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

# Derivative of x⁻ⁿJₙ(x)

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

- Asked in: 2070 Magh, 5 marks; 2068 Bhadra, 5 marks (asked 2 times)
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove Bessel's Function d/dx[x⁻ⁿ Jₙ(x)] = −x⁻ⁿ J_(n+1)(x).

## Short answer

Start from the series for J_n, multiply by x^{-n}, differentiate term by term, then shift index. This yields the series for -x^{-n} J_{n+1}. Hence 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), k = 0, 1, 2, …

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

#### Differentiate term by term
```text
d/dx [x⁻ⁿ Jₙ] = Σ (−1)ᵏ 2k x^(2k−1) / (2^(2k+n) k! Γ(n+k+1))
```
The k = 0 term vanishes. Put k = m + 1 (m = 0, 1, 2, …), and use k! = (m+1)! = (m+1)·m! and Γ(n+k+1) = Γ(n+m+2):

```text
= Σ (−1)^(m+1) 2(m+1) x^(2m+1) / (2^(2m+n+2) (m+1) m! Γ(n+m+2))
= − Σ (−1)^m x^(2m+1) / (2^(2m+n+1) m! Γ(n+m+2))
```
Now take x⁻ⁿ outside: x^(2m+1) = x⁻ⁿ · x^(2m+n+1), so

```text
= − x⁻ⁿ Σ (−1)^m / (m! Γ(n+1+m+1)) · (x/2)^(2m+n+1)
= − x⁻ⁿ J_(n+1)(x)
```
since the last sum is the series of J_(n+1).

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

## Check yourself

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

1. The proof starts from:
   A. The recurrence relation
   B. The series for J_n multiplied by x^(-n)
   C. Rodrigues' formula
   D. The generating function
   Answer: B. Term-by-term differentiation of the series. (Key point: Starting point)

2. Why does the k = 0 term disappear?
   A. Its coefficient is infinite
   B. Gamma function is zero
   C. Its derivative is zero (constant after multiplying by x^(-n))
   D. It cancels with k = 1
   Answer: C. x^0 is constant. (Key point: Vanishing term)

3. The sum is re-indexed by:
   A. k = m + 1
   B. k = m - 1
   C. k = 2m
   D. k = m
   Answer: A. So that it starts at m = 0. (Key point: Index shift)

4. The resulting sum is recognised as:
   A. The series of J_(n-1)
   B. The series of J_(n+1)
   C. The series of J_n
   D. The series of I_n
   Answer: B. Exponent 2m+n+1 and Γ(n+m+2). (Key point: Recognition)

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