Let's solve the given problem, where $y = \sec^3(8x)$, and we are tasked with finding and simplifying $y'$.

Step 1: Differentiate using the chain rule and the power rule

The function $y = \sec^3(8x)$ is a composition of a power function and the secant function. The derivative of $y$ is:

$y' = \frac{d}{dx} \left[ \sec^3(8x) \right]$

Using the chain rule and the power rule, we first bring down the exponent 3:

$y' = 3 \cdot \sec^2(8x) \cdot \frac{d}{dx} \left[ \sec(8x) \right]$

Step 2: Differentiate $\sec(8x)$ with respect to $x$

The derivative of $\sec(8x)$ is:

[ \frac{d}{dx} \left[ \sec(8x) \right] = \sec(8x) \cdot \tan(8x) \cdot \frac{d}{dx}[8x] ]

Since $\frac{d}{dx}[8x] = 8$, this becomes:

$\frac{d}{dx} \left[ \sec(8x) \right] = 8 \cdot \sec(8x) \cdot \tan(8x)$

Step 3: Substitute back into $y'$

Now substitute $\frac{d}{dx} \left[ \sec(8x) \right]$ into the expression for $y'$:

$y' = 3 \cdot \sec^2(8x) \cdot \left[ 8 \cdot \sec(8x) \cdot \tan(8x) \right]$

Simplify:

$y' = 3 \cdot 8 \cdot \sec^3(8x) \cdot \tan(8x)$

$y' = 24 \cdot \sec^3(8x) \cdot \tan(8x)$

Final Answer:

$y' = 24 \cdot \sec^3(8x) \cdot \tan(8x)$

Would you like further clarification on any step or a deeper explanation? Here are some related questions to explore:

1. What is the chain rule, and how does it apply to nested functions like $\sec^3(8x)$?
2. How can you verify a derivative step using software like WolframAlpha or a graphing calculator?
3. Why is the derivative of $\sec(x)$ expressed as $\sec(x) \tan(x)$?
4. What are the common applications of trigonometric derivatives in physics or engineering?
5. Can you generalize this process to find $y'$ for $y = \sec^n(kx)$?

Tip: When working with trigonometric derivatives, always watch for nested functions that require applying the chain rule step by step.