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Leibnitz theorem: y = sin(m sin⁻¹x)

Short answer

Leibnitz's theorem: (uv)_n = ∑_{r=0}^n C(n,r) u_{n-r} v_r. For y=sin(m sin⁻¹x), differentiate (1−x²)y₂−xy₁+m²y=0 n times using the theorem to obtain (1−x²)y_{n+2}−(2n+1)xy_{n+1}+(m²−n²)y_n=0.

State Leibnitz's theorem on higher order derivatives. If y = sin(m sin⁻¹x), then show that (1−x²)y_{n+2} − (2n+1)xy_{n+1} + (m²−n²)y_n = 0.

Leibnitz's theorem. If u and v are functions of x with nth derivatives, then the nth derivative of the product is

(uv)ₙ = uₙv + C(n,1)uₙ₋₁v₁ + C(n,2)uₙ₋₂v₂ + ... + u vₙ

where C(n,r) = n!/(r!(n−r)!).

Proof

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

(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 …(1)

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

Since −n(n−1) − n = −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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