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We are given the function \( f(x) = \cos(\ln(x^3)) \) and are tasked with finding \( f'(1) \).

### Step-by-step solution:

1. **Simplify the argument of the logarithm:**
   \[
   \ln(x^3) = 3\ln(x)
   \]
   Thus, the function becomes:
   \[
   f(x) = \cos(3\ln(x))
   \]

2. **Differentiate \( f(x) \) using the chain rule:**
   \[
   f'(x) = \frac{d}{dx} \left[ \cos(3\ln(x)) \right]
   \]
   Using the chain rule, the derivative of \( \cos(u) \) is \( -\sin(u) \) times the derivative of \( u \). Here, \( u = 3\ln(x) \), so:
   \[
   f'(x) = -\sin(3\ln(x)) \cdot \frac{d}{dx} \left[ 3\ln(x) \right]
   \]

3. **Differentiate \( 3\ln(x) \):**
   \[
   \frac{d}{dx} [3\ln(x)] = \frac{3}{x}
   \]
   Therefore:
   \[
   f'(x) = -\sin(3\ln(x)) \cdot \frac{3}{x}
   \]

4. **Substitute \( x = 1 \) into the derivative:**
   To find \( f'(1) \), we substitute \( x = 1 \):
   \[
   f'(1) = -\sin(3\ln(1)) \cdot \frac{3}{1}
   \]
   Since \( \ln(1) = 0 \), this simplifies to:
   \[
   f'(1) = -\sin(0) \cdot 3 = 0
   \]

### Final answer:
\[
f'(1) = 0
\]

Would you like more details or have any further questions?

Here are five related questions:
1. How do you apply the chain rule in differentiation?
2. What is the derivative of \( \cos(u) \) with respect to \( u \)?
3. How do logarithmic properties simplify expressions like \( \ln(x^3) \)?
4. Why is \( \ln(1) = 0 \)?
5. Can we apply the chain rule in reverse to integrate functions like \( \cos(3\ln(x)) \)?

**Tip:** When differentiating composite functions, always keep track of the inner function and apply the chain rule carefully.

This calculus problem demonstrates how to differentiate f(x) = cos(ln(x^3)) using the chain rule, trigonometric derivatives, and logarithmic properties. The final answer for f'(1) is 0. Ideal for students in Grades 11-12 or early college calculus courses.

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