CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03

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Principal arcsine has no finite derivative at $- 1$ or $1$

Example

The principal arcsine $\arcsin : [- 1, 1] \to [- π / 2, π / 2]$ has no finite derivative at either endpoint $1$ or $- 1$ (with the library's relative, one-sided endpoint convention).

Facts & Assumptions

Given: No hypotheses beyond those quantified in the statement.

[L1]

On $[- 1, 1]$ , $\sin (\arcsin y) = y$ , and $\arcsin (1) = π / 2$ , $\arcsin (- 1) = - π / 2$ (Principal inverse sine and inverse cosine).

[L2]

The chain rule applies to derivatives relative to their domains (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L3]

$(\sin x)^{'} = \cos x$ , while $\cos (π / 2) = \cos (- π / 2) = 0$ (The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

Proof

technique · contradiction

1.1

Suppose $\arcsin$ had a finite derivative at $1$ . Differentiate the identity $\sin (\arcsin y) = y$ at $1$ relative to $[- 1, 1]$ . The derivative of the right side is $1$ , whereas [L2] and [L3] make the derivative of the left side $\cos (π / 2) (\arcsin)^{'} (1) = 0$ , a contradiction.

assume-contraL1L2L3

1.2

The identical argument at $- 1$ gives $1 = \cos (- π / 2) (\arcsin)^{'} (- 1) = 0$ .

assume-contraL1L2L3

2.1

Therefore neither finite endpoint derivative exists.

step 1.1step 1.2discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

NIST Digital Library of Mathematical Functions, §4.24(ii) Derivatives (standard reference, not scraped)

Principal arcsine has no finite derivative at −1−1 or 11