative sign.
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