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Derivatives of y = sin(m sin⁻¹x)

Short answer

From y = sin(m sin⁻¹x), y₁ = m cos(m sin⁻¹x)/√(1−x²). Squaring gives (1−x²)y₁² = m²(1−y²); differentiating yields (1−x²)y₂ − xy₁ + m²y = 0. Applying Leibniz n times to this equation gives (1−x²)y_{n+2} − (2n+1)xy_{n+1} + (m²−n²)y_n = 0.

If y = sin(m sin⁻¹x), prove that a) (1−x²)y₂ − xy₁ + m²y = 0, b) (1−x²)y_{n+2} − (2n+1)xy_{n+1} + (m²−n²)y_n = 0.

Let y = sin(m sin⁻¹x). Then y₁ = m cos(m sin⁻¹x) / √(1 − x²).

(a)

(1 − x²) y₁² = m² cos²(m sin⁻¹x) = m²(1 − y²).

Differentiate: 2(1 − x²) y₁y₂ − 2x y₁² = −2m² y y₁. Divide by 2y₁:

(1 − x²) y₂ − x y₁ + m² y = 0. Proved.

(b)

Differentiate the result of (a) n times by Leibnitz's theorem:

(1−x²)yₙ₊₂ + n(−2x)yₙ₊₁ + [n(n−1)/2](−2)yₙ
      − [x yₙ₊₁ + n yₙ] + m² yₙ = 0

Collect terms: the coefficient of yₙ is −n(n−1) − n + m² = m² − n².

(1 − x²) yₙ₊₂ − (2n+1) x yₙ₊₁ + (m² − n²) yₙ = 0. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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