---
title: "Maclaurin expansion of eˣ/(1+eˣ)"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:maclaurin-ex-over-1-plus-ex"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/maclaurin-ex-over-1-plus-ex
site: BCT Engineering
---

# Maclaurin expansion of eˣ/(1+eˣ)

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2071 Chaitra; 2070 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/maclaurin-ex-over-1-plus-ex
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Assuming the validity of expansion, find the expansion of the function eˣ/(1+eˣ) by Maclaurin's theorem (as far as the term in x³).

## Short answer

Using Maclaurin's series, f(x)=e^x/(1+e^x) gives f(0)=1/2, f'(0)=1/4, f''(0)=0, f'''(0)=-1/8, so the expansion up to x³ is 1/2 + x/4 − x³/48 + ...

## Answer

Maclaurin's series: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + ...

Let f(x) = eˣ/(1+eˣ) = 1 − 1/(1+eˣ). Differentiate:

f′ = eˣ/(1+eˣ)² = f(1 − f)

Since f′ = f − f², we get

```text
f″  = f′ − 2f f′ = f′(1 − 2f)
f‴  = f″(1 − 2f) − 2(f′)²
```
At x = 0: f(0) = 1/2.

```text
f′(0)  = (1/2)(1/2) = 1/4
f″(0)  = (1/4)(1 − 1) = 0
f‴(0)  = 0 − 2(1/16) = −1/8
```
Substitute:

f(x) = 1/2 + x(1/4) + (x²/2)(0) + (x³/6)(−1/8) + ...

**Answer: eˣ/(1+eˣ) = 1/2 + x/4 − x³/48 + ...**

## Check yourself

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

1. For f = e^x/(1+e^x), f' equals:
   A. f/(1+f)
   B. f^2
   C. f(1 - f)
   D. 1 - f
   Answer: C. f' = e^x/(1+e^x)^2 = f(1-f). (Key point: Derivative identity)

2. f(0), f'(0), f''(0) are:
   A. 1, 1/4, 0
   B. 1/2, 1/2, 0
   C. 1/2, 1/4, 0
   D. 1/2, 1/4, 1/2
   Answer: C. f(0) = 1/2, f'(0) = 1/4, f''(0) = (1/4)(1 - 1) = 0. (Key point: Values at 0)

3. f''' (0) is:
   A. -1/8
   B. -1/4
   C. 1/8
   D. 0
   Answer: A. f''' = f''(1-2f) - 2(f')^2 = 0 - 2/16 = -1/8. (Key point: Third derivative)

4. The expansion up to x^3 is:
   A. 1/2 + x/4 + x^3/48
   B. 1/2 + x/4 - x^3/8
   C. 1/2 + x/2 - x^3/6
   D. 1/2 + x/4 - x^3/48
   Answer: D. (x^3/6)(-1/8) = -x^3/48. (Key point: Series)

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