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The question asks for the derivative \( f'(x) \) where:

\[
f(x) = -\frac{2 \cos(x)}{3x^3}
\]

To find \( f'(x) \), we can apply the quotient rule, which states:

\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
\]

Here, \( u = -2 \cos(x) \) and \( v = 3x^3 \).

### Step 1: Differentiate \( u \) and \( v \)

1. \( u = -2 \cos(x) \), so \( u' = 2 \sin(x) \)
2. \( v = 3x^3 \), so \( v' = 9x^2 \)

### Step 2: Apply the quotient rule

\[
f'(x) = \frac{(2 \sin(x))(3x^3) - (-2 \cos(x))(9x^2)}{(3x^3)^2}
\]

### Step 3: Simplify the numerator

\[
= \frac{6x^3 \sin(x) + 18x^2 \cos(x)}{9x^6}
\]

### Step 4: Simplify the expression

\[
= \frac{6x \sin(x) + 18 \cos(x)}{9x^6}
\]

Finally, we can divide each term in the numerator by 9:

\[
f'(x) = \frac{2x \sin(x) + 6 \cos(x)}{3x^6}
\]

Thus, the derivative is:

\[
f'(x) = \frac{2x \sin(x) + 6 \cos(x)}{3x^6}
\]

Would you like further clarification on any steps or additional details? Here are five related questions to expand this concept:

1. How do you apply the product rule for derivatives?
2. What is the derivative of \( \sin(x) \) and \( \cos(x) \)?
3. How does the chain rule work in combination with the quotient rule?
4. Can you explain how to find higher-order derivatives of rational functions?
5. What are common mistakes to avoid when using the quotient rule?

**Tip:** Always simplify your derivative expressions step by step to avoid errors during complex algebraic manipulations.

Find f'(x) where f(x) = -\frac{2 \cos(x)}{3x^3}.

Learn how to find the derivative of f(x) = -2cos(x) / 3x^3 using the quotient rule. This step-by-step solution walks through key calculus concepts such as differentiation, trigonometric functions, and the quotient rule. Suitable for high school calculus students in Grades 11-12.

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