---
title: "Leibnitz theorem: y = log(x+√(a²+x²))"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:leibnitz-log-x-sqrt"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-log-x-sqrt
site: BCT Engineering
---

# Leibnitz theorem: y = log(x+√(a²+x²))

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2071 Shrawan, 5 marks; 2069 Chaitra; 2068 Shrawan (asked 3 times)
- Video: [Leibnitz's Theorem [Q4]  If Y = log(X +root(a^2+x^2) Show that:::: Bachelor's Mathematics Nepal BSC](https://www.youtube.com/watch?v=_GJwprFCMcE) — BSC Math
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-log-x-sqrt
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

If y = log(x + √(a²+x²)), then show that (a²+x²)y_{n+2} + (2n+1)xy_{n+1} + n²y_n = 0.

## 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.

## 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
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):

```text
(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.

## Check yourself

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

1. For y = log(x + sqrt(a^2 + x^2)), what is y_1?
   A. 1/sqrt(a^2 + x^2)
   B. sqrt(a^2 + x^2)
   C. 1/(x + sqrt(a^2 + x^2))
   D. x/sqrt(a^2 + x^2)
   Answer: A. The numerator 1 + x/sqrt(a^2+x^2) equals (x + sqrt(a^2+x^2))/sqrt(a^2+x^2), which cancels with the denominator. (Key point: First derivative)

2. Which relation is obtained after differentiating y_1 = 1/sqrt(a^2+x^2) once more and simplifying?
   A. (a^2+x^2) y_2 - x y_1 = 0
   B. (a^2+x^2) y_2 + x y_1 = 0
   C. y_2 + x y_1 = 0
   D. (a^2+x^2) y_2 + 2x y_1 = 0
   Answer: B. From (a^2+x^2) y_1^2 = 1, differentiating and dividing by 2 y_1 gives (a^2+x^2) y_2 + x y_1 = 0. (Key point: Base relation)

3. When differentiating (a^2+x^2) y_2 + x y_1 = 0 n times, what is the coefficient of y_n from the first term?
   A. 2n
   B. n(2x)
   C. n^2
   D. n(n-1)/2 times 2 = n(n-1)
   Answer: D. The second-derivative term C(n,2)*2*y_n gives n(n-1) y_n; with the n y_n from x y_1 this makes n^2. (Key point: Leibnitz step)

4. Why does the coefficient of y_n become n^2?
   A. n^2 - n + 1
   B. 2n + n = 3n
   C. n(n-1) - n = n^2 - 2n
   D. n(n-1) + n = n^2
   Answer: D. The terms n(n-1) from (a^2+x^2) and n from x y_{n+1}'s derivative add to n^2. (Key point: Final coefficients)

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