---
title: "y'(1) and y''(1) from table"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:derivatives-at-1-table"
source: https://www.bctengineer.com/q/semester_4/numerical_method/derivatives-at-1-table
site: BCT Engineering
---

# y'(1) and y''(1) from table

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2070 Bhadra, 4 marks
- Video: [1. Numerical Differentiation example | Numerical Differentiation in numerical Analysis](https://www.youtube.com/watch?v=hQvmLnyZDuE) — Mathematics Analysis
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/derivatives-at-1-table
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Determine y'(1) and y''(1) from the following data.
| x | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
|---|---|---|---|---|---|
| y | 6 | 3 | 2 | 1.2 | 0.8 |

## Short answer

Apply Newton forward differences with h = 0.5 and p = 1. Substituting into the derivative formulas gives y'(1) approximately -3.0667 and y''(1) approximately 7.3333.

## Answer

The data are equally spaced (h = 0.5), so use Newton's forward-difference formulas with x₀ = 0.5 and p = (x − x₀)/h = (1 − 0.5)/0.5 = 1.

#### Forward difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 0.5 | 6 | -3 | 2 | -1.8 | 2 |
| 1 | 3 | -1 | 0.2 | 0.2 |  |
| 1.5 | 2 | -0.8 | 0.4 |  |  |
| 2 | 1.2 | -0.4 |  |  |  |
| 2.5 | 0.8 |  |  |  |  |

#### Formulas
```text
y'(x) = (1/h)[Δy₀ + (2p−1)/2·Δ²y₀ + (3p²−6p+2)/6·Δ³y₀ + (4p³−18p²+22p−6)/24·Δ⁴y₀]
y''(x) = (1/h²)[Δ²y₀ + (p−1)Δ³y₀ + (6p²−18p+11)/12·Δ⁴y₀]
```

#### Substitute p = 1
```text
y'(1)  = (1/0.5)[ -3 + (1/2)(2) + (−1/6)(-1.8) + (2/24)(2.0) ]
       = 2[ -3 + 1.0000 + 0.3000 + 0.1667 ]
       = -3.0667
y''(1) = (1/0.25)[ 2 + 0·(-1.8) + (−1/12)(2.0) ]
       = 4[ 2 − 0.1667 ]
       = 7.3333
```

**Answer: y'(1) ≈ -3.0667, y''(1) ≈ 7.3333**

Check with the 3-point central formulas (less accurate for this coarse table): y'(1) = (2−6)/(2×0.5) = -4.0000, y''(1) = (2 − 6 + 6)/0.25 = 8.0000.

## Check yourself

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

1. With x0 = 0.5, h = 0.5, what is p at x = 1?
   A. p = 1
   B. p = 0.5
   C. p = 2
   D. p = 0
   Answer: A. p = (1 - 0.5)/0.5 = 1. (Key point: Value of p)

2. What is the Δ²y0 coefficient in y'(x) at p = 1?
   A. 0
   B. (2p-1)/2 = 1/2
   C. 1
   D. -1/2
   Answer: B. (2×1-1)/2 = 0.5. (Key point: Coefficient)

3. What is y'(1)?
   A. -4.0
   B. 7.3333
   C. -3.0667
   D. -1.0
   Answer: C. 2[-3 + 1 + 0.3 + 0.1667] = -3.0667. (Key point: y'(1))

4. What is y''(1) and why does the Δ³ term vanish?
   A. 8.0; Δ³y is zero
   B. 7.3333; Δ³y0 is zero
   C. -3.0667; h cancels
   D. 7.3333; its coefficient (p-1) is 0
   Answer: D. 4[2 + 0 - 2/12] = 7.3333. (Key point: y''(1))

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