---
title: "Laplace transform of derivative/shift property"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transform-of-derivative-shift-property"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transform-of-derivative-shift-property
site: BCT Engineering
---

# Laplace transform of derivative/shift property

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2069 Asar, 5 marks
- Video: [The Laplace Transform of Derivatives and Integrals](https://www.youtube.com/watch?v=zfhyeXbb0d4) — Dr. Trefor Bazett
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transform-of-derivative-shift-property
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

If L[f(t)] = F(s), then prove that L[f'(t)] = sF(s) − f(0).

## Answer

**Assumptions:** f(t) is continuous for t ≥ 0, of exponential order (|f(t)| ≤ M e^(kt)), and f′(t) is piecewise continuous, so that L[f′] exists.

**Proof.** By definition,

```text
L[f′(t)] = ∫₀^∞ e^(−st) f′(t) dt
```

Integrate by parts with u = e^(−st), dv = f′(t) dt:

```text
L[f′(t)] = [ e^(−st) f(t) ]₀^∞ + s ∫₀^∞ e^(−st) f(t) dt
```

Because |f(t)| ≤ M e^(kt), for s > k we have e^(−st) f(t) → 0 as t → ∞. So the upper limit gives 0, and the lower limit gives −f(0):

```text
L[f′(t)] = 0 − f(0) + s F(s)
```

**Hence L[f′(t)] = s F(s) − f(0).** ∎

**Note.** Applying the result again to f″ gives L[f″] = s²F(s) − s f(0) − f′(0). If f has a jump at t = 0 the value f(0) is taken as f(0⁺).

## Check yourself

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

1. How is the proof done?
   A. Integrate by parts with u = e^(-st), dv = f'(t) dt
   B. Differentiate under the integral sign
   C. Use the convolution theorem
   D. Expand e^(-st) in series
   Answer: A. Integration by parts. (Key point: Method)

2. Why does e^(-st) f(t) -> 0 as t -> infinity?
   A. f is bounded by 1
   B. f has exponential order and s > k
   C. f(0) = 0
   D. f is periodic
   Answer: B. |f| <= M e^(kt). (Key point: Upper limit)

3. The lower limit contributes what?
   A. +f(0)
   B. 0
   C. -f(0)
   D. f'(0)
   Answer: C. [e^(-st) f]_0^inf. (Key point: Lower limit)

4. L[f''] = ?
   A. s^2 F(s) - f(0) - f'(0)
   B. s^2 F(s) + s f(0) + f'(0)
   C. s F(s) - f'(0)
   D. s^2 F(s) - s f(0) - f'(0)
   Answer: D. Apply twice. (Key point: Second derivative)

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