---
title: "Leibnitz theorem: y = (sin⁻¹x)²"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:leibnitz-asin-squared"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-asin-squared
site: BCT Engineering
---

# Leibnitz theorem: y = (sin⁻¹x)²

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2073 Shrawan
- Video: [If y=(sin-1x)^2 prove that (1+x^2)yn+2-(2n+1)xyn+1-n^2yn=0 | Leibnitz Theorem](https://www.youtube.com/watch?v=agd8vMJRCvc) — Passion Maths Study
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-asin-squared
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

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

## 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.

## Answer

**Leibnitz's theorem.** If u and v are functions of x, the nth derivative of the product uv 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 = (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):

```text
(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.

## Check yourself

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

1. For y = (sin^-1 x)^2, (1-x^2) y_1^2 =
   A. 2y
   B. y
   C. 4y^2
   D. 4y
   Answer: D. y_1 = 2 sin^-1 x/sqrt(1-x^2). (Key point: Square relation)

2. After differentiating, we get:
   A. (1-x^2) y_2 - x y_1 = 0
   B. (1-x^2) y_2 - x y_1 - 2 = 0
   C. (1-x^2) y_2 + x y_1 - 2 = 0
   D. (1-x^2) y_2 - x y_1 + 2 = 0
   Answer: B. 2(1-x^2) y_1 y_2 - 2x y_1^2 = 4 y_1. (Key point: Second order)

3. What happens to the constant 2 when differentiating n times (n >= 1)?
   A. It becomes y_n
   B. It becomes 0
   C. It stays
   D. It becomes 2n
   Answer: B. Derivative of a constant is zero. (Key point: Constant)

4. Coefficient of y_n is:
   A. -n^2
   B. n^2
   C. -n
   D. -(n^2 + n)
   Answer: A. -n(n-1) - n = -n^2. (Key point: y_n coefficient)

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