Calculus I › Differentiation rules › full formula sheet
The derivative of arcsin x
The inverse sine’s derivative is 1/√(1−x²) — implicit differentiation on sin y = x, with the sign settled by the range.
Notation: arcsin x is the angle in [−π/2, π/2] whose sine is x. Defined for |x| ≤ 1; the derivative for |x| < 1.
Before this lesson: Implicit differentiation · Derivative of sin x
Where it comes from
We need the derivative of an inverse trig function. Write y = arcsin x as sin y = x, then differentiate implicitly — the same flip trick as the ln proof:
Before reading on: y = arcsin x means sin y = x. Differentiate both sides — you’ll get dy/dx = 1/cos y. How do you turn cos y back into x, and why is the sign unambiguous?
The wrong guess to kill: d/dx [arcsin x] = −1/√(1−x²)?? At x = 0, arcsin crosses at 45° — slope +1. The minus version gives −1 ✗. Dead on arrival — arcsin rises, so its derivative is positive.
Derivation
The implicit proof, stated cleanly. The crux is the sign of the square root:
Why the range matters: without restricting arcsin to [−π/2, π/2], cos y could be negative and the sign would be ambiguous. The principal range isn’t bureaucracy — it’s what makes the formula well-defined.
How to use it
The procedure:
- Plain arcsin x → 1/√(1−x²).
- Something inside? Chain: d/dx [arcsin(u)] = u′/√(1−u²). Example: d/dx [arcsin(2x)] = 2/√(1−4x²).
- Watch the domain: |u| < 1 required — at |u| = 1 the denominator is zero (vertical tangent; arcsin has a cusp-like steepness at ±1).
- Don’t confuse with arctan: arcsin′ = 1/√(1−x²) vs. arctan′ = 1/(1+x²). Root vs. no root; minus vs. plus.
Judgment calls
arcsin(2x) needs |2x| < 1, i.e. |x| < 1/2 — the chain doesn’t just change the formula, it shrinks the domain. Memorize via the flip: sin y = x → cos y·y′ = 1 — re-derivable in 30 seconds.
Worked examples
Four problems, easiest first. The chain rule does the heavy lifting.
Example 1 — d/dx [arcsin(2x)]
- Layers: outer arcsin u, inner u = 2x.
- Outside: 1/√(1−(2x)²) = 1/√(1−4x²). (Why (2x)²? The whole inside gets squared.)
- Inside’s derivative: 2. Multiply: = 2/√(1−4x²), for |x| < 1/2.
Your turn: d/dx [arcsin(5x)]
Answer: 5/√(1−25x²), for |x| < 1/5
Layers: outer arcsin u, inner u = 5x. Outside: 1/√(1−(5x)²) = 1/√(1−25x²). Inside’s derivative: 5. Multiply: 5/√(1−25x²).
Example 2 — d/dx [x·arcsin x] (product)
- Structure: multiplied — product rule: f = x, g = arcsin x.
- Derivatives: f′ = 1, g′ = 1/√(1−x²).
- Assemble: = arcsin x + x/√(1−x²).
Your turn: d/dx [x²·arcsin x] (product)
Answer: 2x arcsin x + x²/√(1−x²)
Product rule: f = x², g = arcsin x. f′ = 2x, g′ = 1/√(1−x²). Assemble: 2x arcsin x + x²/√(1−x²).
Example 3 — d/dx [arcsin(√x)] (nested)
- Layers: outer arcsin u, inner u = √x.
- Outside: 1/√(1−(√x)²) = 1/√(1−x).
- Inside’s derivative: 1/(2√x). Multiply: = 1/[2√x·√(1−x)] = 1/(2√(x−x²)), for 0 < x < 1.
Your turn: d/dx [arcsin(x²)]
Answer: 2x/√(1−x⁴), for |x| < 1
Layers: outer arcsin u, inner u = x². Outside: 1/√(1−(x²)²) = 1/√(1−x⁴). Inside’s derivative: 2x. Multiply: 2x/√(1−x⁴).
Example 4 — sanity check at x = 0
- Formula: 1/√(1−0) = 1.
- Geometry: arcsin crosses the origin at 45° (it’s the reflection of sin across y = x, and sin has slope 1 at 0). Matches ✓
- Near x = 1: 1/√(1−x²) → ∞ — the vertical tangent at (1, π/2). The formula predicts the steepness ✓
Your turn: Sanity check at x = √3/2
Answer: 2
Formula: 1/√(1−3/4) = 1/√(1/4) = 1/(1/2) = 2. Steeper than the slope 1 at 0 — the formula predicts the steepening toward the vertical tangent at x = 1.
Memorization tips
- Flip and chain: sin y = x → cos y·y′ = 1. The whole proof in one line — re-derive it live if you blank.
- Range picks the sign: y ∈ [−π/2, π/2] → cos y ≥ 0 → positive root. The range isn’t trivia; it’s the sign.
- Plus root, minus inside: 1/√(1−x²). Say it: “plus on the fraction, minus under the root.”
- Square the whole inside: arcsin(2x) → 1−(2x)² = 1−4x². The #1 algebra slip is 1−2x².
- Domain shrinks with chains: arcsin(2x) needs |x| < 1/2. Check |inside| < 1, not |x| < 1.
- arcsin vs arctan: 1/√(1−x²) vs 1/(1+x²). Root+minus vs no-root+plus — learn them as a contrasting pair.
Final challenge
Five mixed questions — the root’s sign, chains, and the arccos mirror. 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 arcsin x?
d/dx[arcsin x] = 1/√(1−x²), for |x| < 1. Proof: write y = arcsin x as sin y = x, differentiate implicitly to get cos y·dy/dx = 1, so dy/dx = 1/cos y = 1/√(1−x²).
Why is the square root positive?
Because arcsin’s range is [−π/2, π/2], where cosine is nonnegative. So cos y = +√(1−sin²y) = +√(1−x²) — the range settles the sign.
What is d/dx[arcsin(2x)]?
2/√(1−4x²), for |x| < 1/2. Chain rule: 1/√(1−(2x)²) times the inside’s derivative 2. Note (2x)² = 4x², and the domain shrinks.
How is it different from arctan’s derivative?
arcsin′ = 1/√(1−x²) (root, minus inside) vs. arctan′ = 1/(1+x²) (no root, plus). Different inverses, different shapes — learn them as a contrasting pair.
What happens at x = ±1?
The derivative blows up: 1/√(1−x²) → ∞. Geometrically, arcsin has vertical tangents at (±1, ±π/2) — the formula predicts the infinite steepness.
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