Skip to main content

Leibnitz theorem: y = (sin⁻¹x)²

Short answer

Leibnitz theorem: (uv)_n = ∑ C(n,r) u_{n-r} v_r. From y = (sin⁻¹x)² we get (1-x²)y₂ - xy₁ - 2 = 0. Differentiating n times with Leibnitz yields (1-x²)y_{n+2} - (2n+1)xy_{n+1} - n²y_n = 0.

State Leibnitz's theorem. If y = (sin⁻¹x)², show that (1−x²)y_{n+2} − (2n+1)xy_{n+1} − n²y_n = 0.

Leibnitz's theorem. If u and v are functions of x, the nth derivative of the product uv 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⁻¹x)². Then y₁ = 2 sin⁻¹x /√(1 − x²), so

(1 − x²) y₁² = 4 (sin⁻¹x)² = 4y.

Differentiate: 2(1 − x²) y₁y₂ − 2x y₁² = 4y₁. Divide by 2y₁:

(1 − x²) y₂ − x y₁ − 2 = 0 …(1)

Differentiate (1) n times (n ≥ 1; the constant 2 disappears):

(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 − n + n)yₙ = 0

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

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics I chapter 1 questions →