Derivative of arcsin(x)
Arcsine trades its trig identity for a square root — and its derivative encodes exactly why the function only lives on the interval [−1, 1].
Step-by-step solution
Let y = arcsin(x), so sin(y) = x with y restricted to [−π/2, π/2].
cos(y) · y′ = 1, hence y′ = 1/cos(y).
On the restricted range, cos(y) is non-negative, so cos(y) = √(1 − sin²(y)) = √(1 − x²), giving y′ = 1/√(1 − x²).
Why it works
Look at what the formula does at the edges: as x approaches ±1, the denominator √(1 − x²) collapses to zero and the slope blows up to infinity. That's the graph of arcsine going vertical at its endpoints — the inverse-function mirror of sine going horizontal at its peaks. Steep places on a function become flat places on its inverse, and vice versa.
The restriction of y to [−π/2, π/2] in step 3 isn't pedantry; it's what lets us drop the ± when taking the square root of cos²(y). Choose a different branch of the inverse and the sign — and the derivative — changes. Inverse trig functions are only functions at all because of these branch choices, and every derivative formula silently depends on them.
Common mistakes
- Writing a minus sign under the root as (x² − 1) — the formula is 1 − x², and the order matters since the domain is |x| < 1.
- Confusing this with arccosine's derivative, which is the same expression with a leading minus: −1/√(1 − x²).
- Treating arcsin(x) as 1/sin(x) = csc(x) — the inverse-vs-reciprocal notation trap again.
Practice problems
Differentiate arcsin(x/2)
Answer: 1/√(4 − x²) — chain rule, then simplify.
Find the slope of arcsin(x) at x = 0
Answer: 1/√1 = 1.
Differentiate arcsin(x) + arccos(x)
Answer: 0 — the two derivatives cancel, because the sum is the constant π/2.