The problem provided requires us to differentiate the function $h(x) = |7x| \cos(8x)$ using the given formula for the derivative of the absolute value.

The formula given is:

$\frac{d}{dx} |u| = \frac{u'}{|u|}, \text{ for } u \neq 0.$

Step-by-step solution:

1. Let $u(x) = 7x$.

The absolute value part $|7x|$ involves differentiating using the provided formula: $\frac{d}{dx} |7x| = \frac{7}{|7x|}.$

2. Differentiate the second part $\cos(8x)$:

The derivative of $\cos(8x)$ is: $\frac{d}{dx} \cos(8x) = -8 \sin(8x).$

3. Apply the product rule:

Since $h(x) = |7x| \cos(8x)$, we use the product rule for differentiation: $h'(x) = \frac{d}{dx} \left( |7x| \right) \cdot \cos(8x) + |7x| \cdot \frac{d}{dx} \left( \cos(8x) \right).$ Substitute the derivatives: $h'(x) = \frac{7}{|7x|} \cdot \cos(8x) + |7x| \cdot (-8 \sin(8x)).$

4. Simplify the result: $h'(x) = \frac{7 \cos(8x)}{|x|} - 8 |7x| \sin(8x).$ Since $|7x| = 7|x|$, we substitute that into the expression: $h'(x) = \frac{7 \cos(8x)}{|x|} - 56 |x| \sin(8x).$

Thus, the derivative of $h(x) = |7x| \cos(8x)$ is: $h'(x) = \frac{7 \cos(8x)}{|x|} - 56 |x| \sin(8x).$

Would you like further details or clarifications on any of the steps? Here are some related questions:

1. What is the product rule and how is it applied?
2. How is the derivative of $\cos(8x)$ computed?
3. Can you explain the absolute value derivative formula?
4. Why is $|7x|$ treated as $7|x|$?
5. How does the chain rule relate to this problem?

Tip: When differentiating functions involving absolute values, always pay attention to the cases when the function inside the absolute value changes signs!