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Derivatives of y = (sin⁻¹x)²

Short answer

Differentiating (1-x²)y₁² = 4y gives (1-x²)y₂ - xy₁ - 2 = 0. Applying Leibnitz's theorem n times to this produces (1-x²)y_{n+2} - (2n+1)xy_{n+1} - n²y_n = 0.

If y = (sin⁻¹x)², then show that: i) (1−x²)y₂ − xy₁ − 2 = 0, ii) (1−x²)y_{n+2} − (2n+1)xy_{n+1} − n²y_n = 0.

Let y = (sin⁻¹x)². Then y₁ = 2 sin⁻¹x / √(1 − x²), so (1 − x²) y₁² = 4y.

(i)

Differentiate (1 − x²) y₁² = 4y with respect to x:

2(1 − x²) y₁y₂ − 2x y₁² = 4 y₁

Divide by 2y₁:

(1 − x²) y₂ − x y₁ − 2 = 0. Proved.

(ii)

Differentiate the result of (i) n times using Leibnitz's theorem (the constant 2 gives 0):

(1−x²)yₙ₊₂ + n(−2x)yₙ₊₁ + [n(n−1)/2](−2)yₙ − x yₙ₊₁ − n yₙ = 0
(1−x²)yₙ₊₂ − (2n+1)x yₙ₊₁ − n² yₙ = 0

since −n(n−1) − n = −n². Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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