Short on time? This page is the recap. For the derivations, the reference triangles and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: arcsin x = sin⁻¹ x (an angle in radians), which is not 1/sin x.
Recap
- Each inverse trig function undoes a restricted trig function: sin on [−π/2, π/2], cos on [0, π], tan on (−π/2, π/2).
- Derivation pattern: y = arcsin x means sin y = x. Differentiate implicitly: cos y · y′ = 1, so y′ = 1/cos y = 1/√(1 − x²).
- The root is positive because cos y ≥ 0 when −π/2 ≤ y ≤ π/2.
- arctan: tan y = x gives sec²y · y′ = 1, and sec²y = 1 + x².
- arccos: from arcsin x + arccos x = π/2, its derivative is the negative of arcsin’s.
- arcsin and arccos are not differentiable at x = ±1 (vertical tangents). arctan is differentiable everywhere.
Key relationships
| Function | Derivative | Chain-rule form (u a function of x) |
|---|---|---|
| arcsin x | 1/√(1 − x²), −1 < x < 1 | u′/√(1 − u²) |
| arccos x | −1/√(1 − x²), −1 < x < 1 | −u′/√(1 − u²) |
| arctan x | 1/(1 + x²), all x | u′/(1 + u²) |
| Inverse-function rule (3.3) | g′(x) = 1/f′(g(x)) | e.g. arcsin′(x) = 1/cos(arcsin x) |
Useful exact values: arcsin(1/2) = π/6, arccos(1/2) = π/3, arctan 1 = π/4, arctan 0 = 0.
Assumptions behind the formulas
- Angles are in radians.
- The standard ranges are used (they fix the sign of the square root).
- For arcsin u or arccos u, the inner value u must satisfy −1 < u < 1 for the derivative to exist.
Mistakes to avoid
- Treating sin⁻¹ x as 1/sin x.
- Forgetting the minus sign on arccos.
- Missing the chain-rule factor u′.
- Squaring only part of u: arcsin(3x) gives √(1 − 9x²), not √(1 − 3x²).
- Putting a square root in the arctan formula, or leaving it out of arcsin.
- Using degrees for exact values.
Quick self-check
- Differentiate arcsin(3x). (3/√(1 − 9x²))
- Find the derivative of arctan(x³) at x = 1. (3x²/(1 + x⁶) at x = 1 is 3/2)
- Find the derivative of arccos x at x = 0. (−1)
- Is arcsin x differentiable at x = 1? (No. The denominator √(1 − x²) is 0 there, and the graph has a vertical tangent.)
Next: practice questions.