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]
