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

# Leibnitz theorem: y = sin(m sin⁻¹x)

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2071 Chaitra; 2070 Asar (asked 2 times)
- Video: [If y=sin(m sin⁻¹x), then prove that (1-x²)yₙ₊₂-(2n+1)xyₙ₊₁+(m²-n²)yₙ=0](https://www.youtube.com/watch?v=Vzez-eZ5HFQ) — Doctor of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-sin-m-asin
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

State Leibnitz's theorem on higher order derivatives. If y = sin(m sin⁻¹x), then show that (1−x²)y_{n+2} − (2n+1)xy_{n+1} + (m²−n²)y_n = 0.

## Short answer

Leibnitz's theorem: (uv)_n = ∑_{r=0}^n C(n,r) u_{n-r} v_r. For y=sin(m sin⁻¹x), differentiate (1−x²)y₂−xy₁+m²y=0 n times using the theorem to obtain (1−x²)y_{n+2}−(2n+1)xy_{n+1}+(m²−n²)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 = sin(m sin⁻¹x). Then y₁ = m cos(m sin⁻¹x)/√(1−x²), so

(1 − x²) y₁² = m² cos²(m sin⁻¹x) = m²(1 − y²).

Differentiate: 2(1−x²) y₁y₂ − 2x y₁² = −2m² y y₁. Divide by 2y₁:

**(1 − x²) y₂ − x y₁ + m² y = 0** …(1)

Differentiate (1) n times by Leibnitz's theorem:

```text
(1−x²)yₙ₊₂ + n(−2x)yₙ₊₁ + [n(n−1)/2](−2)yₙ
   − [x yₙ₊₁ + n yₙ] + m² yₙ = 0
(1−x²)yₙ₊₂ − (2n+1)x yₙ₊₁ + (m² − n² + n − n) yₙ = 0
```
Since −n(n−1) − n = −n²:

**(1 − x²) yₙ₊₂ − (2n+1) x yₙ₊₁ + (m² − 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(m sin^-1 x), (1-x^2) y_1^2 equals:
   A. 1 - y^2
   B. m^2 y^2
   C. m^2(1 - y^2)
   D. m^2(1 + y^2)
   Answer: C. m^2 cos^2(...) = m^2(1 - sin^2(...)) = m^2(1 - y^2). (Key point: Square relation)

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

3. The coefficient of y_n is -n(n-1) - n + m^2 which equals:
   A. m^2 - n
   B. m^2 - n^2
   C. m^2 + n^2
   D. n^2 - m^2
   Answer: B. -n^2 + n - n + m^2 = m^2 - n^2. (Key point: y_n coefficient)

4. Why can we divide by 2y_1?
   A. It removes the factor common to every term, leaving a linear relation in y, y_1, y_2
   B. It eliminates m
   C. It is required by Leibnitz's theorem
   D. It makes y_1 = 1
   Answer: A. Every term has y_1, and we need a linear equation to apply Leibnitz. (Key point: Why divide by y_1)

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