Answer :

Certainly! Let's evaluate the derivative of the given expression step-by-step:

Given expression:
[tex]\[ \frac{d}{d x}\left[8\left(x^2 + 3|x|\right)\right] \][/tex]

Let's start by differentiating the expression inside the brackets:

1. Expression:
[tex]\[ 8(x^2 + 3|x|) \][/tex]

2. Differentiate each term inside the brackets:
[tex]\[ \frac{d}{d x}[8(x^2 + 3|x|)] = 8 \cdot \frac{d}{d x}[x^2 + 3|x|] \][/tex]

3. Differentiate [tex]\(x^2\)[/tex]:
[tex]\[ \frac{d}{d x}[x^2] = 2x \][/tex]

4. Differentiate [tex]\(3|x|\)[/tex]:

- For [tex]\(x \geq 0\)[/tex], [tex]\(|x| = x\)[/tex]:
[tex]\[ \frac{d}{d x}[3x] = 3 \][/tex]
- For [tex]\(x < 0\)[/tex], [tex]\(|x| = -x\)[/tex]:
[tex]\[ \frac{d}{d x}[-3x] = -3 \][/tex]

This can be represented more compactly using the sign function:
[tex]\[ \frac{d}{d x}[3|x|] = 3 \cdot \frac{d}{d x}[|x|] = 3 \cdot \text{sign}(x) \][/tex]

where [tex]\(\text{sign}(x)\)[/tex] is defined as:
[tex]\[ \text{sign}(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \][/tex]

5. Combine these results:
[tex]\[ \frac{d}{d x}[x^2 + 3|x|] = 2x + 3 \cdot \text{sign}(x) \][/tex]

6. Multiply by 8:
[tex]\[ 8 \cdot \frac{d}{d x}[x^2 + 3|x|] = 8 \cdot (2x + 3 \cdot \text{sign}(x)) \][/tex]

7. Simplify the final expression:
[tex]\[ 8 \cdot (2x + 3 \cdot \text{sign}(x)) = 16x + 24 \cdot \text{sign}(x) \][/tex]

The derivative of the given expression [tex]\(\frac{d}{d x}\left[8\left(x^2 + 3|x|\right)\right]\)[/tex] is:
[tex]\[ 16x + 24 \cdot \text{sign}(x) \][/tex]

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