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Leibnitz theorem: y = log(x+√(a²+x²))

Short answer

First derivative y₁ = 1/√(a²+x²) gives (a²+x²)y₁² = 1, leading to (a²+x²)y₂ + x y₁ = 0. Apply Leibnitz’s theorem n times to obtain (a²+x²)y_{n+2} + (2n+1)x y_{n+1} + n² y_n = 0.

If y = log(x + √(a²+x²)), then show that (a²+x²)y_{n+2} + (2n+1)xy_{n+1} + n²y_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

Given y = log(x + √(a² + x²)). Differentiate:

y₁ = [1 + x/√(a²+x²)] / (x + √(a²+x²)) = 1/√(a² + x²)

So (a² + x²) y₁² = 1. Differentiate again:

2(a² + x²) y₁y₂ + 2x y₁² = 0, and dividing by 2y₁:

(a² + x²) y₂ + x y₁ = 0 …(1)

Differentiate (1) n times using Leibnitz's theorem, taking u = y₂ (or y₁) and v = a² + x² (or x):

(a²+x²)yₙ₊₂ + n(2x)yₙ₊₁ + [n(n−1)/2](2)yₙ  +  x yₙ₊₁ + n yₙ = 0
(a²+x²)yₙ₊₂ + (2n+1)x yₙ₊₁ + (n² − n + n) yₙ = 0

Hence

(a² + x²) yₙ₊₂ + (2n+1) x yₙ₊₁ + n² yₙ = 0. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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