---
title: "Leibnitz theorem: y = e^(x²)"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:leibnitz-exp-x-squared"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-exp-x-squared
site: BCT Engineering
---

# Leibnitz theorem: y = e^(x²)

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2073 Chaitra
- Video: [Successive Differentiation|Leibnitz's Theorem|Lecture 03|ENGINEERING MATHEMATICS|PRADEEP GIRI SIR](https://www.youtube.com/watch?v=PHIGwMmi7iU) — Pradeep Giri Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-exp-x-squared
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

State Leibnitz theorem. If y = e^(x²), then show that y_{n+1} − 2xy_n − 2ny_{n−1} = 0.

## Short answer

Leibniz theorem states (uv)_n = sum_{k=0}^n C(n,k) u_{n-k} v_k. For y = e^(x^2), y_1 = 2x y; differentiating n times by the theorem gives the recurrence y_{n+1} − 2x y_n − 2n y_{n−1} = 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 = e^(x²). Then

y₁ = 2x e^(x²) = 2x y.

Differentiate y₁ = 2x y n times by Leibnitz's theorem, with u = y, v = 2x (v₁ = 2, v₂ = 0):

yₙ₊₁ = 2x yₙ + n(2) yₙ₋₁

**yₙ₊₁ − 2x yₙ − 2n yₙ₋₁ = 0.** Proved.

## Check yourself

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

1. For y = e^(x^2), y_1 =
   A. x^2 y
   B. 2x y
   C. 2 y
   D. x y
   Answer: B. d/dx e^(x^2) = 2x e^(x^2). (Key point: Base)

2. Differentiating y_1 = 2x y n times, with v = 2x, v_2 = 0 means:
   A. only two terms survive
   B. only one term survives
   C. all n terms survive
   D. three terms survive
   Answer: A. v_2 and higher vanish. (Key point: Leibnitz choice)

3. The result is:
   A. y_{n+1} = 2x y_n - 2n y_{n-1}
   B. y_{n+1} = 2x y_n + 2n y_{n-1}
   C. y_{n+1} = 2x y_n + n y_{n-1}
   D. y_{n+1} = x y_n + 2n y_{n-1}
   Answer: B. u_n v + n u_{n-1} v_1 with v_1 = 2. (Key point: Result)

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