---
title: "Derivative of xⁿJₙ(x)"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:derivative-x-j-x-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/derivative-x-j-x-2
site: BCT Engineering
---

# Derivative of xⁿJₙ(x)

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2069 Bhadra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/derivative-x-j-x-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove the Bessel's function d/dx[xⁿ Jₙ(x)] = xⁿ J_(n−1)(x).

## Short answer

Multiply J_n series by x^n, differentiate termwise, simplify using Γ(n+k+1) = (n+k)Γ(n+k). The resulting series is x^n J_{n-1}, so d/dx[x^n J_n] = x^n J_{n-1}.

## Answer

Jₙ(x) = Σₖ (−1)ᵏ / (k! Γ(n + k + 1)) · (x/2)^(2k+n).

#### Multiply by xⁿ
```text
xⁿ Jₙ(x) = Σ (−1)ᵏ x^(2k+2n) / (2^(2k+n) k! Γ(n+k+1))
```

#### Differentiate term by term
```text
d/dx [xⁿ Jₙ] = Σ (−1)ᵏ (2k+2n) x^(2k+2n−1) / (2^(2k+n) k! Γ(n+k+1))
```
Use Γ(n + k + 1) = (n + k)Γ(n + k) and 2k + 2n = 2(n + k):

```text
= Σ (−1)ᵏ 2(n+k) x^(2k+2n−1) / (2^(2k+n) k! (n+k) Γ(n+k))
= Σ (−1)ᵏ x^(2k+2n−1) / (2^(2k+n−1) k! Γ(n+k))
```
Take xⁿ outside: x^(2k+2n−1) = xⁿ · x^(2k+n−1), and note n + k = (n − 1) + k + 1:

```text
= xⁿ Σ (−1)ᵏ / (k! Γ((n−1)+k+1)) · (x/2)^(2k+(n−1))
= xⁿ J_(n−1)(x)
```

**Hence d/dx[xⁿ Jₙ(x)] = xⁿ J_(n−1)(x).**

## Check yourself

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

1. Which series is used for Jₙ(x)?
   A. Σ x^k /(k! Γ(n+k))
   B. Σ (−1)^k x^(2k) /(2k)!
   C. Σ (−1)^k /(k! Γ(n+k+1)) · (x/2)^(2k+n)
   D. Σ (x/2)^(k+n) /(k+n)!
   Answer: C. The proof starts from the standard series for Jₙ(x) and multiplies by xⁿ. (Key point: Start from series)

2. Differentiating x^(2k+2n) term by term gives the factor:
   A. (2k + n) x^(2k+n−1)
   B. (2k + 2n) x^(2k+2n−1)
   C. (2k + 2n) x^(2k+2n+1)
   D. x^(2k+2n)/(2k+2n)
   Answer: B. The power rule gives (2k+2n) x^(2k+2n−1). (Key point: Term-by-term derivative)

3. Why is Γ(n+k+1) = (n+k)Γ(n+k) used?
   A. The factor (n+k) cancels with 2(n+k), leaving Γ(n+k)
   B. It turns the series into a factorial
   C. It removes the factor k!
   D. It makes the series start at k = 1
   Answer: A. 2k+2n = 2(n+k) cancels the (n+k) in the Gamma function, giving Γ(n+k) = Γ((n−1)+k+1), the series of J(n−1). (Key point: Gamma identity)

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