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The given function is \( g(x) = \frac{1}{x^3} \), and we are asked to find \( g'(3) \), the derivative of \( g(x) \) evaluated at \( x = 3 \).

### Step 1: Find the derivative \( g'(x) \)
We can rewrite \( g(x) = \frac{1}{x^3} \) as \( g(x) = x^{-3} \). Now, applying the power rule for derivatives, which states \( \frac{d}{dx} [x^n] = n x^{n-1} \):

\[
g'(x) = \frac{d}{dx} \left( x^{-3} \right) = -3x^{-4} = \frac{-3}{x^4}
\]

### Step 2: Evaluate \( g'(x) \) at \( x = 3 \)
Now we substitute \( x = 3 \) into the derivative expression:

\[
g'(3) = \frac{-3}{3^4} = \frac{-3}{81} = -\frac{1}{27}
\]

### Final Answer:
\[
g'(3) = -\frac{1}{27}
\]

Would you like further details on the derivative process or have any additional questions?

Here are 5 related questions for practice:
1. What is \( g'(x) \) if \( g(x) = \frac{1}{x^n} \) for any positive integer \( n \)?
2. How does the power rule apply to negative exponents?
3. What is the second derivative \( g''(x) \) for \( g(x) = \frac{1}{x^3} \)?
4. Can you compute \( g'(2) \) for \( g(x) = \frac{1}{x^3} \)?
5. How do derivatives change when dealing with fractions like \( \frac{1}{x^n} \)?

**Tip:** When differentiating functions involving powers of \( x \), always remember the power rule: \( \frac{d}{dx} [x^n] = n x^{n-1} \). This rule simplifies finding derivatives of terms like \( x^{-3} \).

This page explains how to differentiate g(x) = 1/x^3 and evaluate the derivative at x = 3. It uses the power rule to solve the problem and provides a step-by-step solution. Suitable for high school students in Grades 11-12.

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