---
title: "Euler's theorem with log function"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:euler-s-theorem-log-function"
source: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-log-function
site: BCT Engineering
---

# Euler's theorem with log function

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2074 Bhadra, 5 marks; 2073 Magh, 5 marks; 2071 Bhadra, 5 marks; 2070 Bhadra, 5 marks (asked 4 times)
- Video: [Euler's Theorem Part 4 I Modified Euler's Theorem I For Inverse and  Logarithmic Functions](https://www.youtube.com/watch?v=DFVsT80WpRc) — JEE Maths PYQS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-log-function
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

If u = log((x²+y²)/(x+y)), then show that x ∂u/∂x + y ∂u/∂y = 1.

## Short answer

Define z = e^u = (x²+y²)/(x+y). Replacing x,y by tx,ty shows z is homogeneous of degree 1. By Euler's theorem, x ∂z/∂x + y ∂z/∂y = z, which becomes x ∂u/∂x + y ∂u/∂y = 1.

## Answer

Use Euler's theorem on a suitable function of u.

#### Euler's theorem
If f(x, y) is homogeneous of degree *n*, then x ∂f/∂x + y ∂f/∂y = n f.

#### Proof of the result
u = log((x² + y²)/(x + y)) is not homogeneous itself, so put

z = eᵘ = (x² + y²)/(x + y).

Replacing x by tx and y by ty: z(tx, ty) = t²(x² + y²)/(t(x + y)) = t z. So z is homogeneous of degree n = 1.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = 1 · z

Since z = eᵘ, ∂z/∂x = eᵘ ∂u/∂x and ∂z/∂y = eᵘ ∂u/∂y. Therefore

```text
eᵘ ( x ∂u/∂x + y ∂u/∂y ) = eᵘ
x ∂u/∂x + y ∂u/∂y = 1
```

Hence proved. ∎

## Check yourself

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

1. Why is Euler's theorem not applied directly to u = log((x²+y²)/(x+y))?
   A. u is not homogeneous, so we use z = e^u instead
   B. u has no partial derivatives
   C. Euler's theorem only works for polynomials
   D. u is homogeneous of degree 0
   Answer: A. u is a logarithm of a degree-1 function and is not homogeneous itself; z = e^u = (x²+y²)/(x+y) is. (Key point: Why not apply directly)

2. What is the degree of homogeneity of z = (x²+y²)/(x+y)?
   A. 2
   B. 3
   C. 1
   D. 0
   Answer: C. z(tx, ty) = t²(...)/(t(...)) = t z, so n = 1. (Key point: Degree of z)

3. After Euler gives x z_x + y z_y = z, how is the result for u obtained?
   A. Put u = 1 in the equation
   B. Take log of both sides of Euler's equation
   C. Square both sides
   D. z_x = e^u u_x, so e^u(x u_x + y u_y) = e^u and e^u cancels
   Answer: D. Since z = e^u, z_x = e^u u_x and z_y = e^u u_y; dividing by e^u leaves x u_x + y u_y = 1. (Key point: Chain step)

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