Derivatives of inverse trig functions

Inverse trigonometric functions are continuous on their domains, and they are differentiated through implicit differentiation of the matching trigonometric relation.

Worked example: y = sin⁻¹x means x = sin y with −π/2 ≤ y ≤ π/2. Differentiating, 1 = cos y · dy/dx, so dy/dx = 1/cos y.

In that range cos y ≥ 0, so cos y = √(1 − sin²y) = √(1 − x²), giving d/dx(sin⁻¹x) = 1/√(1 − x²) for −1 < x < 1. It fails at ±1, where cos y = 0.

In the same way d/dx(cos⁻¹x) = −1/√(1 − x²) on (−1, 1), and d/dx(tan⁻¹x) = 1/(1 + x²) for every real x.

Memory hook: sin⁻¹x + cos⁻¹x = π/2 is a constant, so their derivatives must add to 0, which is why they differ only in sign.

Combine with the chain rule: d/dx tan⁻¹(2x) = 2/(1 + 4x²).

A substitution can simplify first. For x in (−1, 1), sin⁻¹(2x/(1 + x²)) = 2 tan⁻¹x when |x| ≤ 1, so its derivative is 2/(1 + x²).