---
title: "Leibnitz theorem: y = (x²−1)ⁿ"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:leibnitz-x2-minus-1-power-n"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-x2-minus-1-power-n
site: BCT Engineering
---

# Leibnitz theorem: y = (x²−1)ⁿ

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2072 Chaitra; 2069 Asar (asked 2 times)
- Video: [If y= (x²-1)ⁿ, then prove that (x²-1)yn+2 +2xyn+1 -n(n+1)yn= 0 | Leibnitz Theorem](https://www.youtube.com/watch?v=kR7lSJxeBtY) — RK Eduworld (Classic)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-x2-minus-1-power-n
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

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.

## 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.

## Answer

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

```text
(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](2) yₙ

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

Equate:

```text
(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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For y = (x^2 - 1)^n, which relation is used?
   A. y_1 = 2n x y
   B. (x^2-1) y_1 = n y
   C. (x^2-1) y_1 = 2 x y
   D. (x^2-1) y_1 = 2n x y
   Answer: D. y_1 = 2nx(x^2-1)^(n-1); multiplying by (x^2-1) gives 2nxy. (Key point: Base relation)

2. Left y_n coefficient n(n+1), right 2n(n+1). After moving terms, the y_n term is:
   A. 0
   B. -n(n+1) y_n
   C. -2n(n+1) y_n
   D. +n(n+1) y_n
   Answer: B. n(n+1) - 2n(n+1) = -n(n+1). (Key point: Combining)

3. 2(n+1)x - 2n x equals:
   A. 4nx
   B. 2x
   C. 0
   D. 2(n+1)x
   Answer: B. The difference is 2x. (Key point: y_{n+1} coefficient)

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