---
title: "Find f'(3) from table"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:derivative-f-prime-3"
source: https://www.bctengineer.com/q/semester_4/numerical_method/derivative-f-prime-3
site: BCT Engineering
---

# Find f'(3) from table

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2075 Baisakh, 5 marks
- Video: [Numerical Differentiation | Forward, Backward and Central Difference | Numerical Computation](https://www.youtube.com/watch?v=5zwxZ439nTA) — The Infinite Math
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/derivative-f-prime-3
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Find f'(3) from the following table:
| x | 2 | 4 | 8 | 12 | 16 |
|---|---|---|---|---|---|
| f(x) | 20 | 23 | 30 | 35 | 40 |

## Short answer

Construct Newton's divided difference interpolating polynomial for the unequally spaced points, then differentiate and evaluate at x=3. The result is f'(3) ≈ 1.5260.

## Answer

The x values (2, 4, 8, 12, 16) are not equally spaced, so the difference formulas cannot be used. Fit Newton's divided difference polynomial and differentiate it.

#### Divided difference table

| x | f(x) | f[xᵢ,xᵢ₊₁] | 2nd | 3rd | 4th |
|---|---|---|---|---|---|
| 2 | 20 | 1.500000 | 0.041667 | -0.010417 | 0.001116 |
| 4 | 23 | 1.750000 | -0.062500 | 0.005208 |  |
| 8 | 30 | 1.250000 | 0.000000 |  |  |
| 12 | 35 | 1.250000 |  |  |  |
| 16 | 40 |  |  |  |  |

(for example f[2,4] = (23 − 20)/(4 − 2) = 1.5, f[2,4,8] = (1.75 − 1.5)/(8 − 2) = 0.041667)

#### Interpolating polynomial

```text
f(x) = 20 + 1.5(x−2) + 0.041667(x−2)(x−4) + (-0.010417)(x−2)(x−4)(x−8)
       + 0.001116(x−2)(x−4)(x−8)(x−12)
```

#### Differentiate

```text
f′(x) = 1.5 + 0.041667[(x−2) + (x−4)]
        + (-0.010417)[(x−4)(x−8) + (x−2)(x−8) + (x−2)(x−4)]
        + 0.001116[(x−4)(x−8)(x−12) + (x−2)(x−8)(x−12)
                      + (x−2)(x−4)(x−12) + (x−2)(x−4)(x−8)]
```

At x = 3:

```text
(x−2)+(x−4) = 1 − 1 = 0
(x−4)(x−8)+(x−2)(x−8)+(x−2)(x−4) = 5 − 5 − 1 = −1
(x−4)(x−8)(x−12)+(x−2)(x−8)(x−12)+(x−2)(x−4)(x−12)+(x−2)(x−4)(x−8)
   = (−1)(−5)(−9) + (1)(−5)(−9) + (1)(−1)(−9) + (1)(−1)(−5) = −45 + 45 + 9 + 5 = 14

f′(3) = 1.5 + 0.041667(0) + (-0.010417)(−1) + 0.001116(14)
      = 1.5 + 0.010417 + 0.015625
      = 1.5260
```

**Answer: f′(3) ≈ 1.5260**

## Check yourself

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

1. Why are ordinary difference formulas not used for the table x = 2, 4, 8, 12, 16?
   A. The table has too few points
   B. f(x) values are not integers
   C. The x values are not equally spaced
   D. x = 3 lies outside the table
   Answer: C. Forward/backward difference formulas need equal spacing h, so Newton's divided differences are used. (Key point: Method choice)

2. How is f[2, 4] computed?
   A. (23 - 20)/(4 + 2) = 0.5
   B. (23 + 20)/2 = 21.5
   C. (4 - 2)/(23 - 20) = 0.667
   D. (23 - 20)/(4 - 2) = 1.5
   Answer: D. A first divided difference is the change in f divided by the change in x. (Key point: Divided difference)

3. How is f'(3) obtained once the polynomial is built?
   A. Differentiate the Newton polynomial with the product rule and put x = 3
   B. Evaluate the polynomial at x = 3
   C. Take the average of the first divided differences
   D. Use (f(4) - f(2))/2 only
   Answer: A. The polynomial is differentiated term by term (product rule) and evaluated at x = 3. (Key point: Differentiate polynomial)

4. What is the result?
   A. f'(3) ≈ 1.5000
   B. f'(3) ≈ 1.5260
   C. f'(3) ≈ 0.0104
   D. f'(3) ≈ 1.7500
   Answer: B. 1.5 + 0.010417 + 0.015625 = 1.5260. (Key point: Result)

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