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We are given the function \( f(x) = 8 \sec^{-1}(3x) \) and asked to find \( f'(x) \), followed by \( f'(1) \).

### Step 1: Differentiate \( f(x) = 8 \sec^{-1}(3x) \)

To differentiate this function, we need the derivative of the inverse secant function \( \sec^{-1}(x) \). The derivative of \( \sec^{-1}(x) \) is given by:

\[
\frac{d}{dx} \left[ \sec^{-1}(x) \right] = \frac{1}{x \sqrt{x^2 - 1}}
\]

Now, apply the chain rule to \( f(x) = 8 \sec^{-1}(3x) \).

Let \( u(x) = 3x \), then:

\[
f'(x) = 8 \cdot \frac{1}{u(x) \sqrt{u(x)^2 - 1}} \cdot u'(x)
\]

where \( u'(x) = 3 \), since the derivative of \( u(x) = 3x \).

Thus, we get:

\[
f'(x) = 8 \cdot \frac{1}{3x \sqrt{(3x)^2 - 1}} \cdot 3
\]

Simplifying:

\[
f'(x) = \frac{8 \cdot 3}{3x \sqrt{9x^2 - 1}} = \frac{8}{x \sqrt{9x^2 - 1}}
\]

### Step 2: Find \( f'(1) \)

Now we substitute \( x = 1 \) into the derivative \( f'(x) = \frac{8}{x \sqrt{9x^2 - 1}} \):

\[
f'(1) = \frac{8}{1 \cdot \sqrt{9(1)^2 - 1}} = \frac{8}{\sqrt{9 - 1}} = \frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 4\sqrt{2}
\]

So, \( f'(1) = 4\sqrt{2} \).

### Final Answers:
1. \( f'(x) = \frac{8}{x \sqrt{9x^2 - 1}} \)
2. \( f'(1) = 4\sqrt{2} \)

Would you like further clarification on any steps?

Here are 5 related questions:
1. What is the derivative of the inverse tangent function \( \tan^{-1}(x) \)?
2. How do you apply the chain rule when differentiating composite functions?
3. How does the secant function differ from cosine, and how does that affect its derivative?
4. What are the steps for solving derivatives of inverse trigonometric functions?
5. How would the result change if the function were \( f(x) = 8 \sec^{-1}(5x) \)?

**Tip:** Always check if the function inside the inverse trigonometric function has a chain rule application to avoid errors in differentiation.

Let f(x) = 8 sec⁻¹(3x). Find f'(x). Also, find f'(1).

Learn how to differentiate the function f(x) = 8 sec⁻¹(3x) and find f'(x) along with the value at x = 1. This step-by-step guide explores inverse trigonometric derivatives and the chain rule, suitable for advanced calculus students.

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