---
title: "Directional derivative of x²yz+4xz²"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:directional-derivative-x2yz-4xz2"
source: https://www.bctengineer.com/q/semester_2/mathematics/directional-derivative-x2yz-4xz2
site: BCT Engineering
---

# Directional derivative of x²yz+4xz²

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2072 Ashwin, 5 marks
- Video: [Find the directional derivative of x²yz + 4xz² at (1, -2, -1) in the direction of 2i - j - 2k #ve...](https://www.youtube.com/watch?v=rJwDsNbn-Vs) — Gari-Math
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/directional-derivative-x2yz-4xz2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Find the directional derivative of φ = x²yz + 4xz² at the point (1,−2,−1) in the direction of vector 2i−j−2k.

## Short answer

∇φ at (1,-2,-1) = 8i - j - 10k. Unit direction vector is (2i-j-2k)/3. Directional derivative = (16+1+20)/3 = 37/3.

## Answer

The directional derivative of φ along a vector d at a point is ∇φ · d̂, with d̂ = d/|d|.

#### Gradient
φ = x²yz + 4xz²

```text
∂φ/∂x = 2xyz + 4z²
∂φ/∂y = x²z
∂φ/∂z = x²y + 8xz
```
At (1, −2, −1):

```text
∂φ/∂x = 2(1)(−2)(−1) + 4(1) = 4 + 4 = 8
∂φ/∂y = (1)(−1) = −1
∂φ/∂z = (1)(−2) + 8(1)(−1) = −2 − 8 = −10
∇φ = 8i − j − 10k
```

#### Unit vector
d = 2i − j − 2k, |d| = √(4 + 1 + 4) = 3, d̂ = (2i − j − 2k)/3.

#### Directional derivative
∇φ · d̂ = (8·2 + (−1)(−1) + (−10)(−2))/3 = (16 + 1 + 20)/3 = 37/3.

**Answer: 37/3 ≈ 12.33** (φ increases in that direction).

## Check yourself

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

1. ∂φ/∂x = 2xyz + 4z² at (1,−2,−1) equals:
   A. 0
   B. 4
   C. −8
   D. 8
   Answer: D. 4 + 4 = 8. (Key point: Gradient)

2. ∇φ at the point is:
   A. 8i + j − 10k
   B. 8i − j + 10k
   C. 8i − j − 10k
   D. 4i − j − 10k
   Answer: C. φy = x²z = −1; φz = x²y + 8xz = −2 − 8 = −10. (Key point: Full gradient)

3. d̂ for d = 2i − j − 2k is:
   A. (2i − j − 2k)/9
   B. (2i − j − 2k)/3
   C. (2i − j − 2k)/5
   D. (2i − j − 2k)
   Answer: B. |d| = 3. (Key point: Unit vector)

4. The directional derivative is:
   A. 37/3
   B. −37/3
   C. 37/9
   D. 11
   Answer: A. (16 + 1 + 20)/3. (Key point: Result)

## Close variants set in other exams

- [Directional derivative of xy²+yz³](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adirectional-derivative-xy2-yz3) — Mathematics, chapter 5, asked 4×

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