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Leibnitz theorem: y = (x²−1)ⁿ

Short answer

Leibnitz's theorem states that the nth derivative of product uv is (uv)_n = ∑_{r=0}^n C(n,r) u_{n-r} v_r. For y=(x²-1)^n, differentiate (x²-1)y₁=2nxy n+1 times using the theorem to obtain (x²-1)y_{n+2}+2xy_{n+1}−n(n+1)y_n=0.

State Leibnitz's theorem. If y = (x²−1)ⁿ, then prove that (x²−1)y_{n+2} + 2xy_{n+1} − n(n+1)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 = (x² − 1)ⁿ. Then y₁ = n(x² − 1)ⁿ⁻¹ · 2x, so multiplying by (x² − 1):

(x² − 1) y₁ = 2n x y …(1)

Differentiate (1) (n+1) times by Leibnitz's theorem.

Left side, with u = y₁, v = x² − 1: (x² − 1) yₙ₊₂ + (n+1)(2x) yₙ₊₁ + (n+1)n/2 yₙ

Right side, with u = y, v = x: 2n [ x yₙ₊₁ + (n+1) yₙ ]

Equate:

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

Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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