---
title: "Derivatives of y^(1/m) + y^(−1/m) = 2x"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:derivatives-y-power-m-sum"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/derivatives-y-power-m-sum
site: BCT Engineering
---

# Derivatives of y^(1/m) + y^(−1/m) = 2x

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2068 Chaitra
- Video: [If y^1/m + y^-1/m = 2x, prove that (x^2-1)yn+2 +(2n+1)yn+1 +(n^2-m^2)yn =0| Leibniz's Theorem| #26](https://www.youtube.com/watch?v=qYfYgotvSyk) — Maths Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/derivatives-y-power-m-sum
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

If y^(1/m) + y^(−1/m) = 2x, show that: a) (x²−1)y₂ + xy₁ − m²y = 0, b) (x²−1)y_{n+2} + (2n+1)xy_{n+1} + (n²−m²)y_n = 0.

## Short answer

Let t = y^(1/m); then t+1/t=2x ⇒ y = (x+√(x²−1))^m. Differentiate: y₁ = m y/√(x²−1) ⇒ (x²−1)y₁² = m²y². This yields (x²−1)y₂ + xy₁ − m²y = 0. Differentiating n times by Leibniz gives (x²−1)y_{n+2} + (2n+1)xy_{n+1} + (n²−m²)y_n = 0.

## Answer

Given y^(1/m) + y^(−1/m) = 2x. Put t = y^(1/m); then t + 1/t = 2x, so t² − 2xt + 1 = 0 and t = x ± √(x² − 1). Hence

y = (x + √(x² − 1))^m (the other sign gives the same equations below).

Differentiate: y₁ = m (x + √(x²−1))^(m−1) · [1 + x/√(x²−1)] = m y / √(x² − 1).

So **(x² − 1) y₁² = m² y²** …(1)

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

**(x² − 1) y₂ + x y₁ − m² y = 0.** Proved. …(2)

#### (b)
Differentiate (2) n times by Leibnitz's theorem:

```text
(x²−1)yₙ₊₂ + n(2x)yₙ₊₁ + [n(n−1)/2](2)yₙ + x yₙ₊₁ + n yₙ − m² yₙ = 0
```
The coefficient of yₙ is n(n−1) + n − m² = n² − m².

**(x² − 1) yₙ₊₂ + (2n+1) x yₙ₊₁ + (n² − m²) yₙ = 0.** Proved.

## Check yourself

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

1. From y^(1/m) + y^(-1/m) = 2x, we get y =
   A. (x + sqrt(x^2-1))^m
   B. (2x)^m
   C. x^m
   D. (x + sqrt(x^2+1))^m
   Answer: A. t + 1/t = 2x gives t = x +- sqrt(x^2 - 1). (Key point: Solve for y)

2. (x^2-1) y_1^2 =
   A. m^2 y
   B. m y^2
   C. y^2
   D. m^2 y^2
   Answer: D. y_1 = m y/sqrt(x^2-1). (Key point: Square relation)

3. Differentiating gives:
   A. (x^2-1) y_2 + x y_1 - m^2 y = 0
   B. (x^2-1) y_2 + 2x y_1 = 0
   C. (x^2-1) y_2 - x y_1 - m^2 y = 0
   D. (x^2-1) y_2 + x y_1 + m^2 y = 0
   Answer: A. 2(x^2-1) y_1 y_2 + 2x y_1^2 = 2m^2 y y_1. (Key point: Part (a))

4. The y_n coefficient is:
   A. m^2 - n^2
   B. n^2 + m^2
   C. n - m^2
   D. n^2 - m^2
   Answer: D. n(n-1) + n - m^2. (Key point: Part (b))

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